problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
N1. Determine the largest positive integer $n$ that divides $p^{6}-1$ for all primes $p>7$.
|
Solution. Note that
$$
p^{6}-1=(p-1)(p+1)\left(p^{2}-p+1\right)\left(p^{2}+p+1\right)
$$
For $p=11$ we have
$$
p^{6}-1=1771560=2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 37
$$
For $p=13$ we have
$$
p^{6}-1=2^{3} \cdot 3^{2} \cdot 7 \cdot 61 \cdot 157
$$
From the last two calculations we find evidence to tr... | 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 213 |
N2. Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions:
a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and
b) The sum $x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1}$ is divisible by 11 .
|
Solution. The required maximum is 10 .
According to a), the numbers $x_{i}, 1 \leq i \leq m$, are all different $(\bmod 11)$ (1)
Hence, the number of natural numbers satisfying the conditions is at most 11.
If $x_{j} \equiv 0(\bmod 11)$ for some $j$, then
$$
x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 214 |
N3. Find all positive integers $n$ such that the number $A_{n}=\frac{2^{4 n+2}+1}{65}$ is
a) an integer;
b) a prime.
|
Solution. a) Note that $65=5 \cdot 13$.
Obviously, $5=2^{2}+1$ is a divisor of $\left(2^{2}\right)^{2 n+1}+1=2^{4 n+2}+1$ for any positive integer $n$. Since $2^{12} \equiv 1(\bmod 13)$, if $n \equiv r(\bmod 3)$, then $2^{4 n+2}+1 \equiv 2^{4 r+2}+1(\bmod 13)$. Now, $2^{4 \cdot 0+2}+1=5,2^{4 \cdot 1+2}+1=65$, and $2^... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 215 |
N4. Find all triples of integers $(a, b, c)$ such that the number
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$
is a power of 2016 .
|
Solution. Let $z$ be a positive integer such that
$$
(a-b)(b-c)(c-a)+4=2 \cdot 2016^{z}
$$
We set $a-b=-x, b-c=-y$ and we rewrite the equation as
$$
x y(x+y)+4=2 \cdot 2016^{z}
$$
Note that the right hand side is divisible by 7 , so we have that
$$
x y(x+y)+4 \equiv 0 \quad(\bmod 7)
$$
or
$$
3 x y(x+y) \equiv 2... | (,b,)=(k+2,k+1,k),k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 216 |
N5. Determine all four-digit numbers $\overline{a b c d}$ such that
$$
(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\overline{a b c d}
$$
|
Solution. Depending on the parity of $a, b, c, d$, at least two of the factors $(a+b),(a+c)$, $(a+d),(b+c),(b+d),(c+d)$ are even, so that $4 \mid \overline{a b c d}$.
We claim that $3 \mid \overline{a b c d}$.
Assume $a+b+c+d \equiv 2(\bmod 3)$. Then $x+y \equiv 1(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$... | 2016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 217 |
A 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
a+b+c=\frac{a b+b c+c a}{a b c}
$$
Now, from the first condition and the second condition we get
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 218 |
A 2. Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ defined by $a_{1}=9$ and
$$
a_{n+1}=\frac{(n+5) a_{n}+22}{n+3}
$$
for $n \geqslant 1$.
Find all natural numbers $n$ for which $a_{n}$ is a perfect square of an integer.
|
Solution: Define $b_{n}=a_{n}+11$. Then
$$
22=(n+3) a_{n+1}-(n+5) a_{n}=(n+3) b_{n+1}-11 n-33-(n+5) b_{n}+11 n+55
$$
giving $(n+3) b_{n+1}=(n+5) b_{n}$. Then
$b_{n+1}=\frac{n+5}{n+3} b_{n}=\frac{(n+5)(n+4)}{(n+3)(n+2)} b_{n-1}=\frac{(n+5)(n+4)}{(n+2)(n+1)} b_{n-2}=\cdots=\frac{(n+5)(n+4)}{5 \cdot 4} b_{1}=(n+5)(n+4... | n=1orn=8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 219 |
A 3. Find all triples of positive real numbers $(a, b, c)$ so that the expression
$$
M=\frac{(a+b)(b+c)(a+b+c)}{a b c}
$$
gets its least value.
