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G4 Let $S$ be a point inside $\varangle p O q$, and let $k$ be a circle which contains $S$ and touches the legs $O p$ and $O q$ in points $P$ and $Q$ respectively. Straight line $s$ parallel to $O p$ from $S$ intersects $O q$ in a point $R$. Let $T$ be the point of intersection of the ray $(P S$ and circumscribed circle of $\triangle S Q R$ and $T \neq S$. Prove that $O T \| S Q$ and $O T$ is a tangent of the circumscribed circle of $\triangle S Q R$.
|
## Solution
Let $\varangle O P S=\varphi_{1}$ and $\varangle O Q S=\varphi_{2}$. We have that $\varangle O P S=\varangle P Q S=\varphi_{1}$ and $\varangle O Q S=$ $\varangle Q P S=\varphi_{2}$ (tangents to circle $k$ ).
Because $R S \| O P$ we have $\varangle O P S=\varangle R S T=\varphi_{1}$ and $\varangle R Q T=\varangle R S T=\varphi_{1}$ (cyclic quadrilateral $R S Q T$ ). So, we have as follows $\varangle O P T=\varphi_{1}=\varangle R Q T=\varangle O Q T$, which implies that the quadrilateral $O P Q T$ is cyclic. From that we directly obtain $\varangle Q O T=$ $\varangle Q P T=\varphi_{2}=\varangle O Q S$, so $O T \| S Q$. From the cyclic quadrilateral $O P Q T$ by easy calculation we get
$$
\varangle O T R=\varangle O T P-\varangle R T S=\varangle O Q P-\varangle R Q S=\left(\varphi_{1}+\varphi_{2}\right)-\varphi_{2}=\varphi_{1}=\varangle R Q T
$$
Thus, $O T$ is a tangent to the circumscribed circle of $\triangle S Q R$.
### 2.4 Number Theory
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 107
|
NT1 Find all the pairs positive integers $(x, y)$ such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{[x, y]}+\frac{1}{(x, y)}=\frac{1}{2}
$$
where $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$.
|
## Solution
We put $x=d u$ and $y=d v$ where $d=(x, y)$. So we have $(u, v)=1$. From the conclusion we obtain $2(u+1)(v+1)=d u v$. Because $(v, v+1)=1, v$ divides $2(u+1)$.
Case 1. $u=v$. Hence $x=y=[x, y]=(x, y)$, which leads to the solution $x=8$ and $y=8$.
Case 2. $uv$. Because of the symmetry of $u, v$ and $x, y$ respectively we get exactly the symmetrical solutions of case 2 .
Finally the pairs of $(x, y)$ which are solutions of the problem are:
$(8,8),(9,24),(24,9),(5,20),(20,5),(12,15),(15,12),(8,12),(12,8),(6,12),(12,6)$.
|
(8,8),(9,24),(24,9),(5,20),(20,5),(12,15),(15,12),(8,12),(12,8),(6,12),(12,6)
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 108
|
NT2 Prove that the equation $x^{2006}-4 y^{2006}-2006=4 y^{2007}+2007 y$ has no solution in the set of the positive integers.
|
## Solution
We assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have
$$
\begin{gathered}
x^{2006}=4 y^{2007}+4 y^{2006}+2007 y+2006 \\
x^{2006}+1=4 y^{2006}(y+1)+2007(y+1) \\
x^{2006}+1=\left(4 y^{2006}+2007\right)(y+1)
\end{gathered}
$$
But $4 y^{2006}+2007 \equiv 3(\bmod 4)$, so $x^{2006}+1$ will have at least one prime divisor of the type $4 k+3$. It is known (and easily obtainable by using Fermat's Little Theorem) that this is impossible.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 109
|
NT3 Let $n>1$ be a positive integer and $p$ a prime number such that $n \mid(p-1)$ and $p \mid\left(n^{6}-1\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square.
|
## Solution
Since $n \mid p-1$, then $p=1+n a$, where $a \geq 1$ is an integer. From the condition $p \mid n^{6}-1$, it follows that $p|n-1, p| n+1, p \mid n^{2}+n+1$ or $p \mid n^{2}-n+1$.
- Let $p \mid n-1$. Then $n \geq p+1>n$ which is impossible.
- Let $p \mid n+1$. Then $n+1 \geq p=1+n a$ which is possible only when $a=1$ and $p=n+1$, i.e. $p-n=1=1^{2}$.
- Let $p \mid n^{2}+n+1$, i.e. $n^{2}+n+1=p b$, where $b \geq 1$ is an integer.
The equality $p=1+n a$ implies $n \mid b-1$, from where $b=1+n c, c \geq 0$ is an integer. We have
$$
n^{2}+n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \text { or } n+1=a c n+a+c
$$
If $a c \geq 1$ then $a+c \geq 2$, which is impossible. If $a c=0$ then $c=0$ and $a=n+1$. Thus we obtain $p=n^{2}+n+1$ from where $p+n=n^{2}+2 n+1=(n+1)^{2}$.
- Let $p \mid n^{2}-n+1$, i.e. $n^{2}-n+1=p b$ and analogously $b=1+n c$. So
$$
n^{2}-n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \text { or } n-1=a c n+a+c
$$
Similarly, we have $c=0, a=n-1$ and $p=n^{2}-n+1$ from where $p-n=n^{2}-2 n+1=$ $(n-1)^{2}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 110
|
NT4 Let $a, b$ be two co-prime positive integers. A number is called good if it can be written in the form $a x+b y$ for non-negative integers $x, y$. Define the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ as $f(n)=n-n_{a}-n_{b}$, where $s_{t}$ represents the remainder of $s$ upon division by $t$. Show that an integer $n$ is good if and only if the infinite sequence $n, f(n), f(f(n)), \ldots$ contains only non-negative integers.
|
## Solution
If $n$ is good then $n=a x+b y$ also $n_{a}=(b y)_{a}$ and $n_{b}=(a x)_{b}$ so
$$
f(n)=a x-(a x)_{b}+b y-(b y)_{a}=b y^{\prime}+a x^{\prime}
$$
is also good, thus the sequence contains only good numbers which are non-negative.
Now we have to prove that if the sequence contains only non-negative integers then $n$ is good. Because the sequence is non-increasing then the sequence will become constant from some point onwards. But $f(k)=k$ implies that $k$ is a multiple of $a b$ thus some term of the sequence is good. We are done if we prove the following:
Lemma: $f(n)$ is good implies $n$ is good.
Proof of Lemma: $n=2 n-n_{a}-n_{b}-f(n)=a x^{\prime}+b y^{\prime}-a x-b y=a\left(x^{\prime}-x\right)+b\left(y^{\prime}-y\right)$ and $x^{\prime} \geq x$ because $n \geq f(n) \Rightarrow n-n_{a} \geq f(n)-f(n)_{a} \Rightarrow a x^{\prime} \geq a x+b y-(b y)_{a} \geq a x$. Similarly $y^{\prime} \geq y$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 111
|
NT5 Let $p$ be a prime number. Show that $7 p+3^{p}-4$ is not a perfect square.
|
## Solution
Assume that for a prime number $p$ greater than $3, m=7 p+3^{p}-4$ is a perfect square. Let $m=n^{2}$ for some $n \in \mathbb{Z}$. By Fermat's Little Theorem,
$$
m=7 p+3^{p}-4 \equiv 3-4 \equiv-1 \quad(\bmod p)
$$
If $p=4 k+3, k \in \mathbb{Z}$, then again by Fermat's Little Theorem
$$
-1 \equiv m^{2 k+1} \equiv n^{4 k+2} \equiv n^{p-1} \equiv 1 \quad(\bmod p), \text { but } p>3
$$
a contradiction. So $p \equiv 1(\bmod 4)$.
Therefore $m=7 p+3^{p}-4 \equiv 3-1 \equiv 2(\bmod 4)$. But this is a contradiction since 2 is not perfect square in $(\bmod 4)$. For $p=2$ we have $m=19$ and for $p=3$ we have $m=44$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 112
|
## A1
For any real number a, let $\lfloor a\rfloor$ denote the greatest integer not exceeding a. In positive real numbers solve the following equation
$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=2014
$$
|
Solution1. Obviously $n$ must be positive integer. Now note that $44^{2}=19362000$ than $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor>2000+44+12=2056$, a contradiction!
So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$. Plugging that into the original equation we get:
$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=n+44+12=2014
$$
From which we get $n=1956$, which is the only solution.
|
1956
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 113
|
## A2
Let $a, b$ and $c$ be positive real numbers such that abc $=\frac{1}{8}$. Prove the inequality
$$
a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq \frac{15}{16}
$$
When does equality hold?
|
Solution1. By using The Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}= \\
& \quad=\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{c^{2}}{4}+\frac{c^{2}}{4}+\frac{c^{2}}{4}+\frac{c^{2}}{4}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq \\
& \quad \geq 1515 \sqrt{\frac{a^{12} b^{12} c^{12}}{4^{12}}}=15 \sqrt[5]{\left(\frac{a b c}{4}\right)^{4}}=15 \sqrt[5]{\left(\frac{1}{32}\right)^{4}}=\frac{15}{16}
\end{aligned}
$$
as desired. Equality holds if and only if $a=b=c=\frac{1}{2}$.
|
\frac{15}{16}
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 114
|
## A3
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold?
|
Solution1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right) \\
& =\left(a b+1+\frac{a}{c}+a\right)+\left(b c+1+\frac{b}{a}+b\right)+\left(c a+1+\frac{c}{b}+c\right) \\
& =a b+b c+c a+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+3+a+b+c
\end{aligned}
$$
Notice that by AM-GM we have $a b+\frac{b}{a} \geq 2 b, b c+\frac{c}{b} \geq 2 c$, and $c a+\frac{a}{c} \geq 2 a$.
Thus,
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq\left(a b+\frac{b}{a}\right)+\left(b c+\frac{c}{b}\right)+\left(c a+\frac{a}{c}\right)+3+a+b+c \geq 3(a+b+c+1)
$$
The equality holds if and only if $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 115
|
## A4
Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{7+2 b}{1+a}+\frac{7+2 c}{1+b}+\frac{7+2 a}{1+c} \geq \frac{69}{4}
$$
When does equality hold?
|
Solution1. The inequality can be written as: $\frac{5+2(1+b)}{1+a}+\frac{5+2(1+c)}{1+b}+\frac{5+2(1+a)}{1+c} \geq \frac{69}{4}$.
We substitute $1+a=x, 1+b=y, 1+c=z$.
So, we have to prove the inequality
$$
\frac{5+2 y}{x}+\frac{5+2 z}{y}+\frac{5+2 x}{z} \geq \frac{69}{4} \Leftrightarrow 5\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right) \geq \frac{69}{4}
$$
where $x, y, z>1$ real numbers and $x+y+z=4$.
We have
- $\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{4}$
- $\frac{y}{x}+\frac{z}{y}+\frac{x}{z} \geq 3 \cdot \sqrt[3]{\frac{y}{x} \cdot \frac{z}{y} \cdot \frac{x}{z}}=3$
Thus, $\frac{5+2 y}{x}+\frac{5+2 z}{y}+\frac{5+2 x}{z}=5\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right) \geq 5 \cdot \frac{9}{4}+2 \cdot 3=\frac{69}{4}$.
The equality holds, when $\left(x=y=z, \frac{y}{x}=\frac{z}{y}=\frac{x}{z}, x+y+z=4\right)$, thus $x=y=z=\frac{4}{3}$, i.e. $a=b=c=\frac{1}{3}$.
## (Egw)
人s
Let $x, y, z$ be non-negative real numbers satisfying $x+y+z=x y z$. Prove that
$$
2\left(x^{2}+y^{2}+z^{2}\right) \geq 3(x+y+z)
$$
and determine when equality occurs.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 116
|
## A6
Let $a, b, c$ be positive real numbers. Prove that
$$
\left(\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2}\right)\left(\left(3 b^{2}+1\right)^{2}+2\left(1+\frac{3}{c}\right)^{2}\right)\left(\left(3 c^{2}+1\right)^{2}+2\left(1+\frac{3}{a}\right)^{2}\right) \geq 48^{3}
$$
When does equality hold?
|
Solution. Let $x$ be a positive real number. By AM-GM we have $\frac{1+x+x+x}{4} \geq x^{\frac{3}{4}}$, or equivalently $1+3 x \geq 4 x^{\frac{3}{4}}$. Using this inequality we obtain:
$$
\left(3 a^{2}+1\right)^{2} \geq 16 a^{3} \text { and } 2\left(1+\frac{3}{b}\right)^{2} \geq 32 b^{-\frac{3}{2}}
$$
Moreover, by inequality of arithmetic and geometric means we have
$$
f(a, b)=\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2} \geq 16 a^{3}+32 b^{-\frac{3}{2}}=16\left(a^{3}+b^{-\frac{3}{2}}+b^{-\frac{3}{2}}\right) \geq 48 \frac{a}{b}
$$
Therefore, we obtain
$$
f(a, b) f(b, c) f(c, a) \geq 48 \cdot \frac{a}{b} \cdot 48 \cdot \frac{b}{c} \cdot 48 \cdot \frac{c}{a}=48^{3}
$$
Equality holds only when $a=b=c=1$.
## 1 IH $^{\text {th J.M. }} 2014$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 117
|
## A8
Let $x, y$ and $z$ be positive real numbers such that $x y z=1$. Prove the inequality
$$
\frac{1}{x(a y+b)}+\frac{1}{y(a z+b)}+\frac{1}{z(a x+b)} \geq 3 \text {, if: }
$$
a) $a=0$ and $b=1$;
b) $a=1$ and $b=0$;
c) $a+b=1$ for $a, b>0$
When does the equality hold true?
|
Solution. a) The inequality reduces to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3$, which follows directly from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
b) Here the inequality reduces to $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3$, i.e. $x+y+z \geq 3$, which also follows from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
c) Let $m, n$ and $p$ be such that $x=\frac{m}{n}, y=\frac{n}{p}$ и $z=\frac{p}{m}$. The inequality reduces to
$$
\frac{n p}{a m n+b m p}+\frac{p m}{a n p+b n m}+\frac{m n}{a p m+b p n} \geq 3
$$
By substituting $u=n p, v=p m$ and $w=m n$, (1) becomes
$$
\frac{u}{a w+b v}+\frac{v}{a u+b w}+\frac{w}{a v+b u} \geq 3
$$
The last inequality is equivalent to
$$
\frac{u^{2}}{a u w+b u v}+\frac{v^{2}}{a u v+b v w}+\frac{w^{2}}{a v w+b u w} \geq 3
$$
Cauchy-Schwarz Inequality implies
$$
\frac{u^{2}}{a u w+b u v}+\frac{v^{2}}{a u v+b v w}+\frac{w^{2}}{a v w+b u w} \geq \frac{(u+v+w)^{2}}{a u w+b u v+a u v+b v w+a v w+b u w}=\frac{(u+v+w)^{2}}{u w+v u+w v}
$$
Thus, the problem simplifies to $(u+v+w)^{2} \geq 3(u w+v u+w v)$, which is equivalent to $(u-v)^{2}+(v-w)^{2}+(w-u)^{2} \geq 0$.
Equality holds only when $u=v=w$, that is only for $x=y=z=1$.
Remark. The problem can be reformulated:
Let $a, b, x, y$ and $z$ be nonnegative real numbers such that $x y z=1$ and $a+b=1$. Prove the inequality
$$
\frac{1}{x(a y+b)}+\frac{1}{y(a z+b)}+\frac{1}{z(a x+b)} \geq 3
$$
When does the equality hold true?
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 118
|
## A9
Let $n$ be a positive integer, and let $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ be positive real numbers such that $x_{1}+\ldots+x_{n}=y_{1}+\ldots+y_{n}=1$. Show that
$$
\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right| \leq 2-\min _{1 \leq i \leq n} \frac{x_{i}}{y_{i}}-\min _{1 \leq i \leq n} \frac{y_{i}}{x_{i}}
$$
|
Solution. Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\frac{x_{1}}{y_{1}} \leq \ldots \leq \frac{x_{n}}{y_{n}}$. Let $A=\frac{x_{1}}{y_{1}}$ and $B=\frac{x_{n}}{y_{n}}$, and $\mathrm{S}=\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right|$. Our aim is to prove that $S \leq 2-A-\frac{1}{B}$.
First, note that we cannot have $A>1$, since that would imply $x_{i}>y_{i}$ for all $i \leq n$, hence $x_{1}+\ldots+x_{n}>y_{1}+\ldots+y_{n}$. Similarly, we cannot have $B<1$, since that would imply $x_{i}<y_{i}$ for all $i \leq n$, hence $x_{1}+\ldots+x_{n}<y_{1}+\ldots+y_{n}$.
If $n=1$, then $x_{1}=y_{1}=A=B=1$ and $S=0$, hence $S \leq 2-A-\frac{1}{B}$. For $n \geq 2$ let $1 \leq k<n$ be some integer such that $\frac{x_{k}}{y_{k}} \leq 1 \leq \frac{x_{k+1}}{y_{k+1}}$. We define the positive real numbers $X_{1}=x_{1}+\ldots+x_{k}$, $X_{2}=x_{k+1}+\ldots+x_{n}, Y_{1}=y_{1}+\ldots+y_{k}, Y_{2}=y_{k+1}+\ldots+y_{n}$. Note that $Y_{1} \geq X_{1} \geq A Y_{1}$ and $Y_{2} \leq X_{2} \leq B Y_{2}$. Thus, $A \leq \frac{X_{1}}{Y_{1}} \leq 1 \leq \frac{X_{2}}{Y_{2}} \leq B$. In addition, $S=Y_{1}-X_{1}+X_{2}-Y_{2}$.
From $0<X_{2}, Y_{1} \leq 1,0 \leq Y_{1}-X_{1}$ and $0 \leq X_{2}-Y_{2}$, follows
$$
S=Y_{1}-X_{1}+X_{2}-Y_{2}=\frac{Y_{1}-X_{1}}{Y_{1}}+\frac{X_{2}-Y_{2}}{X_{2}}=2-\frac{X_{1}}{Y_{1}}-\frac{Y_{2}}{X_{2}} \leq 2-A-\frac{1}{B}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 119
|
## C1
Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\frac{a b}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:
a) the length of the last remaining segment does not depend on the order of the deletions.
b) for every positive integer $n$, the initial segments on the board can be chosen with distinct integer lengths, such that the last remaining segment has length $n$.
|
Solution. a) Observe that $\frac{1}{\frac{a b}{a+b}}=\frac{1}{a}+\frac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_{1}, a_{2}, \ldots, a_{n}$, and $c$ is the length of the last remaining segment, then $\frac{1}{c}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}$ , proving a).
b) From a) and the equation $\frac{1}{n}=\frac{1}{2 n}+\frac{1}{3 n}+\frac{1}{6 n}$ it follows that if the lengths of the starting segments are $2 n, 3 n$ and $6 n$, then the length of the last remaining segment is $n$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 120
|
## C2
In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.
|
Solution. Denote by $X_{1}, X_{2}, \ldots X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}\left(m_{i}-1\right)$ non-direct routes. Thus $r=m_{1}^{2}+\ldots+m_{n}^{2}$. As each $m_{i}$ is at most $n-1$ and $13 \cdot 12^{2}<2014$, we deduce $n \geq 14$.
Consider $n=14$. As each route appears in two opposite directions, $r$ is even, so $r \geq 2016$. We can achieve $r=2016$ by arranging the 14 cities uniformly on a circle connect (by direct two-way airlines) all of them, except the diametrically opposite pairs. This way, there are exactly $14 \cdot 12^{2}=2016$ routes.
|
2016
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 121
|
## C3
For a given positive integer n, two players $A$ and B play the following game: Given is pile of $\boldsymbol{\Omega}$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the last stone. Assuming perfect play, find the number of values for $S_{\infty}$, for which A cannot win.
|
Solution. Denote by $k$ the sought number and let $\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ be the corresponding values for $a$. We will call each $a_{i}$ a losing number and every other positive integer a winning numbers. Clearly every multiple of $n$ is a winning number.
Suppose there are two different losing numbers $a_{i}>a_{j}$, which are congruent modulo $n$. Then, on his first turn of play, the player $A$ may remove $a_{i}-a_{j}$ stones (since $n \mid a_{i}-a_{j}$ ), leaving a pile with $a_{j}$ stones for B. This is in contradiction with both $a_{i}$ and $a_{j}$ being losing numbers. Therefore there are at most $n-1$ losing numbers, i.e. $k \leq n-1$.