|
Solution. The expression $M$ is homogeneous, therefore we can assume that $a b c=1$. We set $s=a+c$ and $p=a c$ and using $b=\frac{1}{a c}$, we get
$$
M=\left(a+\frac{1}{a c}\right)\left(\frac{1}{a c}+c\right)\left(a+\frac{1}{a c}+c\right)=\left(a+p^{-1}\right)\left(c+p^{-1}\right)\left(s+p^{-1}\right)
$$
Expanding ... | ==\sqrt[3]{\frac{1+\sqrt{5}}{2}},b=\frac{1}{} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 220 |
C 2. Viktor and Natalia bought 2020 buckets of ice-cream and want to organize a degustation schedule with 2020 rounds such that:
- In every round, each one of them tries 1 ice-cream, and those 2 ice-creams tried in a single round are different from each other.
- At the end of the 2020 rounds, each one of them has tri... |
Solution. If we fix the order in which Natalia tries the ice-creams, we may consider 2 types of fair schedules:
1) Her last 1010 ice-creams get assigned as Viktor's first 1010 ice-creams, and vice versa: Viktor's first 1010 ice-creams are assigned as Natalia's last 1010 ice-creams. This generates (1010!) $)^{2}$ dist... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 221 |
C 3. Alice and Bob play the following game: Alice begins by picking a natural number $n \geqslant 2$. Then, with Bob starting first, they alternately choose one number from the set $A=\{1,2, \ldots, n\}$ according to the following condition: The number chosen at each step should be distinct from all the already chosen... |
Solution. Alice has a winning strategy. She initially picks $n=8$. We will give a strategy so that she can end up with $S$ even, or $S=15$, or $S=21$, so she wins.
Case 1: If Bob chooses 1, then the game ends with Alice choosing 2,4,6,8 so $S$ is even (larger than 2) and Alice wins.
Case 2: If Bob chooses 2, then Al... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 222 |
G 1. Let $\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\omega$ at a point $X$ such that $X$ and $B$ are not on the same side ... |
Solution. Denote by $s$ the line $A D$. Let $T$ be the second intersection point of the circumcircles of $\triangle B D X$ and $\triangle C D Y$. Then $T$ is on the line $s$. Note that $C D Y T$ and $B D X T$ are cyclic. Using this and the fact that $A D$ is perpendicular to $B C$ we obtain:
$$
\angle T Y E=\angle T ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 223 |
G 2. Problem: Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c), $\left(c_{1}\right)$ be the circmucircles of the triangles $\triangle A E Z$ and $\triangle... |
Solution. Since the triangles $\triangle A E B$ and $\triangle C A B$ are similar, then
$$
\frac{A B}{E B}=\frac{C B}{A B}
$$
Since $A B=B Z$ we get
$$
\frac{B Z}{E B}=\frac{C B}{B Z}
$$
from which it follows that the triangles $\triangle Z B E$ and $\triangle C B Z$ are also similar. Since $F E B Z$ is cyclic,
!... | proof | Geometry | proof | Yes | Yes | olympiads | false | 224 |
G 3. Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\triangle A E Z$. Let $D$ be the second point of intersection of ... |
Solution. We will first show that $P A$ is tangent to $(c)$ at $A$.
Since $E, D, Z, A$ are concyclic, then $\angle E D C=\angle E A Z=\angle E A B$. Since also the triangles $\triangle A B C$ and $\triangle E B A$ are similar, then $\angle E A B=\angle B C A$, therefore $\angle E D C=\angle B C A$.
Since $\angle F E... | proof | Geometry | proof | Yes | Yes | olympiads | false | 225 |
NT 1. Determine whether there is a natural number $n$ for which $8^{n}+47$ is prime.
|
Solution. The number $m=8^{n}+47$ is never prime.
If $n$ is even, say $n=2 k$, then $m=64^{k}+47 \equiv 1+2 \equiv 0 \bmod 3$. Since also $m>3$, then $m$ is not prime.
If $n \equiv 1 \bmod 4$, say $n=4 k+1$, then $m=8 \cdot\left(8^{k}\right)^{4}+47 \equiv 3+2 \equiv 0 \bmod 5$. Since also $m>3$, then $m$ is not prim... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 226 |
NT 2. Find all positive integers $a, b, c$ and $p$, where $p$ is a prime number, such that
$$
73 p^{2}+6=9 a^{2}+17 b^{2}+17 c^{2}
$$
|
Solution. Since the equation is symmetric with respect to the numbers $b$ and $c$, we assume that $b \geq c$.
If $p \geq 3$, then $p$ is an odd number. We consider the equation modulo 8 . Since,
$$
73 p^{2}+6 \equiv 79 \equiv 7 \quad(\bmod 8)
$$
we get that
$$
a^{2}+b^{2}+c^{2} \equiv 7 \quad(\bmod 8)
$$
This can... | (,b,,p)\in{(1,1,4,2),(1,4,1,2)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 227 |
NT 3. Find the largest integer $k(k \geq 2)$, for which there exists an integer $n(n \geq k)$ such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions:
1. each chosen number is not divisible by 6 , by 7 and by 8 ;
2. the positive diffe... |
Solution. An integer is divisible by 6,7 and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \times 7 \times 4=168$.
Let $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ ( mo... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 228 |
NT 4. Find all prime numbers $p$ such that
$$
(x+y)^{19}-x^{19}-y^{19}
$$
is a multiple of $p$ for any positive integers $x, y$.
|
Solution. If $x=y=1$ then $p$ divides
$$
2^{19}-2=2\left(2^{18}-1\right)=2\left(2^{9}-1\right)\left(2^{9}+1\right)=2 \cdot 511 \cdot 513=2 \cdot 3^{3} \cdot 7 \cdot 19 \cdot 73
$$
If $x=2, y=1$ then
$$
p \mid 3^{19}-2^{19}-1
$$
We will show that $3^{19}-2^{19}-1$ is not a multiple of 73 . Indeed,
$$
3^{19} \equiv... | 2,3,7,19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 229 |
NT 5. The positive integer $k$ and the set $A$ of different integers from 1 to $3 k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2 b=a+c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1,2 k]$ - medium and those in $[2 k+1,3 k]$ - large. Is it always tr... |
Solution. A counterexample for a) is $k=3, A=\{1,2,9\}, x=2$ and $d=8$. A counterexample for c) is $k=3, A=\{1,8,9\}, x=8$ and $d=1$.
We will prove that b) is true.
Suppose the contrary and let $x, d$ have the above properties. We can assume $03 k$, then since the remainder for $x+d$ is medium we have $4 k2 k$. Ther... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 230 |
NT 6. Are there any positive integers $m$ and $n$ satisfying the equation
$$
m^{3}=9 n^{4}+170 n^{2}+289 ?