Suppose there exists an integer $r \in\{1,2, \ldots, n-1\}$, such that $m n+r$ is a winning number for every $m \in \mathbb{N}_{0}$. Let us denote by $u$ the greatest losing number (if $k>0$ ) or 0 (if $k=0$ ), and let $s=\operatorname{LCM}(2,3, \ldots, u+n+1)$. Note that all the numbers $s+2, s+3, \ldots, s+u+n+1$ are composite. Let $m^{\prime} \in \mathbb{N}_{0}$, be such that $s+u+2 \leq m^{\prime} n+r \leq s+u+n+1$. In order for $m^{\prime} n+r$ to be a winning number, there must exist an integer $p$, which is either one, or prime, or a positive multiple of $n$, such that $m^{\prime} n+r-p$ is a losing number or 0 , and hence lesser than or equal to $u$. Since $s+2 \leq m^{\prime} n+r-u \leq p \leq m^{\prime} n+r \leq s+u+n+1, p$ must be a composite, hence $p$ is a multiple of $n$ (say $p=q n$ ). But then $m^{\prime} n+r-p=\left(m^{\prime}-q\right) n+r$ must be a winning number, according to our assumption. This contradicts our assumption that all numbers $m n+r, m \in \mathbb{N}_{0}$ are winning.
Hence there are exactly $n-1$ losing numbers (one for each residue $r \in\{1,2, \ldots, n-1\}$ ).
|
n-1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 122
|
## C4
Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$.
|
Solution. Grouping the elements of the product by ten we get:
$$
\begin{aligned}
& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\
& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\
& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\
& (30 k+19)(15 k+11)(120 k+100)(15 k+14)
\end{aligned}
$$
(We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.)
We denote $P_{k}=(30 k+1)(15 k+2)(30 k+7)(30 k+13)(15 k+8)(30 k+19)(15 k+11)(15 k+14)$. For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even $k, P_{k}$ ends with the same digit as $1 \cdot 2 \cdot 7 \cdot 3 \cdot 8 \cdot 9 \cdot 1 \cdot 4$, i.e. six and for odd $k, P_{k}$ ends with the same digit as $1 \cdot 7 \cdot 7 \cdot 3 \cdot 3 \cdot 9 \cdot 6 \cdot 9$, i.e. six. Thus $P_{0} P_{1} \ldots P_{66}$ ends with six. If we remove one zero from the end of all numbers divisible with five, we get that the last nonzero digit of the given product is the same as the one from $6 \cdot 2011 \cdot 2014 \cdot 4 \cdot 10 \cdot 16 \cdot \ldots .796 \cdot 802$. Considering that $4 \cdot 6 \cdot 2 \cdot 8$ ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in $4 \cdot 4^{26} \cdot 784 \cdot 796 \cdot 802 \cdot 1 \cdot 4 \cdot \ldots \cdot 76 \cdot 79$. Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in $2 \cdot 6 \cdot 6 \cdot 40 \cdot 100 \cdot 160 \cdot 220 \cdot 280 \cdot 61 \cdot 32 \cdot 67 \cdot 73 \cdot 38 \cdot 79$, which is two.
Let $A B C$ be a triangle with $\measuredangle B=\measuredangle C=40^{\circ}$. The bisector of the $\measuredangle B$ meets $A C$ at the point $D$ . Prove that $\overline{B D}+\overline{D A}=\overline{B C}$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 123
|
## G2
Let $A B C$ be an acute triangle with $\overline{A B}<\overline{A C}<\overline{B C}$ and $c(O, R)$ be its circumcircle. Denote with $D$ and $E$ be the points diametrically opposite to the points $B$ and $C$, respectively. The circle $c_{1}(A, \overline{A E})$ intersects $\overline{A C}$ at point $K$, the circle $c_{2}(A, \overline{A D})$ intersects $B A$ at point $L(A$ lies between $B$ and $L$ ). Prove that the lines $E K$ and $D L$ meet on the circle $c$.
|
Solution. Let $\mathrm{M}$ be the point of intersection of the line $D L$ with the circle $c(O, R)$ (we choose $M \equiv D$ if $L D$ is tangent to $c$ and $M$ to be the second intersecting point otherwise). It is
sufficient to prove that the points $E, K$ and $M$ are collinear.
We have that $\measuredangle E A C=90^{\circ}$ (since $E C$ is diameter of the circle $c$ ). The triangle $A E K$ is right-angled and isosceles ( $\overline{A E}$ and $\overline{A K}$ are radii of the circle $c_{1}$ ). Therefore
$$
\measuredangle \mathrm{AEK}=\measuredangle \mathrm{AKE}=45^{\circ} \text {. }
$$
Similarly, we obtain that $\measuredangle B A D=90^{\circ}=\measuredangle D A L$. Since $\overline{A D}=\overline{A L}$ the triangle $A D L$ is right-angled and isosceles, we have
$$
\measuredangle \mathrm{ADL}=\measuredangle \mathrm{A} L D=45^{\circ} .
$$
If $M$ is between $D$ and $L$, then $\measuredangle \mathrm{ADM}=\measuredangle A E M$, because they are inscribed in the circle $c(O, R)$ and they correspond to the same arch $\overparen{A M}$. Hence $\measuredangle \mathrm{AEK}=\measuredangle \mathrm{AEM}=45^{\circ}$ i.e. the points $E, K, M$ are collinear.
If $D$ is between $M$ and $L$, then $\measuredangle \mathrm{ADM}+\measuredangle A E M=180^{\circ}$ as opposite angles in cyclic quadrilateral. Hence $\measuredangle \mathrm{AEK}=\measuredangle \mathrm{AEM}=45^{\circ}$ i.e. the points $E, K, M$ are collinear.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 124
|
## G3
Let $C D \perp A B(D \in A B), D M \perp A C(M \in A C)$ and $D N \perp B C(N \in B C)$ for an acute triangle ABC with area $S$. If $H_{1}$ and $H_{2}$ are the orthocentres of the triangles $M N C$ and MND respectively. Evaluate the area of the quadrilateral $\mathrm{AH}_{1} \mathrm{BH}_{2}$.
|
Solution1. Let $O, P, K, R$ and $T$ be the midpoints of the segments $C D, M N, C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ and $T R \| M C$. Consequently, $\overline{P K}=\overline{T R}$ and $P K \| T R$. Also $O K \| D N$
(from $\triangle C D N$ ) and since $D N \perp B C$ and $M H_{1} \perp B C$, it follows that $T H_{1} \| O K$. Since $O$ is the circumcenter of $\triangle C M N, O P \perp M N$. Thus, $C H_{1} \perp M N$ implies $O P \| C H_{1}$. We conclude $\Delta T R H_{1} \cong \triangle K P O$ (they have parallel sides and $\overline{T R}=\overline{P K}$ ), hence $\overline{R H_{1}}=\overline{P O}$, i.e. $\overline{C H_{1}}=2 \overline{P O}$ and $\mathrm{CH}_{1} \| \mathrm{PO}$.
Analogously, $\overline{D H_{2}}=2 \overline{P O}$ and $D H_{2} \| P O$. From $\overline{C H_{1}}=2 \overline{P O}=\overline{D H_{2}}$ and
$H_{1} H_{2} \| C D$. Therefore the area of the quadrilateral $A H_{1} B H_{2}$ is $\frac{\overline{A B} \cdot \overline{H_{1} H_{2}}}{2}=\frac{\overline{A B} \cdot \overline{C D}}{2}=S$.
|
S
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 125
|
## G4
Let $A B C$ be a triangle such that $\overline{A B} \neq \overline{A C}$. Let $M$ be a midpoint of $\overline{B C}, H$ the orthocenter of $A B C, O_{1}$ the midpoint of $\overline{A H}$ and $O_{2}$ the circumcenter of $B C H$. Prove that $O_{1} A M O_{2}$ is a parallelogram.
|
Solution1. Let $O_{2}^{\prime}$ be the point such that $O_{1} A M O_{2}^{\prime}$ is a parallelogram. Note that $\overrightarrow{M O_{2}}=\overrightarrow{A O_{1}}=\overrightarrow{O_{1} H}$. Therefore, $O_{1} H O_{2}^{\prime} M$ is a parallelogram and $\overrightarrow{M O_{1}}=\overrightarrow{O_{2} H}$.
Since $M$ is the midpoint of $\overline{B C}$ and $O_{1}$ is the midpoint of $\overline{A H}$, it follows that $4 \overrightarrow{M O_{1}}=\overrightarrow{B A}+\overrightarrow{B H}+\overrightarrow{C A}+\overrightarrow{C H}=2(\overrightarrow{C A}+\overrightarrow{B H})$. Moreover, let $B^{\prime}$ be the midpoint of $\overrightarrow{B H}$. Then,
$$
\begin{aligned}
2 \overrightarrow{O_{2}^{\prime B}} \cdot \overrightarrow{B H} & =\left(\overline{O_{2}^{\prime H}}+\overline{O_{2}^{\prime B}}\right) \cdot \overrightarrow{B H}=\left(2 \overline{O_{2}^{\prime} H}+\overrightarrow{H B}\right) \cdot \overrightarrow{B H}= \\
& =\left(2 \overline{M O_{1}}+\overline{H B}\right) \cdot \overline{B H}=(\overline{C A}+\overrightarrow{B H}+\overrightarrow{H B}) \cdot \overrightarrow{B H}=\overline{C A} \cdot \overrightarrow{B H}=0 .
\end{aligned}
$$
By $\vec{a} \cdot \vec{b}$ we denote the inner product of the vectors $\vec{a}$ and $\vec{b}$.
Therefore, $O_{2}^{\prime}$ lies on the perpendicular bisector of $\overline{B H}$. Since $B$ and $C$ play symmetric roles, $\mathrm{O}_{2}^{\prime}$ also lies on the perpendicular bisector of $\overline{\mathrm{CH}}$, hence $\mathrm{O}_{2}^{\prime}$ is the circumcenter of $\triangle B C H$ and $\mathrm{O}_{2}=\mathrm{O}_{2}^{\prime}$.
Note: The condition $\overline{A B} \neq \overline{A C}$ just aims at ensuring that the parallelogram $O_{1} A N O_{2}$ is not degenerate, hence at helping students to focus on the "general" case.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 126
|
## G5
Let $A B C$ be a triangle with $\overline{A B} \neq \overline{B C}$, and let $B D$ be the internal bisector of $\measuredangle A B C(D \in A C)$. Denote the midpoint of the arc $A C$ which contains point BbyM. The circumcircle of the triangle $B D M$ intersects the segment $A B$ at point $K \neq B$, and let $J$ be the reflection of $A$ with respect to $K$. If $D J \cap A M=\{O\}$, prove that the points $J, B, M, O$ belong to the same circle.
|
## Solution1.
Let the circumcircle of the triangle $B D M$ intersect the line segment $B C$ at point $L \neq B$. From $\measuredangle C B D=\measuredangle D B A$ we have $\overline{D L}=\overline{D K}$. Since $\measuredangle L C M=\measuredangle B C M=\measuredangle B A M=\measuredangle K A M, \overline{M C}=\overline{M A}$ and
$$
\measuredangle L M C=\measuredangle L M K-\measuredangle C M K=\measuredangle L B K-\measuredangle C M K=\measuredangle C B A-\measuredangle C M K=\measuredangle C M A-\measuredangle C M K=\measuredangle K M A,
$$
it follows that triangles $M L C$ and $M K A$ are congruent, which implies $\overline{C L}=\overline{A K}=\overline{K J}$. Furthermore, $\measuredangle C L D=180^{\circ}-\measuredangle B L D=\measuredangle D K B=\measuredangle D K J$ and $\overline{D L}=\overline{D K}$, it follows that triangles $D C L$ and $D J K$ are congruent. Hence, $\angle D C L=\angle D J K=\measuredangle B J O$. Then
$$
\measuredangle B J O+\measuredangle B M O=\angle D C L+\angle B M A=\angle B C A+180^{\circ}-\angle B C A=180^{\circ}
$$
so the points $J, B, M, O$ belong to the same circle, q.e.d.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 127
|
## G6
Let $A B C D$ be a quadrilateral whose sides $A B$ and $C D$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_{1}$ and $H_{2}$ the orthocenters of the triangles $O A B$ and OCD, respectively. If $M$ and $N$ are the midpoints of the segments $\overline{A B}$ and $\overline{C D}$, respectively, prove that the lines $M N$ and $\mathrm{H}_{1} \mathrm{H}_{2}$ are parallel if and only if $\overline{A C}=\overline{B D}$.
|
## Solution.
Let $A^{\prime}$ and $B^{\prime}$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $A O B$, and $C^{\prime}$ and $D^{\prime}$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $C O D$. Obviously, $A^{\prime}$ and $D^{\prime}$ belong to the circle $c_{1}$ of diameter $\overline{A D}$, while $B^{\prime}$ and $C^{\prime}$ belong to the circle $c_{2}$ of diameter $\overline{B C}$.
It is easy to see that triangles $H_{1} A B$ and $H_{1} B^{\prime} A^{\prime}$ are similar. It follows that $\overline{H_{1} A} \cdot \overline{H_{1} A^{\prime}}=\overline{H_{1} B} \cdot \overline{H_{1} B^{\prime}}$. (Alternatively, one could notice that the quadrilateral $A B A^{\prime} B^{\prime}$ is cyclic and obtain the previous relation by writing the power of $H_{1}$ with respect to its circumcircle.) It
follows that $H_{1}$ has the same power with respect to circles $c_{1}$ and $c_{2}$. Thus, $H_{1}$ (and similarly, $\mathrm{H}_{2}$ ) is on the radical axis of the two circles.
The radical axis being perpendicular to the line joining the centers of the two circles, one concludes that $\mathrm{H}_{1} \mathrm{H}_{2}$ is perpendicular to $P Q$, where $P$ and $Q$ are the midpoints of the sides $\overline{A D}$ and $\overline{B C}$, respectively. ( $P$ and $Q$ are the centers of circles $c_{1}$ and $c_{2}$.)
The condition $H_{1} H_{2} \| M N$ is equivalent to $M N \perp P Q$. As $M P N Q$ is a parallelogram, we conclude that $H_{1} H_{2} \| M N \Leftrightarrow M N \perp P Q \Leftrightarrow M P N Q$ a rhombus $\Leftrightarrow \overline{M P}=\overline{M Q} \Leftrightarrow \overline{A C}=\overline{B D}$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 128
|
## N1
Each letter of the word OHRID corresponds to a different digit belonging to the set $(1,2,3,4,5)$. Decipher the equality $(O+H+R+I+D)^{2}:(O-H-R+I+D)=O^{H^{R^{I_{D}^{D}}}}$.
|
Solution. Since $O, H, R, I$ and $D$ are distinct numbers from $\{1,2,3,4,5\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D=O+H+R+I+D-2(H+R)15$ and divides 225 , which is only possible for $O^{H^{R^{I^{D}}}}=25$ (must be a power of three or five). This implies that $O=5, H=2$ and $R=1$. It's easy to check that both $I=3, D=4$ and $I=4, D=3$ satisfy the stated equation.
## $\mathbf{N} 2$
Find all triples $(p, q, r)$ of distinct primes $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
O=5,H=2,R=1,I=3,D=4orO=5,H=2,R=1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 129
|
N3
Find the integer solutions of the equation
$$
x^{2}=y^{2}\left(x+y^{4}+2 y^{2}\right)
$$
|
Solution. If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x, y)=(0,0)$ is a solution of the problem. Assume $x \neq 0$ and $y \neq 0$ satisfy the equation. The equation can be transformed in the form $x^{2}-x y^{2}=y^{6}+2 y^{4}$. Then $4 x^{2}-4 x y^{2}+y^{4}=4 y^{6}+9 y^{4}$ and consequently $\left(\frac{2 x}{y^{2}}-1\right)^{2}=4 y^{2}+9 \quad$ (1). Obviously $\frac{2 x}{y^{2}}-1$ is integer. From (1), we get that the numbers $\frac{2 x}{y^{2}}-1,2 y$ and 3 are Pythagorean triplets. It follows that $\frac{2 x}{y^{2}}-1= \pm 5$ and $2 y= \pm 4$. Therefore, $x=3 y^{2}$ or $x=-2 y^{2}$ and $y= \pm 2$. Hence $(x, y)=(12,-2),(x, y)=(12,2),(x, y)=(-8,-2)$ and $(x, y)=(-8,2)$ are the possible solutions. By substituting them in the initial equation we verify that all the 4 pairs are solution. Thus, together with the couple $(x, y)=(0,0)$ the problem has 5 solutions.
|
(x,y)=(0,0),(12,-2),(12,2),(-8,-2),(-8,2)
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 130
|
N4
Prove there are no integers $a$ and $b$ satisfying the following conditions:
i) $16 a-9 b$ is a prime number
ii) $\quad a b$ is a perfect square
iii) $a+b$ is a perfect square
|
Solution. Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
$$
\begin{aligned}
& 16 a-9 b=p \\
& a b=n^{2} \\
& a+b=m^{2}
\end{aligned}
$$
Moreover, let $d=g d c(a, b)$ and $a=d x, b=d y$ for some relatively prime integers $x$ and $y$. Obviously $a \neq 0$ and $b \neq 0, a$ and $b$ are positive (by (2) and (3)).
From (2) follows that $x$ and $y$ are perfect squares, say $x=l^{2}$ and $y=s^{2}$.
From (1), $d \mid p$ and hence $d=p$ or $d=1$. If $d=p$, then $16 x-9 y=1$, and we obtain $x=9 k+4$, $y=16 k+7$ for some nonnegative integer $k$. But then $s^{2}=y \equiv 3(\bmod 4)$, which is a contradiction.
If $d=1$ then $16 l^{2}-9 s^{2}=p \Rightarrow(4 l-3 s)(4 l+3 s)=p \Rightarrow(4 l+3 s=p \wedge 4 l-3 s=1)$.
By adding the last two equations we get $8 l=p+1$ and by subtracting them we get $6 s=p-1$. Therefore $p=24 t+7$ for some integer $t$ and $a=(3 t+1)^{2}$ and $b=(4 t+1)^{2}$ satisfy the conditions (1) and (2). By (3) we have $m^{2}=(3 t+1)^{2}+(4 t+1)^{2}=25 t^{2}+14 t+2$, or equivalently $25 m^{2}=(25 t+7)^{2}+1$.
Since the difference between two nonzero perfect square cannot be 1 , we have a contradiction. As a result there is no solution.
## NS
Find all nonnegative integers $x, y, z$ such that
$$
2013^{x}+2014^{y}=2015^{z}
$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 131
|
## N6
Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:
Vukasin: "Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ( $n$ is perfect if $\sigma(n)=2 n$ ). Sum of the digits of the remaining number is equal to the number of its positive divisors."
Dimitrije:"Each of these three numbers has no more than two digits 1 in its decimal representation."
Dusan:"If we add 11 to one of them, we obtain a square of an integer."
Stefan:"Each of them has exactly one prime divisor less then 10."
Filip:"The 3 numbers are square-free."
Their professor gave the correct answer. Which numbers did he say?
|
Solution. Let the middle number be $n$, so the numbers are $n-1, n$ and $n+1$. Since 4 does not divide any of them, $n \equiv 2(\bmod 4)$. Furthermore, neither 3,5 nor 7 divides $n$. Also $n+1+11 \equiv 2(\bmod 4)$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \mid n+1$ which implies $3 \mid n+10$ (a square), so $9 \mid n+10$ hence $9 \mid n+1$, which is impossible. Thus must be $n+11=m^{2}$.
Further, 7 does not divide $n-1$, nor $n+1$, because $1+11 \equiv 5(\bmod 7)$ and $-1+11 \equiv 3(\bmod 7)$ are quadratic nonresidues modulo 7. This implies $5 \mid n-1$ or $5 \mid n+1$. Again, since $n+11$ is a square, it is impossible $5 \mid n-1$, hence $5 \mid n+1$ which implies $3 \mid n-1$. This yields $n \equiv 4(\bmod 10)$ hence $S(n+1)=S(n)+1=S(n-1)+2(S(n)$ is sum of the digits of $n)$. Since the three numbers are square-free, their numbers of positive divisors are powers of 2 . Thus, we have two even sums of digits - they must be $S(n-1)$ and $S(n+1)$, so $S(n)$ is prime. From $3 \mid n-1$, follows $S(n-1)$ is an even perfect number, and $S(n+1)=2^{p}$. Consequently $S(n)=2^{p}-1$ is a prime, so $p$ is a prime number. One easily verifies $p \neq 2$, so $\mathrm{p}$ is odd implying $3 \mid 2^{p}-2$. Then
$\sigma\left(2^{p}-2\right) \geq\left(2^{p}-2\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=2\left(2^{p}-2\right)$. Since this number is perfect, $\frac{2^{p}-2}{6}$ must be one, i.e. $p=3$ and $S(n-1)=6, S(n)=7$ and $S(n+1)=8$.