$$
|
Solution. We will prove that the answer is no. Note that
$$
m^{3}=9 n^{4}+170 n^{2}+289=\left(9 n^{2}+17\right)\left(n^{2}+17\right)
$$
If $n$ is odd then $m$ is even, therefore $8 \mid m^{3}$. However,
$$
9 n^{4}+170 n^{2}+289 \equiv 9+170+289 \equiv 4(\bmod 8)
$$
which leads to a contradiction. If $n$ is a multi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 231 |
A 1. Let $x, y$ and $z$ be positive numbers. Prove that
$$
\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{7}}}{\sqrt{2 \sqrt{27}}}
$$
|
Solution. Replacing $x=a^{2}, y=b^{2}, z=c^{2}$, where $a, b, c$ are positive numbers, our inequality is equivalent to
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{\sqrt[4]{(a+b+c)^{7}}}{\sqrt{2 \sqrt{27}}}
$$
Using the Cauchy-Schw... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 234 |
A 2. Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>$ $(m+n)^{2}$, we have
$$
m^{3}+n^{3} \geq(m+n)^{2}+k
$$
|
Solution. We see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus
$$
3^{3}+2^{3} \geq(3+2)^{2}+k \Rightarrow k \leq 10
$$
We will show that $k=10$ is the desired maximum. In other words, we have to prove that
$$
m^{3}+n^{3} \geq(m+n)^{2}+10
$$
The last inequality is equivalent to
$$
(m+n)\left(m^{2}... | 10 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 235 |
A 3. Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{1}{a b(b+1)(c+1)}+\frac{1}{b c(c+1)(a+1)}+\frac{1}{c a(a+1)(b+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
|
Solution. The required inequality is equivalent to
$$
\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
or equivalently to,
$$
(1+a b c)^{2}(a b+b c+c a+a+b+c) \geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)
$$
Let $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can be rewritten ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 236 |
A 4. Let $k>1, n>2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_{1}$, $x_{2}, \ldots, x_{n}$ are not all equal and satisfy
$$
x_{1}+\frac{k}{x_{2}}=x_{2}+\frac{k}{x_{3}}=x_{3}+\frac{k}{x_{4}}=\cdots=x_{n-1}+\frac{k}{x_{n}}=x_{n}+\frac{k}{x_{1}}
$$
Find:
a) the product $x_{1} x_{2} \... |
Solution. a) If $x_{i}=x_{i+1}$ for some $i$ (assuming $x_{n+1}=x_{1}$ ), then by the given identity all $x_{i}$ will be equal, a contradiction. Thus $x_{1} \neq x_{2}$ and
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}
$$
Analogously
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}=k^{2} \frac{x_{3}-x_{4}}{\left... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 237 |
A 5. Let $a, b, c, d$ and $x, y, z, t$ be real numbers such that
$$
0 \leq a, b, c, d \leq 1, \quad x, y, z, t \geq 1 \text { and } a+b+c+d+x+y+z+t=8
$$
Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} \leq 28
$$
When does the equality hold?
|
Solution. We observe that if $u \leq v$ then by replacing $(u, v)$ with $(u-\varepsilon, v+\varepsilon)$, where $\varepsilon>0$, the sum of squares increases. Indeed,
$$
(u-\varepsilon)^{2}+(v+\varepsilon)^{2}-u^{2}-v^{2}=2 \varepsilon(v-u)+2 \varepsilon^{2}>0
$$
Then, denoting
$$
E(a, b, c, d, x, y, z, t)=a^{2}+b^... | 28 | Inequalities | proof | Yes | Yes | olympiads | false | 238 |
A 6. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
\frac{a}{\sqrt{a^{3}+5}}+\frac{b}{\sqrt{b^{3}+5}}+\frac{c}{\sqrt{c^{3}+5}} \leq \frac{\sqrt{6}}{2}
$$
|
Solution. From AM-GM inequality we have
$$
a^{3}+a^{3}+1 \geq 3 a^{2} \Rightarrow 2\left(a^{3}+5\right) \geq 3\left(a^{2}+3\right)
$$
Using the condition $a b+b c+c a=3$, we get
$$
\left(a^{3}+5\right) \geq 3\left(a^{2}+a b+b c+c a\right)=3(c+a)(a+b)
$$
therefore
$$
\frac{a}{\sqrt{a^{3}+5}} \leq \sqrt{\frac{2 a^{... | \frac{\sqrt{6}}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 239 |
A 7. Let $A$ be a set of positive integers with the following properties:
(a) If $n$ is an element of $A$ then $n \leqslant 2018$.