Since 4 does not divide $n$, the 2 -digit ending of $n$ must be 14 or 34 . But $n=34$ is impossible, since $n+11=45$ is not a square. Hence, $n=10^{a}+10^{b}+14$ with $a \geq b \geq 2$. If $a \neq b$, then $n$ has three digits 1 in its decimal representation, which is impossible. Therefore $a=b$, and $n=2 \cdot 10^{a}+14$. Now, $2 \cdot 10^{a}+25=m^{2}$, hence $5 \mid m$, say $m=5 t$, and $(t-1)(t+1)=2^{a+1} 5^{a-2}$. Because $\operatorname{gcd}(t-1, t+1)=2$ there are three possibilities:
1) $t-1=2, t+1=2^{a} 5^{a-2}$, which implies $a=2, t=3$;
2) $t-1=2^{a}, t+1=2 \cdot 5^{a-2}$, so $2^{a}+2=2 \cdot 5^{a-2}$, which implies $a=3, t=9$;
3) $t-1=2 \cdot 5^{a-2}, t+1=2^{a}$, so $2 \cdot 5^{a-2}+2=2^{a}$, which implies $a=2, t=3$, same as case 1 ).
From the only two possibilities $(n-1, n, n+1)=(213,214,215)$ and $(n-1, n, n+1)=$ $(2013,2014,2015)$ the first one is not possible, because $S(215)=8$ and $\tau(215)=4$. By checking the conditions, we conclude that the latter is a solution, so the professor said the numbers: 2013, 2014, 2015.
|
2013,2014,2015
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 132
|
A2. Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
|
Solution. The given inequality is equivalent to
$$
\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}\right) \geqslant a+b+c
$$
By the AM-GM Inequality it follows that
$$
a^{3}+b^{3}=\frac{a^{3}+a^{3}+b^{3}}{3}+\frac{b^{3}+b^{3}+a^{3}}{3} \geqslant a^{2} b+b^{2} a=a b(a+b)
$$
Similarly we have
$$
b^{3}+c^{3} \geqslant b c(b+c) \quad \text { and } \quad c^{3}+a^{3} \geqslant c a(c+a)
$$
Summing the three inequalities we get
$$
2\left(a^{3}+b^{2}+c^{3}\right) \geqslant(a b(a+b)+b c(b+c)+c a(c+a))
$$
From the Cauchy-Schwarz Inequality we have
$$
(a b(a+b)+b c(b+c)+c a(c+a))\left(\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}\right) \geqslant(a b+b c+c a)^{2}
$$
We also have
$$
(a b+b c+c a)^{2} \geqslant 3(a b \cdot b c+b c \cdot c a+c a \cdot a b)=3 a b c(a+b+c)=2(a+b+c)
$$
Combining together (2),(3) and (4) we obtain (1) which is the required inequality.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 135
|
A3. Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B=X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A}+P_{B}$ and find all possible equality cases.
|
Solution. For the maximum, we use the fact that $\left(P_{A}-1\right)\left(P_{B}-1\right) \geqslant 0$, to get that $P_{A}+P_{B} \leqslant P_{A} P_{B}+1=11!+1$. Equality holds if and only if $A=\{1\}$ or $B=\{1\}$.
For the minimum observe, first that $P_{A} \cdot P_{B}=11!=c$. Without loss of generality let $P_{A} \leqslant P_{B}$. In this case $P_{A} \leqslant \sqrt{c}$. We write $P_{A}+P_{B}=P_{A}+\frac{c}{P_{A}}$ and consider the function $f(x)=x+\frac{c}{x}$ for $x \leqslant \sqrt{c}$. Since
$$
f(x)-f(y)=x-y+\frac{c(y-x)}{y x}=\frac{(x-y)(x y-c)}{x y}
$$
then $f$ is decreasing for $x \in(0, c]$.
Since $x$ is an integer and cannot be equal with $\sqrt{c}$, the minimum is attained to the closest integer to $\sqrt{c}$. We have $\lfloor\sqrt{11!}\rfloor=\left\lfloor\sqrt{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7 \cdot 11}\right\rfloor=\lfloor 720 \sqrt{77}\rfloor=6317$ and the closest integer which can be a product of elements of $X$ is $6300=2 \cdot 5 \cdot 7 \cdot 9 \cdot 10$.
Therefore the minimum is $f(6300)=6300+6336=12636$ and it is achieved for example for $A=\{2,5,7,9,10\}, B=\{1,3,4,6,8,11\}$.
Suppose now that there are different sets $A$ and $B$ such that $P_{A}+P_{B}=402$. Then the pairs of numbers $(6300,6336)$ and $\left(P_{A}, P_{B}\right)$ have the same sum and the same product, thus the equality case is unique for the numbers 6300 and 6336. It remains to find all possible subsets $A$ with product $6300=2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7$. It is immediate that $5,7,10 \in A$ and from here it is easy to see that all posibilities are $A=\{2,5,7,9,10\},\{1,2,5,7,9,10\},\{3,5,6,7,10\}$ and $\{1,3,5,6,7,10\}$.
|
12636
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 136
|
A4. Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c
$$
Prove that $-\sqrt{c}<a b<0$.
|
Solution. Firstly, we see that
$$
2019(a-b)=a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)
$$
Since $a \neq b$, we get $(a+b)\left(a^{2}+b^{2}\right)=2019$, so $a+b \neq 0$. Thus
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& =-2 a b\left(a^{2}+a b+b^{2}\right)
\end{aligned}
$$
Hence $a b\left(a^{2}+a b+b^{2}\right)=-c-a b$ (the equality does not occur since $a+b \neq 0)$. So
$$
-c=a b\left(a^{2}+a b+b^{2}\right)<-(a b)^{2} \Longrightarrow(a b)^{2}<c \Rightarrow-\sqrt{c}<a b<\sqrt{c}
$$
Therefore, we have $-\sqrt{c}<a b<0$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 137
|
A5. Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove the inequality
$$
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d}+\frac{1}{a+b+c+d^{3}} \leqslant \frac{a+b+c+d}{4}
$$
|
Solution. From the Cauchy-Schwarz Inequality, we obtain
$$
(a+b+c+d)^{2} \leqslant\left(a^{3}+b+c+d\right)\left(\frac{1}{a}+b+c+d\right)
$$
Using this, together with the other three analogous inequalities, we get
$$
\begin{aligned}
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d} & +\frac{1}{a+b+c+d^{3}} \\
& \leqslant \frac{3(a+b+c+d)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}{(a+b+c+d)^{2}}
\end{aligned}
$$
So it suffices to prove that
$$
(a+b+c+d)^{3} \geqslant 12(a+b+c+d)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)
$$
or equivalently, that
$$
\begin{aligned}
& \left(a^{3}+b^{3}+c^{3}+d^{3}\right)+3 \sum a^{2} b+6(a b c+a b d+a c d+b c d) \\
& \quad \geqslant 12(a+b+c+d)+4(a b c+a b d+a c d+b c d)
\end{aligned}
$$
(Here, the sum is over all possible $x^{2} y$ with $x, y \in\{a, b, c, d\}$ and $x \neq y$.) From the AM-GM Inequality we have
$a^{3}+a^{2} b+a^{2} b+a^{2} c+a^{2} c+a^{2} d+a^{2} d+b^{2} a+c^{2} a+d^{2} a+b c d+b c d \geqslant 12 \sqrt[12]{a^{18} b^{6} c^{6} d^{6}}=12 a$.
Similarly, we get three more inequalities. Adding them together gives the inequality we wanted. Equality holds if and only if $a=b=c=d=1$.
Remark by PSC. Alternatively, we can finish off the proof by using the following two inequalities: Firstly, we have $a+b+c+d \geqslant 4 \sqrt[4]{a b c d}=4$ by the AM-GM Inequality, giving
$$
\frac{3}{4}(a+b+c+d)^{3} \geqslant 12(a+b+c+d)
$$
Secondly, by Mclaurin's Inequality, we have
$$
\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant \frac{b c d+a c d+a b d+a b c}{4}
$$
giving
$$
\frac{1}{4}(a+b+c+d)^{3} \geqslant 4(b c d+a c d+a b d+a b c)
$$
Adding those inequlities we get the required result.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 138
|
A6. Let $a, b, c$ be positive real numbers. Prove the inequality
$$
\left(a^{2}+a c+c^{2}\right)\left(\frac{1}{a+b+c}+\frac{1}{a+c}\right)+b^{2}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)>a+b+c
$$
|
Solution. By the Cauchy-Schwarz Inequality, we have
$$
\frac{1}{a+b+c}+\frac{1}{a+c} \geqslant \frac{4}{2 a+b+2 c}
$$
and
$$
\frac{1}{b+c}+\frac{1}{a+b} \geqslant \frac{4}{a+2 b+c}
$$
Since
$$
a^{2}+a c+c^{2}=\frac{3}{4}(a+c)^{2}+\frac{1}{4}(a-c)^{2} \geqslant \frac{3}{4}(a+c)^{2}
$$
then, writing $L$ for the Left Hand Side of the required inequality, we get
$$
L \geqslant \frac{3(a+c)^{2}}{2 a+b+2 c}+\frac{4 b^{2}}{a+2 b+c}
$$
Using again the Cauchy-Schwarz Inequality, we have:
$$
L \geqslant \frac{(\sqrt{3}(a+c)+2 b)^{2}}{3 a+3 b+3 c}>\frac{(\sqrt{3}(a+c)+\sqrt{3} b)^{2}}{3 a+3 b+3 c}=a+b+c
$$
Alternative Question by Proposers. Let $a, b, c$ be positive real numbers. Prove the inequality
$$
\frac{a^{2}}{a+c}+\frac{b^{2}}{b+c}>\frac{a b-c^{2}}{a+b+c}+\frac{a b}{a+b}
$$
Note that both this inequality and the original one are equivalent to
$$
\left(c+\frac{a^{2}}{a+c}\right)+\left(a-\frac{a b-c^{2}}{a+b+c}\right)+\frac{b^{2}}{b+c}+\left(b-\frac{a b}{a+b}\right)>a+b+c
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 139
|
A7. Show that for any positive real numbers $a, b, c$ such that $a+b+c=a b+b c+c a$, the following inequality holds
$$
3+\sqrt[3]{\frac{a^{3}+1}{2}}+\sqrt[3]{\frac{b^{3}+1}{2}}+\sqrt[3]{\frac{c^{3}+1}{2}} \leqslant 2(a+b+c)
$$
|
Solution. Using the condition we have
$$
a^{2}-a+1=a^{2}-a+1+a b+b c+c a-a-b-c=(c+a-1)(a+b-1)
$$
Hence we have
$$
\sqrt[3]{\frac{a^{3}+1}{2}}=\sqrt[3]{\frac{(a+1)\left(a^{2}-a+1\right)}{2}}=\sqrt[3]{\left(\frac{a+1}{2}\right)(c+a-1)(a+b-1)}
$$
Using the last equality together with the AM-GM Inequality, we have
$$
\begin{aligned}
\sum_{\mathrm{cyc}} \sqrt[3]{\frac{a^{3}+1}{2}} & =\sum_{\mathrm{cyc}} \sqrt[3]{\left(\frac{a+1}{2}\right)(c+a-1)(a+b-1)} \\
& \leqslant \sum_{\mathrm{cyc}} \frac{\frac{a+1}{2}+c+a-1+a+b-1}{3} \\
& =\sum_{c y c} \frac{5 a+2 b+2 c-3}{6} \\
& =\frac{3(a+b+c-1)}{2}
\end{aligned}
$$
Hence it is enough to prove that
$$
3+\frac{3(a+b+c-1)}{2} \leqslant 2(a+b+c)
$$
or equivalently, that $a+b+c \geqslant 3$. From a well- known inequality and the condition, we have
$$
(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3(a+b+c)
$$
thus $a+b+c \geqslant 3$ as desired.
Alternative Proof by PSC. Since $f(x)=\sqrt[3]{x}$ is concave for $x \geqslant 0$, by Jensen's Inequality we have
$$
\sqrt[3]{\frac{a^{3}+1}{2}}+\sqrt[3]{\frac{b^{3}+1}{2}}+\sqrt[3]{\frac{c^{3}+1}{2}} \leqslant 3 \sqrt[3]{\frac{a^{3}+b^{3}+c^{3}+3}{6}}
$$
So it is enough to prove that
$$
\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}+3}{6}} \leqslant \frac{2(a+b+c)-3}{3}
$$
We now write $s=a+b+c=a b+b c+c a$ and $p=a b c$. We have
$$
a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)=s^{2}-2 s
$$
and
$$
r=a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}=(a b+b c+c a)(a+b+c)-3 a b c=s^{2}-3 p
$$
Thus,
$$
a^{3}+b^{3}+c^{3}=(a+b+c)^{3}-3 r-6 a b c=s^{3}-3 s^{2}+3 p
$$
So to prove (1), it is enough to show that
$$
\frac{s^{3}-3 s^{2}+3 p+3}{6} \leqslant \frac{(2 s-3)^{3}}{27}
$$
Expanding, this is equivalent to
$$
7 s^{3}-45 s^{2}+108 s-27 p-81 \geqslant 0
$$
By the AM-GM Inequality we have $s^{3} \geqslant 27 p$. So it is enough to prove that $p(s) \geqslant 0$, where
$$
p(s)=6 s^{3}-45 s^{2}+108 s-81=3(s-3)^{2}(2 s-3)
$$
It is easy to show that $s \geqslant 3$ (e.g. as in the first solution) so $p(s) \geqslant 0$ as required.
## COMBINATORICS
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 140
|
C1. Let $S$ be a set of 100 positive integers having the following property:
"Among every four numbers of $S$, there is a number which divides each of the other three or there is a number which is equal to the sum of the other three."
Prove that the set $S$ contains a number which divides each of the other 99 numbers of $S$.
|
Solution. Let $a<b$ be the two smallest numbers of $S$ and let $d$ be the largest number of $S$. Consider any two other numbers $x<y$ of $S$. For the quadruples $(a, b, x, d)$ and $(a, b, y, d)$ we cannot get both of $d=a+b+x$ and $d=a+b+y$, since $a+b+x<a+b+y$. From here, we get $a \mid b$ and $a \mid d$.
Consider any number $s$ of $S$ different from $a, b, d$. From the condition of the problem, we get $d=a+b+s$ or $a$ divides $b, s$ and $d$. But since we already know that $a$ divides $b$ and $d$ anyway, we also get that $a \mid s$, as in the first case we have $s=d-a-b$. This means that $a$ divides all other numbers of $S$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 141
|
C2. In a certain city there are $n$ straight streets, such that every two streets intersect, and no three streets pass through the same intersection. The City Council wants to organize the city by designating the main and the side street on every intersection. Prove that this can be done in such way that if one goes along one of the streets, from its beginning to its end, the intersections where this street is the main street, and the ones where it is not, will apear in alternating order.
|
Solution. Pick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.
On every other street $s_{1}$, exactly one intersection has been organized, namely the one where $s_{1}$ intersects $s$. Call this intersection $I_{1}$. We want to organize the intersections along $s_{1}$ such that they alternate between the two types. Note that, as $I_{1}$ is already organized, we have exactly one way to organize the remaining intersections along $s_{1}$.
For every street $s_{1} \neq s$, we can apply the procedure described above. Now, we only need to show that every intersection not on $s$ is well-organized. More precisely, this means that for every two streets $s_{1}, s_{2} \neq s$ intersecting at $s_{1} \cap s_{2}=A, s_{1}$ is the main street on $A$ if and only if $s_{2}$ is the side street on $A$.
Consider also the intersections $I_{1}=s_{1} \cap s$ and $I_{2}=s_{2} \cap s$. Now, we will define the "role" of the street $t$ at the intersection $X$ as "main" if this street $t$ is the main street on $X$, and "side" otherwise. We will prove that the roles of $s_{1}$ and $s_{2}$ at $A$ are different.
Consider the path $A \rightarrow I_{1} \rightarrow I_{2} \rightarrow A$. Let the number of intersections between $A$ and $I_{1}$ be $u_{1}$, the number of these between $A$ and $I_{2}$ be $u_{2}$, and the number of these between $I_{1}$ and $I_{2}$ be $v$. Now, if we go from $A$ to $I_{1}$, we will change our role $u_{1}+1$ times, as we will encounter $u_{1}+1$ new intersections. Then, we will change our street from $s_{1}$ to $s$, changing our role once more. Then, on the segment $I_{1} \rightarrow I_{2}$, we have $v+1$ new role changes, and after that one more when we change our street from $s_{1}$ to $s_{2}$. The journey from $I_{2}$ to $A$ will induce $u_{2}+1$ new role changes, so in total we have changed our role $u_{1}+1+1+v+1+1+u_{2}+1=u_{1}+v+u_{2}+5$, As we try to show that roles of $s_{1}$ and $s_{2}$ differ, we need to show that the number of role changes is odd, i.e. that $u_{1}+v+u_{2}+5$ is odd.
Obviously, this claim is equivalent to $2 \mid u_{1}+v+u_{2}$. But $u_{1}, v$ and $u_{2}$ count the number of intersections of the triangle $A I_{1} I_{2}$ with streets other than $s, s_{1}, s_{2}$. Since every street other than $s, s_{1}, s_{2}$ intersects the sides of $A I_{1} I_{2}$ in exactly two points, the total number of intersections is even. As a consequence, $2 \mid u_{1}+v+u_{2}$ as required.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 142
|
C3. In a $5 \times 100$ table we have coloured black $n$ of its cells. Each of the 500 cells has at most two adjacent (by side) cells coloured black. Find the largest possible value of $n$.
|
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
We can cover the table by one fragment like the first one on the figure below, 24 fragments like the middle one, and one fragment like the third one.
In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5+24 \cdot 6+1) \cdot 2+2=302$.
|
302
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 143
|
C4. We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25 \%$ have sent a message to $A$. How many possible two-digit values of $n$ are there?
|
Solution. If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4 k+4 k=8 k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$.
- In order to obtain $n=7 m+1$, arrange the kids in a circle and let each kid send a message to the first $4 m$ kids to its right and hence receive a message from the first $4 m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
- In order to obtain $n=7 m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7 m-1$ kids in a circle and let each kid on the circle send a message to the first $4 m-1$ kids to its right and hence receive a message from the first $4 m-1$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
There are 26 two-digit numbers with remainder 0 or 1 modulo 7 . (All numbers of the form $7 m$ and $7 m+1$ with $2 \leqslant m \leqslant 14$.)
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 144
|
G1. Let $A B C$ be a right-angled triangle with $\hat{A}=90^{\circ}$ and $\hat{B}=30^{\circ}$. The perpendicular at the midpoint $M$ of $B C$ meets the bisector $B K$ of the angle $\hat{B}$ at the point $E$. The perpendicular bisector of $E K$ meets $A B$ at $D$. Prove that $K D$ is perpendicular to $D E$.
|
Solution. Let $I$ be the incenter of $A B C$ and let $Z$ be the foot of the perpendicular from $K$ on $E C$. Since $K B$ is the bisector of $\hat{B}$, then $\angle E B C=15^{\circ}$ and since $E M$ is the perpendicular bisector of $B C$, then $\angle E C B=\angle E B C=15^{\circ}$. Therefore $\angle K E C=30^{\circ}$. Moreover, $\angle E C K=60^{\circ}-15^{\circ}=45^{\circ}$. This means that $K Z C$ is isosceles and thus $Z$ is on the perpendicular bisector of $K C$.
Since $\angle K I C$ is the external angle of triangle $I B C$, and $I$ is the incenter of triangle $A B C$, then $\angle K I C=15^{\circ}+30^{\circ}=45^{\circ}$. Thus, $\angle K I C=\frac{\angle K Z C}{2}$. Since also $Z$ is on the perpendicular bisector of $K C$, then $Z$ is the circumcenter of $I K C$. This means that $Z K=Z I=Z C$. Since also $\angle E K Z=60^{\circ}$, then the triangle $Z K I$ is equilateral. Moreover, since $\angle K E Z=30^{\circ}$, we have that $Z K=\frac{E K}{2}$, so $Z K=I K=I E$.
Therefore $D I$ is perpendicular to $E K$ and this means that $D I K A$ is cyclic. So $\angle K D I=$ $\angle I A K=45^{\circ}$ and $\angle I K D=\angle I A D=45^{\circ}$. Thus $I D=I K=I E$ and so $K D$ is perpendicular to $D E$ as required.
Alternative Question by Proposers. We can instead ask to prove that $E D=2 A D$. (After proving $K D \perp D E$ we have that the triangle $E D K$ is right angled and isosceles, therefore $E D=D K=2 A D$.) This alternative is probably more difficult because the perpendicular relation is hidden.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 146
|
G2. Let $A B C$ be a triangle and let $\omega$ be its circumcircle. Let $\ell_{B}$ and $\ell_{C}$ be two parallel lines passing through $B$ and $C$ respectively. The lines $\ell_{B}$ and $\ell_{C}$ intersect with $\omega$ for the second time at the points $D$ and $E$ respectively, with $D$ belonging on the arc $A B$, and $E$ on the arc $A C$. Suppose that $D A$ intersects $\ell_{C}$ at $F$, and $E A$ intersects $\ell_{B}$ at $G$. If $O, O_{1}$ and $O_{2}$ are the circumcenters of the triangles $A B C, A D G$ and $A E F$ respectively, and $P$ is the center of the circumcircle of the triangle $O O_{1} O_{2}$, prove that $O P$ is parallel to $\ell_{B}$ and $\ell_{C}$.
|
Solution. We write $\omega_{1}, \omega_{2}$ and $\omega^{\prime}$ for the circumcircles of $A G D, A E F$ and $O O_{1} O_{2}$ respectively. Since $O_{1}$ and $O_{2}$ are the centers of $\omega_{1}$ and $\omega_{2}$, and because $D G$ and $E F$ are parallel, we get that
$$
\angle G A O_{1}=90^{\circ}-\frac{\angle G O_{1} A}{2}=90^{\circ}-\angle G D A=90^{\circ}-\angle E F A=90^{\circ}-\frac{\angle E O_{2} A}{2}=\angle E A O_{2}
$$
So, because $G, A$ and $E$ are collinear, we come to the conclusion that $O_{1}, A$ and $O_{2}$ are also collinear.
Let $\angle D F E=\varphi$. Then, as a central angle $\angle A O_{2} E=2 \varphi$. Because $A E$ is a common chord of both $\omega$ and $\omega_{2}$, the line $O O_{2}$ that passes through their centers bisects $\angle A O_{2} E$, thus $\angle A O_{2} O=\varphi$. By the collinearity of $O_{1}, A, O_{2}$, we get that $\angle O_{1} O_{2} O=\angle A O_{2} O=\varphi$. As a central angle in $\omega^{\prime}$, we have $\angle O_{1} P O=2 \varphi$, so $\angle P O O_{1}=90^{\circ}-\varphi$. Let $Q$ be the point of intersection of $D F$ and $O P$. Because $A D$ is a common chord of $\omega$ and $\omega_{1}$, we have that $O O_{1}$ is perpendicular to $D A$ and so $\angle D Q P=90^{\circ}-\angle P O O_{1}=\varphi$. Thus, $O P$ is parallel to $\ell_{C}$ and so to $\ell_{B}$ as well.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 147
|
G3. Let $A B C$ be a triangle with incenter $I$. The points $D$ and $E$ lie on the segments $C A$ and $B C$ respectively, such that $C D=C E$. Let $F$ be a point on the segment $C D$. Prove that the quadrilateral $A B E F$ is circumscribable if and only if the quadrilateral $D I E F$ is cyclic.
|
Solution. Since $C D=C E$ it means that $E$ is the reflection of $D$ on the bisector of $\angle A C B$, i.e. the line $C I$. Let $G$ be the reflection of $F$ on $C I$. Then $G$ lies on the segment $C E$, the segment $E G$ is the reflection of the segment $D F$ on the line $C I$. Also, the quadraliteral $D E G F$ is cyclic since $\angle D F E=\angle E G D$.
Suppose that the quadrilateral $A B E F$ is circumscribable. Since $\angle F A I=\angle B A I$ and $\angle E B I=\angle A B I$, then $I$ is the centre of its inscribed circle. Then $\angle D F I=\angle E F I$ and since segment $E G$ is the reflection of segment $D F$ on the line $C I$, we have $\angle E F I=$ $\angle D G I$. So $\angle D F I=\angle D G I$ which means that quadrilateral $D I G F$ is cyclic. Since the quadrilateral $D E G F$ is also cyclic, we have that the quadrilateral $D I E F$ is cyclic.
Suppose that the quadrilateral $D I E F$ is cyclic. Since quadrilateral $D E G F$ is also cyclic, we have that the pentagon $D I E G F$ is cyclic. So $\angle I E B=180^{\circ}-\angle I E G=\angle I D G$ and since segment $E G$ is the reflection of segment $D F$ on the line $C I$, we have $\angle I D G=$ $\angle I E F$. Hence $\angle I E B=\angle I E F$, which means that $E I$ is the angle bisector of $\angle B E F$. Since $\angle I F A=\angle I F D=\angle I G D$ and since the segment $E G$ is the reflection of segment $D F$ on the line $C I$, we have $\angle I G D=\angle I F E$, hence $\angle I F A=\angle I F E$, which means that $F I$ is the angle bisector of $\angle E F A$. We also know that $A I$ and $B I$ are the angle bisectors of $\angle F A B$ and $\angle A B E$. So all angle bisectors of the quadrilateral $A B E F$ intersect at $I$, which means that it is circumscribable.
Comment by PSC. There is no need for introducing the point $G$. One can show that triangles $C I D$ and $C I E$ are equal and also that the triangles $C D M$ and $C E M$ are equal, where $M$ is the midpoint of $D E$. From these, one can deduce that $\angle C D I=\angle C E I$ and $\angle I D E=\angle I E D$ and proceed with similar reasoning as in the solution.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 148
|
G4. Let $A B C$ be a triangle such that $A B \neq A C$, and let the perpendicular bisector of the side $B C$ intersect lines $A B$ and $A C$ at points $P$ and $Q$, respectively. If $H$ is the orthocenter of the triangle $A B C$, and $M$ and $N$ are the midpoints of the segments $B C$ and $P Q$ respectively, prove that $H M$ and $A N$ meet on the circumcircle of $A B C$.
|
Solution. We have
$$
\angle A P Q=\angle B P M=90^{\circ}-\angle M B P=90^{\circ}-\angle C B A=\angle H C B
$$
and
$$
\angle A Q P=\angle M Q C=90^{\circ}-\angle Q C M=90^{\circ}-\angle A C B=\angle C B H
$$
From these two equalities, we see that the triangles $A P Q$ and $H C B$ are similar. Moreover, since $M$ and $N$ are the midpoints of the segments $B C$ and $P Q$ respectively, then the triangles $A Q N$ and $H B M$ are also similar. Therefore, we have $\angle A N Q=\angle H M B$.
Let $L$ be the intersection of $A N$ and $H M$. We have
$\angle M L N=180^{\circ}-\angle L N M-\angle N M L=180^{\circ}-\angle L M B-\angle N M L=180^{\circ}-\angle N M B=90^{\circ}$.
Now let $D$ be the point on the circumcircle of $A B C$ diametrically oposite to $A$. It is known that $D$ is also the relfection of point $H$ over the point $M$. Therefore, we have that $D$ belongs on $M H$ and that $\angle D L A=\angle M L A=\angle M L N=90^{\circ}$. But, as $D A$ is the diameter of the circumcirle of $A B C$, the condition that $\angle D L A=90^{\circ}$ is enough to conclude that $L$ belongs on the circumcircle of $A B C$.
Remark by PSC. There is a spiral similarity mapping $A Q P$ to $H B C$. Since the similarity maps $A N$ to $H M$, it also maps $A H$ to $N M$, and since these two lines are parallel, the centre of the similarity is $L=A N \cap H M$. Since the similarity maps $B C$ to $Q P$, its centre belongs on the circumcircle of $B C X$, where $X=B Q \cap P C$. But $X$ is the reflection of $A$ on $Q M$ and so it must belong on the circumcircle of $A B C$. Hence so must $L$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 149
|
G5. Let $P$ be a point in the interior of a triangle $A B C$. The lines $A P, B P$ and $C P$ intersect again the circumcircles of the triangles $P B C, P C A$, and $P A B$ at $D, E$ and $F$ respectively. Prove that $P$ is the orthocenter of the triangle $D E F$ if and only if $P$ is the incenter of the triangle $A B C$.
|
Solution. If $P$ is the incenter of $A B C$, then $\angle B P D=\angle A B P+\angle B A P=\frac{\hat{A}+\hat{B}}{2}$, and $\angle B D P=\angle B C P=\frac{\hat{C}}{2}$. From triangle $B D P$, it follows that $\angle P B D=90^{\circ}$, i.e. that $E B$ is one of the altitudes of the triangle $D E F$. Similarly, $A D$ and $C F$ are altitudes, which means that $P$ is the orhocenter of $D E F$.
Notice that $A P$ separates $B$ from $C, B$ from $E$ and $C$ from $F$. Therefore $A P$ separates $E$ from $F$, which means that $P$ belongs to the interior of $\angle E D F$. It follows that $P \in$ $\operatorname{Int}(\triangle D E F)$.
If $P$ is the orthocenter of $D E F$, then clearly $D E F$ must be acute. Let $A^{\prime} \in E F, B^{\prime} \in D F$ and $C^{\prime} \in D E$ be the feet of the altitudes. Then the quadrilaterals $B^{\prime} P A^{\prime} F, C^{\prime} P B^{\prime} D$, and $A^{\prime} P C^{\prime} E$ are cyclic, which means that $\angle B^{\prime} F A^{\prime}=180^{\circ}-\angle B^{\prime} P A^{\prime}=180^{\circ}-\angle B P A=$ $\angle B F A$. Similarly, one obtains that $\angle C^{\prime} D B^{\prime}=\angle C D B$, and $\angle A^{\prime} E C^{\prime}=\angle A E C$.
- If $B \in \operatorname{Ext}(\triangle F P D)$, then $A \in \operatorname{Int}(\triangle E P F), C \in \operatorname{Ext}(\triangle D P E)$, and thus $B \in$ $\operatorname{Int}(\triangle F P D)$, contradiction.
- If $B \in \operatorname{Int}(\triangle F P D)$, then $A \in \operatorname{Ext}(\triangle E P F), C \in \operatorname{Int}(\triangle D P E)$, and thus $B \in$ $\operatorname{Ext}(\triangle F P D)$, contradiction.
This leaves us with $B \in F D$. Then we must have $A \in E F, C \in D E$, which means that $A=A^{\prime}, B=B^{\prime}, C=C^{\prime}$. Thus $A B C$ is the orthic triangle of triangle $D E F$ and it is well known that the orthocenter of an acute triangle $D E F$ is the incenter of its orthic triangle.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 150
|
G6. Let $A B C$ be a non-isosceles triangle with incenter $I$. Let $D$ be a point on the segment $B C$ such that the circumcircle of $B I D$ intersects the segment $A B$ at $E \neq B$, and the circumcircle of $C I D$ intersects the segment $A C$ at $F \neq C$. The circumcircle of $D E F$ intersects $A B$ and $A C$ at the second points $M$ and $N$ respectively. Let $P$ be the point of intersection of $I B$ and $D E$, and let $Q$ be the point of intersection of $I C$ and $D F$. Prove that the three lines $E N, F M$ and $P Q$ are parallel.
|
Solution. Since $B D I E$ is cyclic, and $B I$ is the bisector of $\angle D B E$, then $I D=I E$. Similarly, $I D=I F$, so $I$ is the circumcenter of the triangle $D E F$. We also have
$$
\angle I E A=\angle I D B=\angle I F C
$$
which implies that $A E I F$ is cyclic. We can assume that $A, E, M$ and $A, N, F$ are collinear in that order. Then $\angle I E M=\angle I F N$. Since also $I M=I E=I N=I F$, the two isosceles triangles $I E M$ and $I N F$ are congruent, thus $E M=F N$ and therefore $E N$ is parallel to $F M$. From that, we can also see that the two triangles $I E A$ and $I N A$ are congruent, which implies that $A I$ is the perpendicular bisector of $E N$ and $M F$.
Note that $\angle I D P=\angle I D E=\angle I B E=\angle I B D$, so the triangles $I P D$ and $I D B$ are similar, which implies that $\frac{I D}{I B}=\frac{I P}{I D}$ and $I P \cdot I B=I D^{2}$. Similarly, we have $I Q \cdot I C=I D^{2}$, thus $I P \cdot I B=I Q \cdot I C$. This implies that $B P Q C$ is cyclic, which leads to
$$
\angle I P Q=\angle I C B=\frac{\hat{C}}{2}
$$
But $\angle A I B=90^{\circ}+\frac{\hat{C}}{2}$, so $A I$ is perpendicular to $P Q$. Hence, $P Q$ is parallel to $E N$ and FM.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 151
|
G7. Let $A B C$ be a right-angled triangle with $\hat{A}=90^{\circ}$. Let $K$ be the midpoint of $B C$, and let $A K L M$ be a parallelogram with centre $C$. Let $T$ be the intersection of the line $A C$ and the perpendicular bisector of $B M$. Let $\omega_{1}$ be the circle with centre $C$ and radius $C A$ and let $\omega_{2}$ be the circle with centre $T$ and radius $T B$. Prove that one of the points of intersection of $\omega_{1}$ and $\omega_{2}$ is on the line $L M$.
|
Solution. Let $M^{\prime}$ be the symmetric point of $M$ with respect to $T$. Observe that $T$ is equidistant from $B$ and $M$, therefore $M$ belongs on $\omega_{2}$ and $M^{\prime} M$ is a diameter of $\omega_{2}$. It suffices to prove that $M^{\prime} A$ is perpendicular to $L M$, or equivalently, to $A K$. To see this, let $S$ be the point of intersection of $M^{\prime} A$ with $L M$. We will then have $\angle M^{\prime} S M=90^{\circ}$ which shows that $S$ belongs on $\omega_{2}$ as $M^{\prime} M$ is a diameter of $\omega_{2}$. We also have that $S$ belongs on $\omega_{1}$ as $A L$ is diameter of $\omega_{1}$.
Since $T$ and $C$ are the midpoints of $M^{\prime} M$ and $K M$ respectively, then $T C$ is parallel to $M^{\prime} K$ and so $M^{\prime} K$ is perpendicular to $A B$. Since $K A=K B$, then $K M^{\prime}$ is the perpendicular bisector of $A B$. But then the triangles $K B M^{\prime}$ and $K A M^{\prime}$ are equal, showing that $\angle M^{\prime} A K=\angle M^{\prime} B K=\angle M^{\prime} B M=90^{\circ}$ as required.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 152
|
N1. Find all prime numbers $p$ for which there are non-negative integers $x, y$ and $z$ such that the number
$$
A=x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers.
|
Solution. For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Working modulo 2 and modulo 3 we see that $A$ is divisible by both 2 and 3. Moreover, by Fermat's Little Theorem, we have
$$
x^{p}+y^{p}+z^{p}-x-y-z \equiv x+y+z-x-y-z=0 \bmod p
$$
Therefore, by the given condition, we have to solve the equation
$$
x^{p}+y^{p}+z^{p}-x-y-z=6 p
$$
If one of the numbers $x, y$ and $z$ is bigger than or equal to 2 , let's say $x \geqslant 2$, then
$$
6 p \geqslant x^{p}-x=x\left(x^{p-1}-1\right) \geqslant 2\left(2^{p-1}-1\right)=2^{p}-2
$$
It is easy to check by induction that $2^{p}-2>6 p$ for all primes $p \geqslant 7$. This contradiction shows that there are no more values of $p$ which satisfy the required property.
Remark by PSC. There are a couple of other ways to prove that $2^{p}-2>6 p$ for $p \geqslant 7$. For example, we can use the Binomial Theorem as follows:
$$
2^{p}-2 \geqslant 1+p+\frac{p(p-1)}{2}+\frac{p(p-1)(p-2)}{6}-2 \geqslant 1+p+3 p+5 p-2>6 p
$$
We can also use Bernoulli's Inequality as follows:
$$
2^{p}-2=8(1+1)^{p-3}-2 \geqslant 8(1+(p-3))-2=8 p-18>6 p
$$
The last inequality is true for $p \geqslant 11$. For $p=7$ we can see directly that $2^{p}-2>6 p$.
|
proof
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 153
|
N2. Find all triples $(p, q, r)$ of prime numbers such that all of the following numbers are integers
$$
\frac{p^{2}+2 q}{q+r}, \quad \frac{q^{2}+9 r}{r+p}, \quad \frac{r^{2}+3 p}{p+q}
$$
|
Solution. We consider the following cases:
1st Case: If $r=2$, then $\frac{r^{2}+3 p}{p+q}=\frac{4+3 p}{p+q}$. If $p$ is odd, then $4+3 p$ is odd and therefore $p+q$ must be odd. From here, $q=2$ and $\frac{r^{2}+3 p}{p+q}=\frac{4+3 p}{p+2}=3-\frac{2}{p+2}$ which is not an integer. Thus $p=2$ and $\frac{r^{2}+3 p}{p+q}=\frac{10}{q+2}$ which gives $q=3$. But then $\frac{q^{2}+9 r}{r+p}=\frac{27}{4}$ which is not an integer. Therefore $r$ is an odd prime.
2nd Case: If $q=2$, then $\frac{q^{2}+9 r}{r+p}=\frac{4+9 r}{r+p}$. Since $r$ is odd, then $4+9 r$ is odd and therefore $r+p$ must be odd. From here $p=2$, but then $\frac{r^{2}+3 p}{p+q}=\frac{r^{2}+6}{4}$ which is not integer. Therefore $q$ is an odd prime.
Since $q$ and $r$ are odd primes, then $q+r$ is even. From the number $\frac{p^{2}+2 q}{q+r}$ we get that $p=2$. Since $\frac{p^{2}+2 q}{q+r}=\frac{4+2 q}{q+r}<2$, then $4+2 q=q+r$ or $r=q+4$. Since
$$
\frac{r^{2}+3 p}{p+q}=\frac{(q+4)^{2}+6}{2+q}=q+6+\frac{10}{2+q}
$$
is an integer, then $q=3$ and $r=7$. It is easy to check that this triple works. So the only answer is $(p, q, r)=(2,3,7)$.
|
(2,3,7)
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 154
|
N3. Find all prime numbers $p$ and nonnegative integers $x \neq y$ such that $x^{4}-y^{4}=$ $p\left(x^{3}-y^{3}\right)$.
|
Solution. If $x=0$ then $y=p$ and if $y=0$ then $x=p$. We will show that there are no other solutions.
Suppose $x, y>0$. Since $x \neq y$, we have
$$
p\left(x^{2}+x y+y^{2}\right)=(x+y)\left(x^{2}+y^{2}\right)
$$
If $p$ divides $x+y$, then $x^{2}+y^{2}$ must divide $x^{2}+x y+y^{2}$ and so it must also divide $x y$. This is a contradiction as $x^{2}+y^{2} \geqslant 2 x y>x y$.
Thus $p$ divides $x^{2}+y^{2}$, so $x+y$ divides $x^{2}+x y+y^{2}$. As $x+y$ divides $x^{2}+x y$ and $y^{2}+x y$, it also divides $x^{2}, x y$ and $y^{2}$. Suppose $x^{2}=a(x+y), y^{2}=b(x+y)$ and $x y=c(x+y)$. Then $x^{2}+x y+y^{2}=(a+b+c)(x+y), x^{2}+y^{2}=(a+b)(x+y)$, while $(x+y)^{2}=x^{2}+y^{2}+2 x y=(a+b+2 c)(x+y)$ yields $x+y=a+b+2 c$.
Substituting into $(*)$ gives
$$
p(a+b+c)=(a+b+2 c)(a+b)
$$
Now let $a+b=d m$ and $c=d c_{1}$, where $\operatorname{gcd}\left(m, c_{1}\right)=1$. Then
$$
p\left(m+c_{1}\right)=\left(m+2 c_{1}\right) d m
$$
If $m+c_{1}$ and $m$ had a common divisor, it would divide $c_{1}$, a contradiction. So $\operatorname{gcd}(m, m+$ $\left.c_{1}\right)=1$. and similarly, $\operatorname{gcd}\left(m+c_{1}, m+2 c_{1}\right)=1$. Thus $m+2 c_{1}$ and $m$ divide $p$, so $m+2 c_{1}=p$ and $m=1$. Then $m+c_{1}=d$ so $c \geqslant d=a+b$. Now
$$
x y=c(x+y) \geqslant(a+b)(x+y)=x^{2}+y^{2}
$$
again a contradiction.
|
proof
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 155
|
N4. Find all integers $x, y$ such that
$$
x^{3}(y+1)+y^{3}(x+1)=19
$$
|
Solution. Substituting $s=x+y$ and $p=x y$ we get
$$
2 p^{2}-\left(s^{2}-3 s\right) p+19-s^{3}=0
$$
This is a quadratic equation in $p$ with discriminant $D=s^{4}+2 s^{3}+9 s^{2}-152$.
For each $s$ we have $D0$.
For $s \geqslant 11$ and $s \leqslant-8$ we have $D>\left(s^{2}+s+3\right)^{2}$ as this is equivalent to $2 s^{2}-6 s-161>0$, and thus also to $2(s+8)(s-11)>-15$.
We have the following cases:
- If $s \geqslant 11$ or $s \leqslant-8$, then $D$ is a perfect square only when $D=\left(s^{2}+s+4\right)^{2}$, or equivalently, when $s=-21$. From (1) we get $p=232$ (which yields no solution) or $p=20$, giving the solutions $(-1,-20)$ and $(-20,-1)$.
- If $-7 \leqslant s \leqslant 10$, then $D$ is directly checked to be perfect square only for $s=3$. Then $p= \pm 2$ and only $p=2$ gives solutions, namely $(2,1)$ and $(1,2)$.
Remark by PSC. In the second bullet point, one actually needs to check 18 possible values of $s$ which is actually quite time consuming. We did not see many possible shortcuts. For example, $D$ is always a perfect square modulo 2 and modulo 3, while modulo 5 we can only get rid the four cases of the form $s \equiv 0 \bmod 5$.
|
(2,1),(1,2),(-1,-20),(-20,-1)
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 156
|
N5. Find all positive integers $x, y, z$ such that
$$
45^{x}-6^{y}=2019^{z}
$$
|
Solution. We define $v_{3}(n)$ to be the non-negative integer $k$ such that $3^{k} \mid n$ but $3^{k+1} \nmid n$. The equation is equivalent to
$$
3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}=3^{z} \cdot 673^{z}
$$
We will consider the cases $y \neq 2 x$ and $y=2 x$ separately.
Case 1. Suppose $y \neq 2 x$. Since $45^{x}>45^{x}-6^{y}=2019^{z}>45^{z}$, then $x>z$ and so $2 x>z$. We have
$$
z=v_{3}\left(3^{z} \cdot 673^{z}\right)=v_{3}\left(3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}\right)=\min \{2 x, y\}
$$
as $y \neq 2 x$. Since $2 x>z$, we get $z=y$. Hence the equation becomes $3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}=$ $3^{y} \cdot 673^{y}$, or equivalently,
$$
3^{2 x-y} \cdot 5^{x}=2^{y}+673^{y}
$$
Case 1.1. Suppose $y=1$. Doing easy manipulations we have
$$
3^{2 x-1} \cdot 5^{x}=2+673=675=3^{3} \cdot 5^{2} \Longrightarrow 45^{x-2}=1 \Longrightarrow x=2
$$
Hence one solution which satisfies the condition is $(x, y, z)=(2,1,1)$.
Case 1.2. Suppose $y \geqslant 2$. Using properties of congruences we have
$$
1 \equiv 2^{y}+673^{y} \equiv 3^{2 x-y} \cdot 5^{y} \equiv(-1)^{2 x-y} \bmod 4
$$
Hence $2 x-y$ is even, which implies that $y$ is even. Using this fact we have
$$
0 \equiv 3^{2 x-y} \cdot 5^{y} \equiv 2^{y}+673^{y} \equiv 1+1 \equiv 2 \bmod 3
$$
which is a contradiction.
Case 2. Suppose $y=2 x$. The equation becomes $3^{2 x} \cdot 5^{x}-3^{2 x} \cdot 2^{2 x}=3^{z} \cdot 673^{z}$, or equivalently,
$$
5^{x}-4^{x}=3^{z-2 x} \cdot 673^{z}
$$
Working modulo 3 we have
$$
(-1)^{x}-1 \equiv 5^{x}-4^{x} \equiv 3^{z-2 x} \cdot 673^{z} \equiv 0 \bmod 3
$$
hence $x$ is even, say $x=2 t$ for some positive integer $t$. The equation is now equivalent to
$$
\left(5^{t}-4^{t}\right)\left(5^{t}+4^{t}\right)=3^{z-4 t} \cdot 673^{z}
$$
It can be checked by hand that $t=1$ is not possible. For $t \geqslant 2$, since 3 and 673 are the only prime factors of the right hand side, and since, as it is easily checked $\operatorname{gcd}\left(5^{t}-4^{t}, 5^{t}+4^{t}\right)=1$ and $5^{t}-4^{t}>1$, the only way for this to happen is when $5^{t}-4^{t}=3^{z-4 t}$ and $5^{t}+4^{t}=673^{z}$ or $5^{t}-4^{t}=673^{z}$ and $5^{t}+4^{t}=3^{z-4 t}$. Adding together we have
$$
2 \cdot 5^{t}=3^{z-4 t}+673^{z}
$$
Working modulo 5 we have
$$
0 \equiv 2 \cdot 5^{t} \equiv 3^{z-4 t}+673^{z} \equiv 3^{4 t} \cdot 3^{z-4 t}+3^{z} \equiv 2 \cdot 3^{z} \bmod 5
$$
which is a contradiction. Hence the only solution which satisfies the equation is $(x, y, z)=$ $(2,1,1)$.
|
(2,1,1)
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 157
|
N6. Find all triples $(a, b, c)$ of nonnegative integers that satisfy
$$
a!+5^{b}=7^{c}
$$
|
Solution. We cannot have $c=0$ as $a!+5^{b} \geqslant 2>1=7^{0}$.
Assume first that $b=0$. So we are solving $a!+1=7^{c}$. If $a \geqslant 7$, then $7 \mid a!$ and so $7 \nmid a!+1$. So $7 \nmid 7^{c}$ which is impossible as $c \neq 0$. Checking $a0$. In this case, if $a \geqslant 5$, we have $5 \mid a$ !, and since $5 \mid 5^{b}$, we have $5 \mid 7^{c}$, which obviously cannot be true. So we have $a \leqslant 4$. Now we consider the following cases:
Case 1. Suppose $a=0$ or $a=1$. In this case, we are solving the equation $5^{b}+1=7^{c}$. However the Left Hand Side of the equation is always even, and the Right Hand Side is always odd, implying that this case has no solutions.
Case 2. Suppose $a=2$. Now we are solving the equation $5^{b}+2=7^{c}$. If $b=1$, we have the solution $(a, b, c)=(2,1,1)$. Now assume $b \geqslant 2$. We have $5^{b}+2 \equiv 2 \bmod 25$ which implies that $7^{c} \equiv 2 \bmod 25$. However, by observing that $7^{4} \equiv 1 \bmod 25$, we see that the only residues that $7^{c}$ can have when divided with 25 are $7,24,18,1$. So this case has no more solutions.
Case 3. Suppose $a=3$. Now we are solving the equation $5^{b}+6=7^{c}$. We have $5^{b}+6 \equiv 1 \bmod 5$ which implies that $7^{c} \equiv 1 \bmod 5$. As the residues of $7^{c}$ modulo 5 are $2,4,3,1$, in that order, we obtain $4 \mid c$.
Viewing the equation modulo 4 , we have $7^{c} \equiv 5^{b}+6 \equiv 1+2 \equiv 3 \bmod 4$. But as $4 \mid c$, we know that $7^{c}$ is a square, and the only residues that a square can have when divided by 4 are 0,1 . This means that this case has no solutions either.
Case 4. Suppose $a=4$. Now we are solving the equation $5^{b}+24=7^{c}$. We have $5^{b} \equiv 7^{c}-24 \equiv 1-24 \equiv 1 \bmod 3$. Since $5 \equiv 2 \bmod 3$, we obtain $2 \mid b$. We also have $7^{c} \equiv 5^{b}+24 \equiv 4 \bmod 5$, and so we obtain $c \equiv 2 \bmod 4$. Let $b=2 m$ and $c=2 n$. Observe that
$$
24=7^{c}-5^{b}=\left(7^{n}-5^{m}\right)\left(7^{n}+5^{m}\right)
$$
Since $7^{n}+5^{m}>0$, we have $7^{n}-5^{m}>0$. There are only a few ways to express $24=$ $24 \cdot 1=12 \cdot 2=8 \cdot 3=6 \cdot 4$ as a product of two positive integers. By checking these cases we find one by one, the only solution in this case is $(a, b, c)=(4,2,2)$.
Having exhausted all cases, we find that the required set of triples is
$$
(a, b, c) \in\{(3,0,1),(1,2,1),(4,2,2)\}
$$
|
(,b,)\in{(3,0,1),(1,2,1),(4,2,2)}
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 158
|
N7. Find all perfect squares $n$ such that if the positive integer $a \geqslant 15$ is some divisor of $n$ then $a+15$ is a prime power.
|
Solution. We call positive a integer $a$ "nice" if $a+15$ is a prime power.
From the definition, the numbers $n=1,4,9$ satisfy the required property. Suppose that for some $t \in \mathbb{Z}^{+}$, the number $n=t^{2} \geqslant 15$ also satisfies the required property. We have two cases:
1. If $n$ is a power of 2 , then $n \in\{16,64\}$ since
$$
2^{4}+15=31, \quad 2^{5}+15=47, \quad \text { and } \quad 2^{6}+15=79
$$
are prime, and $2^{7}+15=143=11 \cdot 13$ is not a prime power. (Thus $2^{7}$ does not divide $n$ and therefore no higher power of 2 satisfies the required property.)
2. Suppose $n$ has some odd prime divisor $p$. If $p>3$ then $p^{2} \mid n$ and $p^{2}>15$ which imply that $p^{2}$ must be a nice number. Hence
$$
p^{2}+15=q^{m}
$$
for some prime $q$ and some $m \in \mathbb{Z}^{+}$. Since $p$ is odd, then $p^{2}+15$ is even, thus we can conclude that $q=2$. I.e.
$$
p^{2}+15=2^{m}
$$
Considering the above modulo 3 , we can see that $p^{2}+15 \equiv 0,1 \bmod 3$, so $2^{m} \equiv$ $1 \bmod 3$, and so $m$ is even. Suppose $m=2 k$ for some $k \in \mathbb{Z}^{+}$. So we have $\left(2^{k}-p\right)\left(2^{k}+p\right)=15$ and $\left(2^{k}+p\right)-\left(2^{k}-p\right)=2 p \geqslant 10$. Thus
$$
2^{k}-p=1 \quad \text { and } \quad 2^{k}+p=15
$$
giving $p=7$ and $k=3$. Thus we can write $n=4^{x} \cdot 9^{y} \cdot 49^{z}$ for some non-negative integers $x, y, z$.
Note that 27 is not nice, so $27 \nmid n$ and therefore $y \leqslant 1$. The numbers 18 and 21 are also not nice, so similarly, $x, y$ and $y, z$ cannot both positive. Hence, we just need to consider $n=4^{x} \cdot 49^{z}$ with $z \geqslant 1$.
Note that $7^{3}$ is not nice, so $z=1$. By checking directly, we can see that $7^{2}+15=$ $2^{6}, 2 \cdot 7^{2}+15=113,4 \cdot 7^{2}+15=211$ are nice, but $8 \cdot 7^{2}$ is not nice, so only $n=49,196$ satisfy the required property.
Therefore, the numbers $n$ which satisfy the required property are $1,4,9,16,49,64$ and 196 .
Remark by PSC. One can get rid of the case $3 \mid n$ by noting that in that case, we have $9 \mid n$. But then $n^{2}+15$ is a multiple of 3 but not a multiple of 9 which is impossible. This simplifies a little bit the second case.
|
1,4,9,16,49,64,196
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 159
|
NT1 Solve in positive integers the equation $1005^{x}+2011^{y}=1006^{z}$.
|
## Solution
We have $1006^{z}>2011^{y}>2011$, hence $z \geq 2$. Then $1005^{x}+2011^{y} \equiv 0(\bmod 4)$.
But $1005^{x} \equiv 1(\bmod 4)$, so $2011^{y} \equiv-1(\bmod 4) \Rightarrow y$ is odd, i.e. $2011^{y} \equiv-1(\bmod 1006)$.
Since $1005^{x}+2011^{y} \equiv 0(\bmod 1006)$, we get $1005^{x} \equiv 1(\bmod 1006) \Rightarrow x$ is even.
Now $1005^{x} \equiv 1(\bmod 8)$ and $2011^{y} \equiv 3(\bmod 8)$, hence $1006^{z} \equiv 4(\bmod 8) \Rightarrow z=2$.
It follows that $y<2 \Rightarrow y=1$ and $x=2$. The solution is $(x, y, z)=(2,1,2)$.
|
(2,1,2)
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 160
|
NT2 Find all prime numbers $p$ such that there exist positive integers $x, y$ that satisfy the relation $x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p$.
|
## Solution
The given equation is equivalent to $(x+y)(x y-p)=5 p$. Obviously $x+y \geq 2$.
We will consider the following three cases:
Case 1: $x+y=5$ and $x y=2 p$. The equation $x^{2}-5 x+2 p=0$ has at least a solution, so we must have $0 \leq \Delta=25-8 p$ which implies $p \in\{2,3\}$. For $p=2$ we obtain the solutions $(1,4)$ and $(4,1)$ and for $p=3$ we obtain the solutions $(2,3)$ and $(3,2)$.
Case 2: $x+y=p$ and $x y=p+5$. We have $x y-x-y=5$ or $(x-1)(y-1)=6$ which implies $(x, y) \in\{(2,7) ;(3,4) ;(4,3) ;(7,2)\}$. Since $p$ is prime, we get $p=7$.
Case 3: $x+y=5 p$ and $x y=p+1$. We have $(x-1)(y-1)=x y-x-y+1=2-4 p<0$. But this is impossible since $x, y \geq 1$.
Finally, the equation has solutions in positive integers only for $p \in\{2,3,7\}$.
|
p\in{2,3,7}
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 161
|
NT3 Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\left(x^{2}+x y+3 y\right)$ has at least a solution $(x, y)$ in positive integers.
|
## Solution
Clearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \neq y$. We have $03$ and $y>x$. Take $d=\operatorname{gcd}(x+y-$ $\left.3 ; x^{2}+x y+3 y\right)$. Then $d$ divides $x^{2}+x y+3 y-x(x+y-3)=3(x+y)$. Then $d$ also divides $3(x+y)-3(x+y-3)=9$, hence $d \in\{1,3,9\}$. As $n-1=\frac{\frac{x+y-3}{d}(y-x)}{\frac{x^{2}+x y+3 y}{d}}$ and $\operatorname{gcd}\left(\frac{x+y-3}{d} ; \frac{x^{2}+x y+3 y}{d}\right)=1$, it follows that $\frac{x^{2}+x y+3 y}{d}$ divides $y-x$, which leads to $x^{2}+x y+3 y \leq d y-d x \Leftrightarrow x^{2}+d x \leq(d-3-x) y$. It is necessary that
$d-3-x>0 \Rightarrow d>3$, therefore $d=9$ and $x<6$. Take $x+y-3=9 k, k \in \mathbb{N}^{*}$ since $d \mid x+y-3$ and we get $y=9 k+3-x$. Hence $n-1=\frac{k(9 k+3-2 x)}{k(x+3)+1}$. Because $k$ and $k(x+3)+1$ are relatively prime, the number $t=\frac{9 k+3-2 x}{k(x+3)+1}$ must be integer for some positive integers $x<6$. It remains to consider these values of $x$ :
1) For $x=1$, then $t=\frac{9 k+1}{4 k+1}$ and since $1<t<3$, we get $t=2, k=1, y=11$, so $n=3$.
2) For $x=2$, then $t=\frac{9 k-1}{5 k+1}$ and since $1<t<2$, there are no solutions in this case.
3) For $x=3$, then $t=\frac{9 k-3}{6 k+1}$ and since $1 \neq t<2$, there are no solutions in this case.
4) For $x=4$, then $t=\frac{9 k-5}{7 k+1}<2$,i.e. $t=1$ which leads to $k=3, y=26$, so $n=4$.
5) For $x=5$, then $t=\frac{9 k-7}{8 k+1}<2$,i.e. $t=1$ which leads to $k=8, y=70$, so $n=9$.
Finally, the answer is $n \in\{1,3,4,9\}$.
|
n\in{1,3,4,9}
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 162
|
NT4 Find all prime positive integers $p, q$ such that $2 p^{3}-q^{2}=2(p+q)^{2}$.
|
## Solution 1
The given equation can be rewritten as $2 p^{2}(p-1)=q(3 q+4 p)$.
Hence $p\left|3 q^{2}+4 p q \Rightarrow p\right| 3 q^{2} \Rightarrow p \mid 3 q$ (since $p$ is a prime number) $\Rightarrow p \mid 3$ or $p \mid q$. If $p \mid q$, then $p=q$. The equation becomes $2 p^{3}-9 p^{2}=0$ which has no prime solution. If $p \mid 3$, then $p=3$. The equation becomes $q^{2}+4 q-12=0 \Leftrightarrow(q-2)(q+6)=0$.
Since $q>0$, we get $q=2$, so we have the solution $(p, q)=(3,2)$.
|
(p,q)=(3,2)
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 163
|
NT5 Find the least positive integer such that the sum of its digits is 2011 and the product of its digits is a power of 6 .
|
## Solution
Denote this number by $N$. Then $N$ can not contain the digits $0,5,7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then $x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+6 x_{6}+8 x_{8}+9 x_{9}=2011$. (1) The product of digits of the number $N$ is a power of 6 when we have the relation $x_{2}+2 x_{4}+x_{6}+3 x_{8}=x_{3}+x_{6}+2 x_{9}$, hence $x_{2}-x_{3}+2 x_{4}+3 x_{8}-2 x_{9}=0$.
Denote by $S$ the number of digits of $N\left(S=x_{1}+x_{2}+\ldots+x_{9}\right)$. In order to make the coefficients of $x_{8}$ and $x_{9}$ equal, we multiply relation (1) by 5 , then we add relation (2). We get $43 x_{9}+43 x_{8}+30 x_{6}+22 x_{4}+14 x_{3}+11 x_{2}+5 x_{1}=10055 \Leftrightarrow 43 S=10055+13 x_{6}+$ $21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. Then $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ is a multiple of 43 not less than 10055. The least such number is 10062 , but the relation $10062=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ means that among $x_{1}, x_{2}, \ldots, x_{6}$ there is at least one positive, so $10062=10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1} \geq 10055+13=10068$ which is obviously false. The next multiple of 43 is 10105 and from relation $10105=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ we get $50=13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. By writing as $13 x_{6}+21\left(x_{4}-1\right)+29\left(x_{3}-1\right)+32 x_{2}+38 x_{1}=0$, we can easily see that the only possibility is $x_{1}=x_{2}=x_{6}=0$ and $x_{3}=x_{4}=1$. Then $S=235, x_{8}=93, x_{9}=140$. Since $S$ is strictly minimal, we conclude that $N=34 \underbrace{88 \ldots 8}_{93} \underbrace{99 \ldots 9}_{140}$.
|
34\underbrace{88\ldots8}_{93}\underbrace{99\ldots9}_{140}
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 164
|
G1 Let $A B C$ be an isosceles triangle with $A B=A C$. On the extension of the side $[C A]$ we consider the point $D$ such that $A D<A C$. The perpendicular bisector of the segment $[B D]$ meets the internal and the external bisectors of the angle $\widehat{B A C}$ at the points $E$ and $Z$, respectively. Prove that the points $A, E, D, Z$ are concyclic.
|
Solution 1
In $\triangle A B D$ the ray $[A Z$ bisects the angle $\widehat{D A B}$ and the line $Z E$ is the perpendicular bisector of the side $[B D]$. Hence $Z$ belongs to the circumcircle of $\triangle A B D$.
Therefore the points $A, B, D, Z$ are concyclic.
In $\triangle B C D, A E$ and $Z E$ are the perpendicular bisectors $[B C]$ and $[B D]$, respectively. Hence, $E$ is the circumcenter of $\triangle B C D$ and therefore $\widehat{D E Z}=\widehat{B E D} / 2=\widehat{A C B}$. Since $B D \perp Z E$, we conclude that: $\widehat{B D E}=90^{\circ}-\widehat{D E Z}=90^{\circ}-\widehat{A C B}=\widehat{B A E}$. Hence the quadrilateral $A E B D$ is cyclic, that is the points $A, B, D, E$ are concyclic. Therefore, since $A, B, D, Z$ are also concyclic, we conclude that $A E Z D$ is cyclic.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 165
|
G2 Let $A D, B F$ and $C E$ be the altitudes of $\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\triangle A B C$, find the angle $\widehat{C G H}$.
|
## Solution 1
We can see easily that points $C, D, H, F$ lies on a circle of diameter $[C H]$.
Take $\left\{F, G^{\prime}\right\}=\odot(C H F) \cap E F$. We have $\widehat{E F H}=\widehat{B A D}=\widehat{B C E}=\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisector of $\widehat{E F D}$, so $H$ is the midpoint of the arc $D G^{\prime}$. It follows that $D G^{\prime} \perp C H$ since $[C H]$ is a diameter. Therefore $D G^{\prime} \| A B$ and $G \equiv G^{\prime}$. Finally $G$ lies on the circle $\odot(C F H)$, so $\widehat{H G C}=90^{\circ}$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 166
|
G3 Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\widehat{A B C}(L \in A C), A H$ is an altitude of $\triangle A B C(H \in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\triangle A B C$.
|
## Solution
Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \| B C$ and $L H \| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and the altitude $A H$ are also medians of $\triangle A B C$. This shows that $\triangle A B C$ is an equilateral one with all internal angles measuring $60^{\circ}$.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 167
|
G4 Point $D$ lies on the side $[B C]$ of $\triangle A B C$. The circumcenters of $\triangle A D C$ and $\triangle B A D$ are $O_{1}$ and $O_{2}$, respectively and $O_{1} O_{2} \| A B$. The orthocenter of $\triangle A D C$ is $H$ and $A H=O_{1} O_{2}$. Find the angles of $\triangle A B C$ if $2 m(<C)=3 m(<B)$.
|
## Solution 1
As $A D$ is the radical axis of the circumcircles of $\triangle A D C$ and $\triangle B A D$, we have that $O_{1} O_{2} \perp A D$, therefore $\widehat{D A B}=90^{\circ}$. Let $F$ be the midpoint of $[C D]$ and $[C E]$ be a diameter of the circumcircle of $\triangle A D C$. Then $E D \perp C D$ and $E A \perp C A$, so $E D \| A H$ and $E A \| D H$ since $A H \perp C D$ and $D H \perp A C$ (H is the orthocenter of $\triangle A D C$ ) and hence $E A H D$ is a parallelogram. Therefore $O_{1} O_{2}=A H=E D=2 O_{1} F$, so in $\triangle F O_{1} O_{2}$ with $\widehat{O_{1} F O_{2}}=90^{\circ}$ we have $O_{1} O_{2}=2 F O_{1} \Rightarrow \widehat{A B C}=\widehat{O_{1} O_{2} F}=30^{\circ}$.
Then we get $\widehat{A C B}=45^{\circ}$ and $\widehat{B A C}=105^{\circ}$.
|
\widehat{BAC}=105,\widehat{ABC}=30,\widehat{ACB}=45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 168
|
G5 Inside the square $A B C D$, the equilateral triangle $\triangle A B E$ is constructed. Let $M$ be an interior point of the triangle $\triangle A B E$ such that $M B=\sqrt{2}, M C=\sqrt{6}, M D=\sqrt{5}$ and $M E=\sqrt{3}$. Find the area of the square $A B C D$.
|
## Solution
Let $K, F, H, Z$ be the projections of point $M$ on the sides of the square.
Then by Pythagorean Theorem we can prove that $M A^{2}+M C^{2}=M B^{2}+M D^{2}$.
From the given condition we obtain $M A=1$.
With center $A$ and angle $60^{\circ}$, we rotate $\triangle A M E$, so we construct the triangle $A N B$.
Since $A M=A N$ and $\widehat{M A N}=60^{\circ}$, it follows that $\triangle A M N$ is equilateral and $M N=1$. Hence $\triangle B M N$ is right-angled because $B M^{2}+M N^{2}=B N^{2}$.
So $m(\widehat{B M A})=m(\widehat{B M N})+m(\widehat{A M N})=150^{\circ}$.
Applying Pythagorean Generalized Theorem in $\triangle A M B$, we get:
$A B^{2}=A M^{2}+B M^{2}-2 A M \cdot B M \cdot \cos 150^{\circ}=1+2+2 \sqrt{2} \cdot \sqrt{3}: 2=3+\sqrt{6}$.
We conclude that the area of the square $A B C D$ is $3+\sqrt{6}$.
|
3+\sqrt{6}
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 169
|
G6 Let $A B C D$ be a convex quadrilateral, $E$ and $F$ points on the sides $A B$ and $C D$, respectively, such that $\frac{A B}{A E}=\frac{C D}{D F}=n$. Denote by $S$ the area of the quadrilateral $A E F D$. Prove that $S \leq \frac{A B \cdot C D+n(n-1) \cdot D A^{2}+n \cdot A D \cdot B C}{2 n^{2}}$.
|
## Solution
By Ptolemy's Inequality in $A E F D$, we get $S=\frac{A F \cdot D E \cdot \sin (\widehat{A F, D E})}{2} \leq \frac{A F \cdot D E}{2} \leq$ $\frac{A E \cdot D F+A D \cdot E F}{2}=\frac{A B \cdot C D+n^{2} \cdot D A \cdot E F}{2 n^{2}}$.
Let $G$ be a point on diagonal $B D$ such that $\frac{D B}{D G}=n$. By Thales's Theorem we get $G E=\frac{(n-1) A D}{n}$ and $G F=\frac{B C}{n}$. Applying the inequality of triangle in $\triangle E G F$ we get $E F \leq E G+G F=\frac{(n-1) A D+B C}{n}$. Now, we get:
$S \leq \frac{A B \cdot C D+n^{2} A D \cdot E F}{2 n^{2}} \leq \frac{A B \cdot C D+n(n-1) \cdot D A^{2}+n \cdot A D \cdot B C}{2 n^{2}}$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 170
|
NT1. Let $a, b, p, q$ be positive integers such that $a$ and $b$ are relatively prime, $a b$ is even and $p, q \geq 3$. Prove that
$$
2 a^{p} b-2 a b^{q}
$$
$$
a \text { is even }
$$
cannot be a square of an integer number.
|
Solution. Without loss of
Let $a=2 a^{\prime}$. If $\vdots$. Without loss of generality, assume that $a$ is even and consequently $b$ is odd.
$$
2 a^{p} b-2 a b^{q}=4 a^{\prime} b\left(a^{p-1}-b^{q-1}\right)
$$
is a square, then $a^{\prime}, b$ and $a^{p-1}-b^{q-1}$ are pairwise coprime.
On the other hand, $a^{p-1}$ is divisible by 4 and $b^{q-1}$ gives the remainder 1 when divided by 4. It follows that $a^{p-1}-b^{q-1}$ has the form $4 k+3$, a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 171
|
NT3. Find all positive integers $n, n \geq 3$, such that $n \mid(n-2)$ !.
|
Solution. For $n=3$ and $n=4$ we easily check that $n$ does not divide $(n-2)!$.
If $n$ is prime, $n \geq 5$, then $(n-2)!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(n-2)$ is not divided by $n$, since $n$ is a prime not included in the set of factors of ( $n-2)$ !.
If $n$ is composite, $n \geq 6$, then $n=p m$, where $p$ is prime and $m>1$. Since $p \geq 2$, we have $n=p m \geq 2 m$, and consequently $m \leq \frac{n}{2}$. Moreover $m4$.
Therefore, if $p \neq m$, both $p$ and $m$ are distinct factors of ( $n-2)$ ! and so $n$ divides $(n-2)!$.
If $p=m$, then $n=p^{2}$ and $p>2$ (since $n>4$ ). Hence $2 p4$ ). Therefore, both $p$ and $2 p$ are distinct factors of $(n-2)$ ! and so $2 p^{2} \mid(n-2)$ !. Hence $p^{2} \mid(n-2)$ !, i.e. $n \mid(n-2)$ !. So we conclude that $n \mid(n-2)!$ if and only if $n$ is composite, $n \geq 6$.
|
proof
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 173
|
NT4. If the positive integers $x$ and $y$ are such that both $3 x+4 y$ and $4 x+3 y$ are perfect squares, prove that both $x$ and $y$ are multiples of 7 .
|
Solution. Let
$$
3 x+4 y=m^{2}, \quad 4 x+3 y=n^{2}
$$
Then
$$
7(x+y)=m^{2}+n^{2} \Rightarrow 7 \mid m^{2}+n^{2}
$$
Considering $m=7 k+r, \quad r \in\{0,1,2,3,4,5,6\}$ we find that $m^{2} \equiv u(\bmod 7), \quad u \in$ $\{0,1,2,4\}$ and similarly $n^{2} \equiv v(\bmod 7), \quad v \in\{0,1,2,4\}$. Therefore we have either $m^{2}+n^{2} \equiv 0 \quad(\bmod 7)$, when $u=v=0$, or $m^{2}+n^{2} \equiv w(\bmod 7), w \in\{1,2,3,4,5,6\}$. However, from (2) we have that $m^{2}+n^{2} \equiv 0(\bmod \tau)$ and hence $u=v=0$ and
$$
m^{2}+n^{2} \equiv 0 \quad\left(\bmod \tau^{2}\right) \Rightarrow 7(x+y) \equiv 0 \quad\left(\bmod \tau^{2}\right)
$$
and cousequently
$$
x+y \equiv 0 \quad(\bmod 7)
$$
Moreover. from (1) we have $x-y=n^{2}-m^{2}$ and $n^{2}-m^{2} \equiv 0\left(\bmod i^{2}\right)$ (since $\left.u=c=0\right)$, so
$$
x-y \equiv 0 \quad(\bmod 7) .
$$
From (3) and (4) we have tliat $x+y=7 k, x-y=7 l$, where $k$ and $l$ are positive incegers. Hence
$$
2 x=7(k+l), 2 y=7(k-l)
$$
where $k+l$ and $k-l$ are positive integers. It follows that $i \mid 2 x$ and $i \mid 2 y$, and finally $\pi \mid x$ and i|y.
## ALGEBRA
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 174
|
A1. Prove that
$$
(1+a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 3+a+b+c
$$
for any real numbers $a, b, c \geq 1$.
|
Solution. The inequality rewrites as
$$
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+(b c+a c+a b) \geq 3+a+b+c
$$
or
$$
\left(\frac{\frac{1}{a}+\frac{1}{b}}{2}+\frac{\frac{1}{b}+\frac{1}{c}}{2}+\frac{\frac{1}{a}+\frac{1}{c}}{2}\right)+(b c+a c+a b) \geq 3+a+b+c
$$
which is equivalent to
$$
\frac{(2 a b-(a+b))(a b-1)}{2 a b}+\frac{(2 b c-(b+c))(b c-1)}{2 b c}+\frac{(2 c a-(c+a))(c a-1)}{2 c a} \geq 0
$$
The last inequality is true due to the obvious relations
$$
2 x y-(x+y)=x(y-1)+y(x-1) \geq 0
$$
for any two real numbers $x, y \geq 1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 175
|
A2. Prove that, for all real numbers $x, y, z$ :
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq(x+y+z)^{2}
$$
When the equality holds?
|
Solution. For $x=y=z=0$ the equality is valid.
Since $(x+y+z)^{2} \geq 0$ it is enongh to prove that
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq 0
$$
which is equivalent to the inequality
$$
\frac{x^{2}-y^{2}}{x^{2}+\frac{1}{2}}+\frac{y^{2}-z^{2}}{y^{2}+\frac{1}{2}}+\frac{z^{2}-x^{2}}{z^{2}+\frac{1}{2}} \leq 0
$$
Dellote
$$
a=x^{2}+\frac{1}{2}, b=y^{2}+\frac{1}{2}, c=z^{2}+\frac{1}{2}
$$
Then (1) is equivalent to
$$
\frac{a-b}{a}+\frac{b-c}{b}+\frac{c-a}{c} \leq 0
$$
From very well known $A G$ inequality follows that
$$
a^{2} b+b^{2} c+c^{2} a \geq 3 a b c
$$
Fron the equivalencies
$$
a^{2} b+b^{2} c+c^{2} a \geq 3 a b c \Leftrightarrow \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geq-3 \Leftrightarrow \frac{a-b}{a}+\frac{b-c}{b}+\frac{c-a}{c} \leq 0
$$
follows thait the inequality (2) is valid, for positive real numbers $a, b, c$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 176
|
A3. Prove that for all real $x, y$
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{2 \sqrt{2}}{\sqrt{x^{2}+y^{2}}}
$$
|
Solution. The inequality rewrites as
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{\sqrt{2\left(x^{2}+y^{2}\right)}}{\frac{x^{2}+y^{2}}{2}}
$$
Now it is enough to prove the next two simple inequalities:
$$
x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}, \quad x^{2}-x y+y^{2} \geq \frac{x^{2}+y^{2}}{2}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 177
|
A4. Prove that if $0<\frac{a}{b}<b<2 a$ then
$$
\frac{2 a b-a^{2}}{7 a b-3 b^{2}-2 a^{2}}+\frac{2 a b-b^{2}}{7 a b-3 a^{2}-2 b^{2}} \geq 1+\frac{1}{4}\left(\frac{a}{b}-\frac{b}{a}\right)^{2}
$$
|
Solution. If we denote
$$
u=2-\frac{a}{b}, \quad v=2-\frac{b}{a}
$$
then the inequality rewrites as
$$
\begin{aligned}
& \frac{u}{v+u v}+\frac{v}{u+u v} \geq 1+\frac{1}{4}(u-u)^{2} \\
& \frac{(u-v)^{2}+u v(1-u v)}{u v(u v+u+v+1)} \geq \frac{(u-u)^{2}}{4}
\end{aligned}
$$
$\mathrm{Or}$
Since $u>0, v>0, u+v \leq 2, u v \leq 1, u v(u+v+u v+1) \leq 4$, the result is clear.
## GEOMETRY
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 178
|
G1. Two circles $k_{1}$ and $k_{2}$ intersect a.t points $A$ and $B$. A circle $k_{3}$ centered at $A$ meet $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $M P$ and $A B>A M$.
Prove rhat the angles $\angle M B Q$ and $\angle N B P$ are equal.
|
Solution As $A M=A P$, we have
$$
\angle M B A=\frac{1}{2} \operatorname{arcAM}=\frac{1}{2} \operatorname{arc} A P=\angle A B P
$$
and likewise
$$
\angle Q B A=\frac{1}{2} \operatorname{arc} A Q=\frac{1}{2} \operatorname{arc} c A N=\angle A B N
$$
Summing these equalities yields $\angle M B Q=\angle N B P$ as needed.
Q2. Let $E$ and $F$ be two distinct points inside of a parallelogram $A B C D$. Find the maximum number of triangles with the same area and having the vertices in three of the following five points: $A, B, C, D, E, F$.
|
notfound
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 179
|
G5. Let $A B C$ be a triangle with $\angle C=90^{\circ}$ and $D \in C .4, E \in C^{\prime} B$, and $k_{1}, k_{2}, k_{3}, k_{4}$ semicircles with diameters $C A, C B, C D, C E$ respectively, which have common part with the triangle $A B C$. Let also,
$$
k_{1} \cap k_{2}=\{C, K\}, k_{3} \cap k_{4}=\{C, M\}, k_{2} \cap k_{3}=\{C, L\}, k_{1} \cap k_{4}=\{C, N\}
$$
Prove that $K, L, M$ and $N$ are cocyclic points.
|
Solution. The points $K, L, M, N$ belong to the segments $A B, B D, D E, E A$ respectively, where $C K \perp A B, C L \perp B D, C M \perp D E, C N \perp A E$. Then quadrilaterals $C D L M$ and $C E N M$ are inscribed. Let $\angle C A E=\varphi, \angle D C L=\theta$. Then $\angle E M N=\angle E C N=\varphi$ and $\angle D M L=\angle D C L=\theta$. So $\angle D M L+\angle E M N=\varphi+\theta$ and therefore $\angle L M N=180^{\circ}-\varphi-\theta$. The quadrilaterals $C B K L$ and $C A K N$ are also inscribed and hence $\angle L K C=\angle L B C=\theta$, $\angle C K N=\angle C A N=\varphi$, and so $\angle L K N=\varphi+\theta$, while $\angle L M N=180^{\circ}-\varphi-\theta$, which means that $K L M N$ is inscribed.
Note that $K L M N$ is convex because $L, N$ lie in the interior of the convex quadrilateral $A D E B$.
## COMBINATORICS
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 181
|
C1. A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that have in common with the poligon only vertices.
Prove that there are 2 more black triangles than white ones.
|
Solution. Denote by $b, r, w$ the number of black, red white triangles respectively.
It is easy to prove that the polygon is divided into $n-2$ triangles, hence
$$
b+r+w=n-2
$$
Each side of the polygon is a side of exactly one triangle of the decomposition, and thus
$$
2 b+r=n
$$
Subtracting the two relations yields $w=b-2$, as needed. allowed:
|
b-2
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 182
|
C2. Given $m \times n$ table, each cell signed with "-". The following operations are
(i) to change all the signs in entire row to the opposite, i. e. every "-" to "+", and every "+" to "-";
(ii) to change all the signs in entire column to the opposite, i. e. every "-" to "+" and every "+" to " -".
(a) Prove that if $m=n=100$, using the above operations one can not obtain 2004 signs "t".
(b) If $m=1004$, find the least $n>100$ for which 2004 signs " + " can be obtained.
|
Solution. If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$
(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le
$$
50 l+50 k-1 k=1002
$$
Rewrite the lasc equation as
$$
(50-l)(50-h)=2.500-100.2=1498
$$
Since $1498=2 \cdot 7 \cdot 107$, this equation has no solitions in natural numbers.
(b) Let $n=101$. Then we have
$$
(100-k) l+(101-l) k=2004
$$
OI
$$
100 l+101 k-2 l k=2004
$$
l.e.
$$
101 k=2004-100 l+2 l k \div 2(1002-50 l+l k)
$$
Hence $s=2 t$ and we have $101 t=501-25 l+2 l t$. From here we have
$$
t=\frac{501-25 l}{101-2 l}=4+\frac{97-17 l}{101-2 l}
$$
Since $t$ is natural number and $97-17 l<101-2 l$, this is a contradiction, Hence $n \neq 101$. Let $n=1.02$. Then we have
$$
(100-k) l+(102-l) k=2004
$$
or
$$
100 l+102 k-2 l k=2004
$$
$$
50 l+51 k-l k=1002
$$
Rewrite the last equation as
$$
(51-l)(50-k)=25.50-1002=1.548
$$
Since $145 S=2 \cdot 2 \cdot 3 \cdot 3 \cdot 43$ we have $51-l=36$ and $50-k=43$. From here obtain $l=15$ and $k=7$. Indeed,
$$
(100-\bar{\imath}) \cdot 15+(102-1.5) \cdot \overline{7}=93 \cdot 15+87 \cdot 7=1395+609=2004
$$
Hence, the least $n$ is 102 .
|
102
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 183
|
Problem A1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0
$$
implying
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}
$$
It follows that $(a+b)(b+c)(c+a)=0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $0=2+3$, a contradiction). Thus $a+b+c+d \neq 0$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 184
|
Problem A2. Determine all four digit numbers $\overline{a b c d}$ such that
$$
a(a+b+c+d)\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\right)=\overline{a b c d}
$$
|
Solution. From $\overline{a b c d}\overline{1 b c d}=(1+b+c+d)\left(1+b^{2}+c^{2}+d^{2}\right)\left(1+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $(b+1)\left(b^{2}+1\right)\left(2 b^{6}+1\right)$, so $b \leq 2$. Similarly one gets $c\overline{2 b c d}=2(2+b+c+d)\left(4+b^{2}+c^{2}+d^{2}\right)\left(64+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $2(b+2)\left(b^{2}+4\right)\left(2 b^{6}+64\right)$, imposing $b \leq 1$. In the same way one proves $c<2$ and $d<2$. By direct check, we find out that 2010 is the only solution.
|
2010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 185
|
Problem A3. Find all pairs $(x, y)$ of real numbers such that $|x|+|y|=1340$ and $x^{3}+y^{3}+2010 x y=670^{3}$.
|
Solution. Answer: $(-670 ;-670),(1005 ;-335),(-335 ; 1005)$.
To prove this, let $z=-670$. We have
$$
0=x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right)
$$
Thus either $x+y+z=0$, or $x=y=z$. In the latter case we get $x=y=-670$, which satisfies both the equations. In the former case we get $x+y=670$. Then at least one of $x, y$ is positive, but not both, as from the second equation we would get $x+y=1340$. If $x>0 \geq y$, we get $x-y=1340$, which together with $x+y=670$ yields $x=1005, y=-335$. If $y>0 \geq x$ we get similarly $x=-335, y=1005$.
|
(-670,-670),(1005,-335),(-335,1005)
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 186
|
Problem A4. Let $a, b, c$ be positive real numbers such that $a b c(a+b+c)=3$. Prove the inequality
$$
(a+b)(b+c)(c+a) \geq 8
$$
and determine all cases when equality holds.
|
Solution. We have
$A=(a+b)(b+c)(c+a)=\left(a b+a c+b^{2}+b c\right)(c+a)=(b(a+b+c)+a c)(c+a)$,
so by the given condition
$$
A=\left(\frac{3}{a c}+a c\right)(c+a)=\left(\frac{1}{a c}+\frac{1}{a c}+\frac{1}{a c}+a c\right)(c+a)
$$
Aplying the AM-GM inequality for four and two terms respectively, we get
$$
A \geq 4 \sqrt[4]{\frac{a c}{(a c)^{3}}} \cdot 2 \sqrt{a c}=8
$$
From the last part, it is easy to see that inequality holds when $a=c$ and $\frac{1}{a c}=a c$, i.e. $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 187
|
Problem A5. The real positive numbers $x, y, z$ satisfy the relations $x \leq 2$, $y \leq 3, x+y+z=11$. Prove that $\sqrt{x y z} \leq 6$.
|
Solution. For $x=2, y=3$ and $z=6$ the equality holds.
After the substitutions $x=2-u, y=3-v$ with $u \in[0,2), v \in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes
$$
(2-u)(3-v)(6+u+v) \leqslant 36
$$
We shall need the following lemma.
Lemma. If real numbers $a$ and $b$ satisfy the relations $00$ and $y \geq 0$.
The equality in (2) holds if $y=0$. The lemma is proved.
By using the lemma we can write the following inequalities:
$$
\begin{gathered}
\frac{6}{6+u} \geqslant \frac{2-u}{2} \\
\frac{6}{6+v} \geqslant \frac{3-v}{3} \\
\frac{6+u}{6+u+v} \geqslant \frac{6}{6+v}
\end{gathered}
$$
By multiplying the inequalities (3), (4) and (5) we obtain:
$$
\begin{gathered}
\frac{6 \cdot 6 \cdot(6+u)}{(6+u)(6+v)(6+u+v)} \geqslant \frac{6(2-u)(3-v)}{2 \cdot 3(6+v)} \Leftrightarrow \\
(2-u)(3-v)(6+u+v) \leqslant 2 \cdot 3 \cdot 6=36 \Leftrightarrow \quad(1)
\end{gathered}
$$
By virtue of lemma, the equality holds if and only if $u=v=0$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 188
|
Problem G1. Consider a triangle $A B C$ with $\angle A C B=90^{\circ}$. Let $F$ be the foot of the altitude from $C$. Circle $\omega$ touches the line segment $F B$ at point $P$, the altitude $C F$ at point $Q$ and the circumcircle of $A B C$ at point $R$. Prove that points $A, Q, R$ are collinear and $A P=A C$.
|
Solution. Let $M$ be the midpoint of $A B$ and let $N$ be the center of $\omega$. Then $M$ is the circumcenter of triangle $A B C$, so points $M, N$ and $R$ are collinear. From $Q N \| A M$ we get $\angle A M R=\angle Q N R$. Besides that, triangles $A M R$ and $Q N R$ are isosceles, therefore $\angle M R A=\angle N R Q$; thus points $A, Q, R$ are collinear.
Right angled triangles $A F Q$ and $A R B$ are similar, which implies $\frac{A Q}{A B}=\frac{A F}{A R}$, that is $A Q \cdot A R=A F \cdot A B$. The power of point $A$ with respect to $\omega$ gives $A Q \cdot A R=A P^{2}$. Also, from similar triangles $A B C$ and $A C F$ we get $A F \cdot A B=$ $A C^{2}$. Now, the claim follows from $A C^{2}=A F \cdot A B=A Q \cdot A R=A P^{2}$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 189
|
Problem G2. Consider a triangle $A B C$ and let $M$ be the midpoint of the side $B C$. Suppose $\angle M A C=\angle A B C$ and $\angle B A M=105^{\circ}$. Find the measure of $\angle A B C$.
|
Solution. The angle measure is $30^{\circ}$.
Let $O$ be the circumcenter of the triangle $A B M$. From $\angle B A M=105^{\circ}$ follows $\angle M B O=15^{\circ}$. Let $M^{\prime}, C^{\prime}$ be the projections of points $M, C$ onto the line $B O$. Since $\angle M B O=15^{\circ}$, then $\angle M O M^{\prime}=30^{\circ}$ and consequently $M M^{\prime}=\frac{M O}{2}$. On the other hand, $M M^{\prime}$ joins the midpoints of two sides of the triangle $B C C^{\prime}$, which implies $C C^{\prime}=M O=A O$.
The relation $\angle M A C=\angle A B C$ implies $C A$ tangent to $\omega$, hence $A O \perp A C$. It follows that $\triangle A C O \equiv \triangle O C C^{\prime}$, and furthermore $O B \| A C$.
Therefore $\angle A O M=\angle A O M^{\prime}-\angle M O M^{\prime}=90^{\circ}-30^{\circ}=60^{\circ}$ and $\angle A B M=$ $\frac{\angle A O M}{2}=30^{\circ}$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 190
|
Problem G3. Let $A B C$ be an acute-angled triangle. A circle $\omega_{1}\left(O_{1}, R_{1}\right)$ passes through points $B$ and $C$ and meets the sides $A B$ and $A C$ at points $D$ and $E$, respectively. Let $\omega_{2}\left(O_{2}, R_{2}\right)$ be the circumcircle of the triangle $A D E$. Prove that $O_{1} O_{2}$ is equal to the circumradius of the triangle $A B C$.
|
Solution. Recall that, in every triangle, the altitude and the diameter of the circumcircle drawn from the same vertex are isogonal. The proof offers no difficulty, being a simple angle chasing around the circumcircle of the triangle.
Let $O$ be the circumcenter of the triangle $A B C$. From the above, one has $\angle O A E=90^{\circ}-B$. On the other hand $\angle D E A=B$, for $B C D E$ is cyclic. Thus $A O \perp D E$, implying that in the triangle $A D E$ cevians $A O$ and $A O_{2}$ are isogonal. So, since $A O$ is a radius of the circumcircle of triangle $A B C$, one obtains that $A O_{2}$ is an altitude in this triangle.
Moreover, since $O O_{1}$ is the perpendicular bisector of the line segment $B C$, one has $O O_{1} \perp B C$, and furthermore $A O_{2} \| O O_{1}$.
Chord $D E$ is common to $\omega_{1}$ and $\omega_{2}$, hence $O_{1} O_{2} \perp D E$. It follows that $A O \|$ $O_{1} O_{2}$, so $A O O_{1} O_{2}$ is a parallelogram. The conclusion is now obvious.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 191
|
Problem G4. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C(L \in B C, K \in A C)$. The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N \| M K$. Prove that $L N=N A$.
|
Solution. The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=$ $\angle N B L=\angle C B K=\angle N L A$. Now it follows that $L N=N A$.
## Combinatorics
|
LN=NA
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 192
|
Problem C1. There are two piles of coins, each containing 2010 pieces. Two players A and B play a game taking turns (A plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if they both play in the best possible way?
|
Solution. B wins.
In fact, we will show that A will lose if the total number of coins is a multiple of 3 and the two piles differ by not more than one coin (call this a balanced position). To this end, firstly notice that it is not possible to move from one balanced position to another. The winning strategy for B consists in returning A to a balanced position (notice that the initial position is a balanced position).
There are two types of balanced positions; for each of them consider the moves of $\mathrm{A}$ and the replies of B.
If the number in each pile is a multiple of 3 and there is at least one coin:
- if A takes $3 n$ coins from one pile, then $\mathrm{B}$ takes $3 n$ coins from the other one.
- if A takes $3 n+1$ coins from one pile, then $\mathrm{B}$ takes $3 n+2$ coins from the other one.
- if A takes $3 n+2$ coins from one pile, then $\mathrm{B}$ takes $3 n+1$ coins from the other one.
- if A takes a coin from each pile, then $\mathrm{B}$ takes one coin from one pile.
If the numbers are not multiples of 3 , then we have $3 m+1$ coins in one pile and $3 m+2$ in the other one. Hence:
- if A takes $3 n$ coins from one pile, then $\mathrm{B}$ takes $3 n$ coins from the other one.
- if A takes $3 n+1$ coins from the first pile $(n \leq m)$, then $\mathrm{B}$ takes $3 n+2$ coins from the second one.
- if A takes $3 n+2$ coins from the second pile $(n \leq m)$, then $\mathrm{B}$ takes $3 n+1$ coins from the first one.
- if A takes $3 n+2$ coins from the first pile $(n \leq m-1)$, then $\mathrm{B}$ takes $3 n+4$ coins from the second one.
- if A takes $3 n+1$ coins from the second pile $(n \leq m)$, then $\mathrm{B}$ takes $3 n-1$ coins from the first one. This is impossible if A has taken only one coin from the second pile; in this case $\mathrm{B}$ takes one coin from each pile.
- if A takes a coin from each pile, then B takes one coin from the second pile.
In all these cases, the position after B's move is again a balanced position. Since the number of coins decreases and $(0 ; 0)$ is a balanced position, after a finite number of moves, there will be no coins left after B's move. Thus, B wins.
|
Bwins
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 193
|
Problem N1. Find all positive integers $n$ such that $n 2^{n+1}+1$ is a perfect square.
|
Solution. Answer: $n=0$ and $n=3$.
Clearly $n 2^{n+1}+1$ is odd, so, if this number is a perfect square, then $n 2^{n+1}+1=$ $(2 x+1)^{2}, x \in \mathbb{N}$, whence $n 2^{n-1}=x(x+1)$.
The integers $x$ and $x+1$ are coprime, so one of them must divisible by $2^{n-1}$, which means that the other must be at most $n$. This shows that $2^{n-1} \leqslant n+1$.
An easy induction shows that the above inequality is false for all $n \geqslant 4$, and a direct inspection confirms that the only convenient values in the case $n \leqslant 3$ are 0 and 3 .
|
n=0n=3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 195
|
Problem N2. Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.
|
Solution. Answer: $n=1$.
Among each four consecutive integers there is a multiple of 4 . As $36^{n}-6$ is not a multiple of 4 , it must be the product of two or three consecutive positive integers.
Case I. If $36^{n}-6=x(x+1)$ (all letters here and below denote positive integers), then $4 \cdot 36^{n}-23=(2 x+1)^{2}$, whence $\left(2 \cdot 6^{n}+2 x+1\right)\left(2 \cdot 6^{n}-2 x-1\right)=23$. As 23 is prime, this leads to $2 \cdot 6^{n}+2 x+1=23,2 \cdot 6^{n}-2 x-1=1$. Subtracting these yields $4 x+2=22, x=5, n=1$, which is a solution to the problem.
Case II. If $36^{n}-6=(y-1) y(y+1)$, then
$$
36^{n}=y^{3}-y+6=\left(y^{3}+8\right)-(y+2)=(y+2)\left(y^{2}-2 y+3\right)
$$
Thus each of $y+2$ and $y^{2}-2 y+3$ can have only 2 and 3 as prime factors, so the same is true for their GCD. This, combined with the identity $y^{2}-2 y+3=$ $(y+2)(y-4)+11$ yields $\operatorname{GCD}\left(y+2 ; y^{2}-2 y+3\right)=1$. Now $y+2<y^{2}-2 y+3$ and the latter number is odd, so $y+2=4^{n}, y^{2}-2 y+3=9^{n}$. The former identity implies $y$ is even and now by the latter one $9^{n} \equiv 3(\bmod 4)$, while in fact $9^{n} \equiv 1(\bmod 4)$ - a contradiction. So, in this case there is no such $n$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 196
|
A1. Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 6
$$
|
Solution. We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that
$$
(a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)
$$
so that
$$
\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8
$$
Applying AM-GM, we conclude:
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 3 \cdot \sqrt[3]{\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}}=6
$$
Alternatively, we can write LHS as
$$
\frac{b c(a b+4)}{2(b c+4)}+\frac{a c(b c+4)}{2(a c+4)}+\frac{a b(c a+4)}{2(a b+4)}
$$
and then apply AM-GM.
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 197
|
A2. Given positive real numbers $a, b, c$, prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(a+c)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$ and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$. Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b c} \geq \frac{4}{\left(a^{2}+b^{2}\right)(c+1)}
$$
Using the AM-GM inequality, we have
$$
\frac{4}{\left(a^{2}+b^{2}\right)(c+1)}+\frac{a^{2}+b^{2}}{2} \geq 2 \sqrt{\frac{2}{c+1}}=\frac{4}{\sqrt{2(c+1)}}
$$
respectively
$$
\frac{c+3}{8}=\frac{(c+1)+2}{8} \geq \frac{\sqrt{2(c+1)}}{4}
$$
We conclude that
$$
\frac{4}{\left(a^{2}+b^{2}\right)(c+1)}+\frac{a^{2}+b^{2}}{2} \geq \frac{8}{c+3}
$$
and finally
$\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(a+c)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 198
|
A3. Determine the number of pairs of integers $(m, n)$ such that
$$
\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q}
$$
|
Solution. Let $r=\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}}$. Then
$$
n+m+2 \sqrt{n+\sqrt{2016}} \cdot \sqrt{m-\sqrt{2016}}=r^{2}
$$
and
$$
(m-n) \sqrt{2106}=\frac{1}{4}\left(r^{2}-m-n\right)^{2}-m n+2016 \in \mathbb{Q}
$$
Since $\sqrt{2016} \notin \mathbb{Q}$, it follows that $m=n$. Then
$$
\sqrt{n^{2}-2016}=\frac{1}{2}\left(r^{2}-2 n\right) \in \mathbb{Q}
$$
Hence, there is some nonnegative integer $p$ such that $n^{2}-2016=p^{2}$ and (1) becomes $2 n+2 p=r^{2}$.
It follows that $2(n+p)=r^{2}$ is the square of a rational and also an integer, hence a perfect square. On the other hand, $2016=(n-p)(n+p)$ and $n+p$ is a divisor of 2016, larger than $\sqrt{2016}$. Since $n+p$ is even, so is also $n-p$, and $r^{2}=2(n+p)$ is a divisor of $2016=2^{5} \cdot 3^{2} \cdot 7$, larger than $2 \sqrt{2016}>88$. The only possibility is $r^{2}=2^{4} \cdot 3^{2}=12^{2}$. Hence, $n+p=72$ and $n-p=28$, and we conclude that $n=m=50$. Thus, there is only one such pair.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 199
|
A4. If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:
$$
\frac{x+1}{\sqrt{x^{5}+x+1}}+\frac{y+1}{\sqrt{y^{5}+y+1}}+\frac{z+1}{\sqrt{z^{5}+z+1}} \geq 3
$$
When does the equality hold?
|
Solution. First we factor $x^{5}+x+1$ as follows:
$$
\begin{aligned}
x^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\left(x^{3}-1\right)+x^{2}+x+1=x^{2}(x-1)\left(x^{2}+x+1\right)+x^{2}+x+1 \\
& =\left(x^{2}+x+1\right)\left(x^{2}(x-1)+1\right)=\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)
\end{aligned}
$$
Using the $A M-G M$ inequality, we have
$$
\sqrt{x^{5}+x+1}=\sqrt{\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)} \leq \frac{x^{2}+x+1+x^{3}-x^{2}+1}{2}=\frac{x^{3}+x+2}{2}
$$
and since
$x^{3}+x+2=x^{3}+1+x+1=(x+1)\left(x^{2}-x+1\right)+x+1=(x+1)\left(x^{2}-x+1+1\right)=(x+1)\left(x^{2}-x+2\right)$,
then
$$
\sqrt{x^{5}+x+1} \leq \frac{(x+1)\left(x^{2}-x+2\right)}{2}
$$
Using $x^{2}-x+2=\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}>0$, we obtain $\frac{x+1}{\sqrt{x^{5}+x+1}} \geq \frac{2}{x^{2}-x+2}$ Applying the CauchySchwarz inequality and the given condition, we get
$$
\sum_{c y c} \frac{x+1}{\sqrt{x^{5}+x+1}} \geq \sum_{c y c} \frac{2}{x^{2}-x+2} \geq \frac{18}{\sum_{c y c}\left(x^{2}-x+2\right)}=\frac{18}{6}=3
$$
which is the desired result.
For the equality both conditions: $x^{2}-x+2=y^{2}-y+2=z^{2}-z+2$ (equality in CBS) and $x^{3}-x^{2}+1=x^{2}+x+1$ (equality in AM-GM) have to be satisfied.
By using the given condition it follows that $x^{2}-x+2+y^{2}-y+2+z^{2}-z+2=6$, hence $3\left(x^{2}-x+2\right)=6$, implying $x=0$ or $x=1$. Of these, only $x=0$ satisfies the second condition. We conclude that equality can only hold for $x=y=z=0$.
It is an immediate check that indeed for these values equality holds.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 200
|
A5. Let $x, y, z$ be positive real numbers such that $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
a) Prove the inequality
$$
x+y+z \geq \sqrt{\frac{x y+1}{2}}+\sqrt{\frac{y z+1}{2}}+\sqrt{\frac{z x+1}{2}}
$$
b) (Added by the problem selecting committee) When does the equality hold?
|
## Solution.
a) We rewrite the inequality as
$$
(\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1})^{2} \leq 2 \cdot(x+y+z)^{2}
$$
and note that, from CBS,
$$
\text { LHS } \leq\left(\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}\right)(x+y+z)
$$
But
$$
\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2(x+y+z)
$$
which proves (1).
b) The equality occurs when we have equality in CBS, i.e. when
$$
\frac{x y+1}{x^{2}}=\frac{y z+1}{y^{2}}=\frac{z x+1}{z^{2}}\left(=\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\right)
$$
Since we can also write
$$
(\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1})^{2} \leq\left(\frac{x y+1}{y}+\frac{y z+1}{z}+\frac{z x+1}{x}\right)(y+z+x)=2(x+y+z)^{2}
$$
the equality implies also
$$
\frac{x y+1}{y^{2}}=\frac{y z+1}{z^{2}}=\frac{z x+1}{x^{2}}\left(=\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\right)
$$
But then $x=y=z$, and since $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, we conclude that $x=\frac{1}{x}=1=y=z$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 201
|
C1. Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1}+\frac{1}{3}$. Find the least positive integer $k$ making the number $k!\cdot S_{2016}$ an integer.
|
## Solution.
We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$.
Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence
$$
S_{999}=300 \cdot\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{9}\right)
$$
For the numbers in the interval $1000 \rightarrow 1999$, compared to $0 \rightarrow 999$, there are precisely 1000 more digits 1 . We get
$$
S_{1999}-S_{999}=1000+S_{999} \Longrightarrow S_{1999}=1000+600 \cdot\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{9}\right)
$$
Finally, in the interval $2000 \rightarrow 2016$, the digit 1 appears twice as unit and seven times as a ten, the digit 2 twice as a unit and 17 times as a thousand, the digits $3,4,5$, and 6 each appear exactly twice as units, and the digits 7,8 , and 9 each appear exactly once as a unit. Hence
$$
S_{2016}-S_{1999}=9 \cdot 1+19 \cdot \frac{1}{2}+2 \cdot\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)+1 \cdot\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)
$$
In the end, we get
$$
\begin{aligned}
S_{2016} & =1609 \cdot 1+619 \cdot \frac{1}{2}+602 \cdot\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)+601 \cdot\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right) \\
& =m+\frac{1}{2}+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+\frac{2}{6}+\frac{6}{7}+\frac{1}{8}+\frac{7}{9}=n+\frac{p}{2^{3} \cdot 3^{2} \cdot 5 \cdot 7}
\end{aligned}
$$
where $m, n$, and $p$ are positive integers, $p$ coprime to $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Then $k!\cdot S_{2016}$ is an integer precisely when $k$ ! is a multiple of $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Since $7 \mid k!$, it follows that $k \geq 7$. Also, $7!=2^{4} \cdot 3^{2} \cdot 5 \cdot 7$, implying that the least $k$ satisfying $k!\cdot S_{2016} \in \mathbb{Z}$ is $k=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 202
|
C2. The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?
|
Solution. Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2 , so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each pair is a prime:
$$
\begin{aligned}
& (1,2),(3,4),(5,6),(7,10),(8,9),(11,12),(13,16),(14,15),(17,20) \\
& (18,19),(21,22),(23,24),(25,28),(26,27),(29,30),(31,36),(32,35) \\
& (33,34),(37,42),(38,41),(39,40),(43,46),(44,45),(47,50),(48,49)
\end{aligned}
$$
Since at least one number from each pair has to be deleted, the minimal number is 25 .
|
25
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 203
|
C3. Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total sums that may be obtained in this way.
|
Solution. We will prove that the maximum number of total sums is 60 .
The proof is based on the following claim.
Claim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.
Proof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then, in row $R$ we can find three of the numbers in consecutive positions, let $x, y, z$ be the numbers in consecutive positions(where $\{x, y, s, z\}=\{a, b, c, d\}$ ). Due to our hypothesis that in every $2 \times 2$ subarray each number is used exactly once, in the row above $\mathrm{R}$ (if there is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order. And above them will be the numbers $x, y, z$ in this order. The same happens in the rows below $R$ (see at the following figure).
$$
\left(\begin{array}{lllll}
\bullet & x & y & z & \bullet \\
\bullet & z & t & x & \bullet \\
\bullet & x & y & z & \bullet \\
\bullet & z & t & x & \bullet \\
\bullet & x & y & z & \bullet
\end{array}\right)
$$
Completing all the array, it easily follows that each column contains exactly two of the numbers and our claim has been proven.
Rotating the matrix (if it is necessary), we may assume that each row contains exactly two of the numbers. If we forget the first row and column from the array, we obtain a $4 \times 4$ array, that can be divided into four $2 \times 2$ subarrays, containing thus each number exactly four times, with a total sum of $4(a+b+c+d)$. It suffices to find how many different ways are there to put the numbers in the first row $R_{1}$ and the first column $C_{1}$.
Denoting by $a_{1}, b_{1}, c_{1}, d_{1}$ the number of appearances of $a, b, c$, and respectively $d$ in $R_{1}$ and $C_{1}$, the total sum of the numbers in the entire $5 \times 5$ array will be
$$
S=4(a+b+c+d)+a_{1} \cdot a+b_{1} \cdot b+c_{1} \cdot c+d_{1} \cdot d
$$
If the first, the third and the fifth row contain the numbers $x, y$, with $x$ denoting the number at the entry $(1,1)$, then the second and the fourth row will contain only the numbers $z, t$, with $z$ denoting the number at the entry $(2,1)$. Then $x_{1}+y_{1}=7$ and $x_{1} \geqslant 3$, $y_{1} \geqslant 2, z_{1}+t_{1}=2$, and $z_{1} \geqslant t_{1}$. Then $\left\{x_{1}, y_{1}\right\}=\{5,2\}$ or $\left\{x_{1}, y_{1}\right\}=\{4,3\}$, respectively $\left\{z_{1}, t_{1}\right\}=\{2,0\}$ or $\left\{z_{1}, t_{1}\right\}=\{1,1\}$. Then $\left(a_{1}, b_{1}, c_{1}, d_{1}\right)$ is obtained by permuting one of the following quadruples:
$$
(5,2,2,0),(5,2,1,1),(4,3,2,0),(4,3,1,1)
$$
There are a total of $\frac{4!}{2!}=12$ permutations of $(5,2,2,0)$, also 12 permutations of $(5,2,1,1)$, 24 permutations of $(4,3,2,0)$ and finally, there are 12 permutations of $(4,3,1,1)$. Hence, there are at most 60 different possible total sums.
We can obtain indeed each of these 60 combinations: take three rows ababa alternating
with two rows $c d c d c$ to get $(5,2,2,0)$; take three rows ababa alternating with one row $c d c d c$ and a row $(d c d c d)$ to get $(5,2,1,1)$; take three rows $a b a b c$ alternating with two rows $c d c d a$ to get $(4,3,2,0)$; take three rows abcda alternating with two rows $c d a b c$ to get $(4,3,1,1)$. By choosing for example $a=10^{3}, b=10^{2}, c=10, d=1$, we can make all these sums different. Hence, 60 is indeed the maximum possible number of different sums.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 204
|
C4. A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \geq 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles.
|
Solution. The required maximum is $\lceil n / 3\rceil$, the least integer greater than or equal to $n / 3$. To describe a planar $n$-gon splitting into this many triangles, write $n=3 m-r$, where $m$ is a positive integer and $r=0,1,2$, and consider $m$ coplanar equilateral triangles $A_{3 i} A_{3 i+1} A_{3 i+2}$, $i=0, \ldots, m-1$, where the $A_{i}$ are pairwise distinct, the $A$ 's of rank congruent to 0 or 2 modulo 3 are all collinear, $A_{2}, A_{3}, A_{5}, \ldots, A_{3 m-3}, A_{3 m-1}, A_{0}$, in order, and the line $A_{0} A_{2}$ separates $A_{1}$ from the remaining $A$ 's of rank congruent to 1 modulo 3 . The polygon $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5} \ldots A_{3 m-3} A_{3 m-2} A_{3 m-1}$ settles the case $r=0$; removal of $A_{3}$ from the list settles the case $r=1$; and removal of $A_{3}$ and $A_{3 m-1}$ settles the case $r=2$.
Next, we prove that no planar $n$-gon splits into less than $n / 3$ triangles. Alternatively, but equivalently, if a planar polygon splits into $t$ triangles, then its boundary has (combinatorial) length at most $3 t$. Proceed by induction on $t$. The base case $t=1$ is clear, so let $t>1$.
The vertices of the triangles in the splitting may subdivide the boundary of the polygon, making it into a possibly combinatorially longer simple loop $\Omega$. Clearly, it is sufficient to prove that the length of $\Omega$ does not exceed $3 t$.
To this end, consider a triangle in the splitting whose boundary $\omega$ meets $\Omega$ along at least one of its edges. Trace $\Omega$ counterclockwise and let $\alpha_{1}, \ldots, \alpha_{k}$, in order, be the connected components of $\Omega-\omega$. Each $\alpha_{i}$ is a path along $\Omega$ with distinct end points, whose terminal point is joined to the starting point of $\alpha_{i+1}$ by a (possibly constant) path $\beta_{i}$ along $\omega$. Trace $\omega$ clockwise from the terminal point of $\alpha_{i}$ to its starting point to obtain a path $\alpha_{i}^{\prime}$ of positive length, and notice that $\alpha_{i}+\alpha_{i}^{\prime}$ is the boundary of a polygon split into $t_{i}<t$ triangles. By the induction hypothesis, the length of $\alpha_{i}+\alpha_{i}^{\prime}$ does not exceed $3 t_{i}$, and since $\alpha_{i}^{\prime}$ has positive length, the length of $\alpha_{i}$ is at most $3 t_{i}-1$. Consequently, the length of $\Omega-\omega$ does not exceed $\sum_{i=1}^{k}\left(3 t_{i}-1\right)=3 t-3-k$.
Finally, we prove that the total length of the $\beta_{i}$ does not exceed $k+3$. Begin by noticing that no $\beta_{i}$ has length greater than 4 , at most one has length greater than 2 , and at most three have length 2 . If some $\beta_{i}$ has length 4 , then the remaining $k-1$ are all of length at most 1 , so the total length of the $\beta$ 's does not exceed $4+(k-1)=k+3$. Otherwise, either some $\beta_{i}$ has length 3 , in which case at most one other has length 2 and the remaining $k-2$ all have length at most 1 , or the $\beta_{i}$ all have length less than 3 , in which case there are at most three of length 2 and the remaining $k-3$ all have length at most 1 ; in the former case, the total length of the $\beta$ 's does not exceed $3+2+(k-2)=k+3$, and in the latter, the total length of the $\beta$ 's does not exceed $3 \cdot 2+(k-3)=k+3$. The conclusion follows.
## Chapter 3
## Geometry
|
\lceiln/3\rceil
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
| 205
|
G1. Let $A B C$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be points on the sides $B C, A C, A B$, respectively. The circle $\left(c_{1}\right)$ of radius $F A$, centred at $F$, crosses the segment $(O A)$ at $A^{\prime}$ and the circumcircle (c) of the triangle $A B C$ again at $K$. Similarly, the circle $\left(c_{2}\right)$ of radius $D B$, centred at $D$, crosses the segment $(O B)$ at $B^{\prime}$ and the circle (c) again at $L$. Finally, the circle $\left(c_{3}\right)$ of radius $E C$, centred at $E$, crosses the segment $(O C)$ at $C^{\prime}$ and the circle (c) again at $M$. Prove that the quadrilaterals $B K F A^{\prime}$, $C L D B^{\prime}$ and $A M E C^{\prime}$ are all cyclic, and their circumcircles share a common point.
|
Solution. We will prove that the quadrilateral $B K F A^{\prime}$ is cyclic and its circumcircle passes through the center $O$ of the circle (c).
The triangle $A F K$ is isosceles, so $m(\widehat{K F B})=2 m(\widehat{K A B})=m(\widehat{K O B})$. It follows that the quadrilateral $B K F O$ is cyclic.
The triangles $O F K$ and $O F A$ are congruent (S.S.S.), hence $m(\widehat{O K F})=m(\widehat{O A F})$. The triangle $F A A^{\prime}$ is isosceles, so $m\left(\widehat{F A^{\prime} A}\right)=m(\widehat{O A F})$. Therefore $m\left(\widehat{F A^{\prime} A}\right)=m(\widehat{O K F})$, so the quadrilateral $O K F A^{\prime}$ is cyclic.
(1) and (2) prove the initial claim.
Along the same lines, we can prove that the points $C, D, L, B^{\prime}, O$ and $A, M, E, C^{\prime}, O$ are concylic, respectively, so their circumcircles also pass through $O$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 206
|
G2. Let $A B C$ be a triangle with $m(\widehat{B A C})=60^{\circ}$. Let $D$ and $E$ be the feet of the perpendiculars from $A$ to the external angle bisectors of $\widehat{A B C}$ and $\widehat{A C B}$, respectively. Let $O$ be the circumcenter of the triangle $A B C$. Prove that the circumcircles of the triangles $\triangle A D E$ and $\triangle B O C$ are tangent to each other.
|
Solution. Let $X$ be the intersection of the lines $B D$ and $C E$.
We will prove that $X$ lies on the circumcircles of both triangles $\triangle A D E$ and $\triangle B O C$ and then we will prove that the centers of these circles and the point $X$ are collinear, which is enough for proving that the circles are tangent to each other.
In this proof we will use the notation $(M N P)$ to denote the circumcircle of the triangle $\triangle M N P$.
Obviously, the quadrilateral $A D X E$ is cyclic, and the circle $(D A E)$ has $[A X]$ as diameter. (1)
Let $I$ be the incenter of triangle $A B C$. So, the point $I$ lies on the segment $[A X]$ (2), and the quadrilateral $X B I C$ is cyclic because $I C \perp X C$ and $I B \perp X B$. So, the circle (BIC) has $[I X]$ as diameter.
Finally, $m(\widehat{B I C})=90^{\circ}+\frac{1}{2} m(\widehat{B A C})=120^{\circ}$ and $m(\widehat{B O C})=2 m(\widehat{B A C})=120^{\circ}$.
So, the quadrilateral $B O I C$ is cyclic and the circle $(B O C)$ has $[I X]$ as diameter. (3)
(1), (2), (3) imply the conclusion.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 207
|
G3. A trapezoid $A B C D(A B \| C D, A B>C D)$ is circumscribed. The incircle of triangle $A B C$ touches the lines $A B$ and $A C$ at $M$ and $N$, respectively. Prove that the incenter of the trapezoid lies on the line $M N$.
|
Solution. Let $I$ be the incenter of triangle $A B C$ and $R$ be the common point of the lines $B I$ and $M N$. Since
$$
m(\widehat{A N M})=90^{\circ}-\frac{1}{2} m(\widehat{M A N}) \quad \text { and } \quad m(\widehat{B I C})=90^{\circ}+\frac{1}{2} m(\widehat{M A N})
$$
the quadrilateral $I R N C$ is cyclic.
It follows that $m(\widehat{B R C})=90^{\circ}$ and therefore
$$
m(\widehat{B C R})=90^{\circ}-m(\widehat{C B R})=90^{\circ}-\frac{1}{2}\left(180^{\circ}-m(\widehat{B C D})\right)=\frac{1}{2} m(\widehat{B C D})
$$
So, $(C R$ is the angle bisector of $\widehat{D C B}$ and $R$ is the incenter of the trapezoid.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 208
|
G4. Let $A B C$ be an acute angled triangle whose shortest side is $[B C]$. Consider a variable point $P$ on the side $[B C]$, and let $D$ and $E$ be points on $(A B]$ and $(A C]$, respectively, such that $B D=B P$ and $C P=C E$. Prove that, as $P$ traces $[B C]$, the circumcircle of the triangle $A D E$ passes through a fixed point.
|
Solution. We claim that the fixed point is the center of the incircle of $A B C$.
Let $I$ be the center of the incircle of $A B C$. Since $B D=B P$ and $[B I$ is the bisector of $\widehat{D B P}$, the line $B I$ is the perpendicular bisector of $[D P]$. This yields $D I=P I$. Analogously we get $E I=P I$. So, the point $I$ is the circumcenter of the triangle $D E P$.
This means $m(\widehat{D I E})=2 m(\widehat{D P E})$.
On the other hand
$$
\begin{aligned}
m(\widehat{D P E}) & =180^{\circ}-m(\widehat{D P B})-m(\widehat{E P C}) \\
& =180^{\circ}-\left(90^{\circ}-\frac{1}{2} m(\widehat{D B P})\right)-\left(90^{\circ}-\frac{1}{2} m(\widehat{E C P})\right) \\
& =90^{\circ}-\frac{1}{2} m(\widehat{B A C})
\end{aligned}
$$
So, $m(\widehat{D I E})=2 m(\widehat{D P E})=180^{\circ}-m(\widehat{D A E})$, which means that the points $A, D, E$ and $I$ are cocyclic.
Remark. The fact that the incentre $I$ of the triangle $A B C$ is the required fixed point could be guessed by considering the two extremal positions of $P$. Thus, if $P=B$, then $D=D_{B}=B$ as well, and $C E=C E_{B}=B C$, so $m(\angle A E B)=m(\angle C)+m(\angle E B C)=$ $m(\angle C)+\frac{180^{\circ}-m(\angle C)}{2}=90^{\circ}+\frac{m(\angle C)}{2}=m(\angle A I B)$. Hence the points $A, E=E_{B}, I, D_{B}=B$ are cocyclic. Similarly, letting $P=C$, the points $A, D=D_{C}, I, E_{C}=C$ are cocyclic. Consequently, the circles $A D_{B} E_{B}$ and $A D_{C} E_{C}$ meet again at $I$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 209
|
G5. Let $A B C$ be an acute angled triangle with orthocenter $H$ and circumcenter $O$. Assume the circumcenter $X$ of $B H C$ lies on the circumcircle of $A B C$. Reflect $O$ across $X$ to obtain $O^{\prime}$, and let the lines $X H$ and $O^{\prime} A$ meet at $K$. Let $L, M$ and $N$ be the midpoints of $[X B],[X C]$ and $[B C]$, respectively. Prove that the points $K, L, M$ and $N$ are cocyclic.
|
Solution. The circumcircles of $A B C$ and $B H C$ have the same radius. So, $X B=$ $X C=X H=X O=r$ (where $r$ is the radius of the circle $A B C$ ) and $O^{\prime}$ lies on $C(X, r)$. We conclude that $O X$ is the perpendicular bisector for $[B C]$. So, $B O X$ and $C O X$ are equilateral triangles.
It is known that $A H=2 O N=r$. So, $A H O^{\prime} X$ is parallelogram, and $X K=K H=r / 2$. Finally, $X L=X K=X N=X M=r / 2$. So, $K, L, M$ and $N$ lie on the circle $c(X, r / 2)$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 210
|
G6. Given an acute triangle $A B C$, erect triangles $A B D$ and $A C E$ externally, so that $m(\widehat{A D B})=m(\widehat{A E C})=90^{\circ}$ and $\widehat{B A D} \equiv \widehat{C A E}$. Let $A_{1} \in B C, B_{1} \in A C$ and $C_{1} \in A B$ be the feet of the altitudes of the triangle $A B C$, and let $K$ and $L$ be the midpoints of $\left[B C_{1}\right]$ and $\left[C B_{1}\right]$, respectively. Prove that the circumcenters of the triangles $A K L, A_{1} B_{1} C_{1}$ and $D E A_{1}$ are collinear.
|
Solution. Let $M, P$ and $Q$ be the midpoints of $[B C],[C A]$ and $[A B]$, respectively.
The circumcircle of the triangle $A_{1} B_{1} C_{1}$ is the Euler's circle. So, the point $M$ lies on this circle.
It is enough to prove now that $\left[A_{1} M\right]$ is a common chord of the three circles $\left(A_{1} B_{1} C_{1}\right)$, $(A K L)$ and $\left(D E A_{1}\right)$.
The segments $[M K]$ and $[M L]$ are midlines for the triangles $B C C_{1}$ and $B C B_{1}$ respectively, hence $M K \| C C_{1} \perp A B$ and $M L \| B B_{1} \perp A C$. So, the circle $(A K L)$ has diameter $[A M]$ and therefore passes through $M$.
Finally, we prove that the quadrilateral $D A_{1} M E$ is cyclic.
From the cyclic quadrilaterals $A D B A_{1}$ and $A E C A_{1}, \widehat{A A_{1} D} \equiv \widehat{A B D}$ and $\widehat{A A_{1} E} \equiv \widehat{A C E} \equiv$ $\widehat{A B D}$, so $m\left(\widehat{D A_{1} E}\right)=2 m(\widehat{A B D})=180^{\circ}-2 m(\widehat{D A B})$.
We notice now that $D Q=A B / 2=M P, Q M=A C / 2=P E$ and
$$
\begin{aligned}
& m(\widehat{D Q M})=m(\widehat{D Q B})+m(\widehat{B Q M})=2 m(\widehat{D A B})+m(\widehat{B A C}) \\
& m(\widehat{E P M})=m(\widehat{E P C})+m(\widehat{C P M})=2 m(\widehat{E A C})+m(\widehat{C A B})
\end{aligned}
$$
so $\triangle M P E \equiv \triangle D Q M$ (S.A.S.). This leads to $m(\widehat{D M E})=m(\widehat{D M Q})+m(Q M P)+$ $m(P M E)=m(\widehat{D M Q})+m(B Q M)+m(Q D M)=180^{\circ}-m(\widehat{D Q B})=180^{\circ}-2 m(\widehat{D A B})$. Since $m\left(\widehat{D A_{1} E}\right)=m(\widehat{D M E})$, the quadrilateral $D A_{1} M E$ is cyclic.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 211
|
G7. Let $[A B]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment $(A B)$ such that $A K<B K$. Two circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $K L$ and the circle (c), and let the lines $A B$ and $L O$ meet at $M$. Prove that the line $M P$ is tangent to the circle $(c)$.
|
Solution. Let $\left(c_{1}\right)$ and $\left(c_{2}\right)$ be circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $\left(c_{1}\right)$ and $(c)$ meet the common tangent to $\left(c_{2}\right)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles $\left(c_{1}\right),\left(c_{2}\right)$ and $(c)$, and the line $K L$ passes through $Q$.
We have $m(\widehat{Q L B})=m(\widehat{Q B K})=m(\widehat{Q B A})=\frac{1}{2} m(\overparen{B A})=m(\widehat{Q O B})$. So, the quadrilateral $O B Q L$ is cyclic. We conclude that $m(\widehat{Q L O})=90^{\circ}$ and the points $O, B, Q, A$ and $L$ are cocyclic on a circle $(k)$.
In the sequel, we will denote $\mathcal{P}_{\omega}(X)$ the power of the point $X$ with respect of the circle $\omega$. The first continuation.
From $M O^{2}-O P^{2}=\mathcal{P}_{c}(M)=M A \cdot M B=\mathcal{P}_{k}(M)=M L \cdot M O=(M O-O L) \cdot M O=$ $M O^{2}-O L \cdot M O$ follows that $O P^{2}=O L \cdot O M$. Since $P L \perp O M$, this shows that the triangle $M P O$ is right at point $P$. Thus, the line $M P$ is tangent to the circle (c).
The second continuation.
Let $R \in(c)$ be so that $B R \perp M O$. The triangle $L B R$ is isosceles with $L B=L R$, so $\widehat{O L R} \equiv \widehat{O L B} \equiv \widehat{O Q B} \equiv \widehat{O Q A} \equiv \widehat{M L A}$. We conclude that the points $A, L$ and $R$ are collinear.
Now $m(\widehat{A M R})+m(\widehat{A O R})=m(\widehat{A M R})+2 m(\widehat{A B R})=m(\widehat{A M R})+m(\widehat{A B R})+m(\widehat{M R B})=$ $180^{\circ}$, since the triangle $M B R$ is isosceles. So, the quadrilateral $M A O R$ is cyclic.
This yields $L M \cdot L O=-\mathcal{P}_{(\text {MAOR })}(L)=L A \cdot L R=-\mathcal{P}_{c}(L)=L P^{2}$, which as above, shows that $O P \perp P M$.
The third continuation.
$\widehat{K L A} \equiv \widehat{K A Q} \equiv \widehat{K L B}$ and $m(\widehat{M L K})=90^{\circ}$ show that $[L K$ and $[L M$ are the internal and external bisectors af the angle $\widehat{A L B}$, so $(M, K)$ and $(A, B)$ are harmonic conjugates. So, $L K$ is the polar line of $M$ in the circle $(c)$.
## Chapter 4
## Number Theory
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
| 212
|
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