(b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $n, m$ of $S$ with $|n-m| \geqslant \sqrt{n}+\sqrt{m}$.
What is the maximum number of elements that $A$ can have?
|
Solution. Assuming $n>m$ we have
$$
\begin{aligned}
|n-m| \geqslant \sqrt{n}+\sqrt{m} & \Leftrightarrow(\sqrt{n}-\sqrt{m})(\sqrt{n}+\sqrt{m}) \geqslant \sqrt{n}+\sqrt{m} \\
& \Leftrightarrow \sqrt{n} \geqslant \sqrt{m}+1 .
\end{aligned}
$$
Let $A_{k}=\left\{k^{2}, k^{2}+1, \ldots,(k+1)^{2}-1\right\}$. Note that each... | 88 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 240 |
C 1. A set $S$ is called neighbouring if it has the following two properties:
a) $S$ has exactly four elements
b) for every element $x$ of $S$, at least one of the numbers $x-1$ or $x+1$ belongs to $S$.
Find the number of all neighbouring subsets of the set $\{1,2, \ldots, n\}$.
|
Solution. Let us denote with $a$ and $b$ the smallest and the largest element of a neighbouring set $S$, respectively. Since $a-1 \notin S$, we have that $a+1 \in S$. Similarly, we conclude that $b-1 \in S$. So, every neighbouring set has the following form $\{a, a+1, b-1, b\}$ for $b-a \geq 3$. The number of the neig... | \frac{(n-3)(n-2)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 241 |
C 2. A set $T$ of $n$ three-digit numbers has the following five properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers ... |
Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of $6 A$ 's (which means that we add 1 to the current digit) and 2 G's (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 242 |
G 1. Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C>A C$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $C B$ intersects $\Gamma$ at the point $D$, which is on the arc $A B$ not containing $C$. The circle with centre $C$ and radius $C A$ intersects the segment $C D$ at the po... |
Solution. We use standard notation for the angles of triangle $A B C$. Let $P$ be the midpoint of $C H$ and $O$ the centre of $\Gamma$. As
$$
\alpha=\angle B A C=\angle B D C=\angle D K L
$$
the quadrilateral $A C K L$ is cyclic. From the relation $C B=C D$ we get $\angle B C D=180^{\circ}-2 \alpha$, so
$$
\angle A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 243 |
G 2. Let $A B C$ be a right angled triangle with $\angle A=90^{\circ}$ and $A D$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $A B$ and $A C$ respectively. The parallel line from $C$ to $E Z$ intersects the line $A B$ at the po... |
Solution. Suppose that the line $A A^{\prime}$ intersects the lines $E Z, B C$ and $C N$ at the points $L, M$, $F$ respectively. The line $I K$ being diagonal of the rectangle $K A^{\prime} I A$ passes through $L$, which by construction of $A^{\prime}$, is the middle of the other diagonal $A A^{\prime}$. The triangles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 244 |
G 3. Let $A B C$ be an acute triangle, $A^{\prime}, B^{\prime}, C^{\prime}$ the reflexions of the vertices $A, B$ and $C$ with respect to $B C, C A$, and $A B$, respectively, and let the circumcircles of triangles $A B B^{\prime}$ and $A C C^{\prime}$ meet again at $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined simil... |
Solution. Let $O_{1}, O_{2}$ and $O$ be the circumcenters of triangles $A B B^{\prime}, A C C^{\prime}$ and $A B C$ respectively. As $A B$ is the perpendicular bisector of the line segment $C C^{\prime}, O_{2}$ is the intersection of the perpendicular bisector of $A C$ with $A B$. Similarly, $O_{1}$ is the intersectio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 245 |
G 4. Let $A B C$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be it's incenter. Let $P_{1}, P_{2}$ and $P_{3}$ be the areas of the triangles $A B I, B C I$ and $C A I$, respectively. Prove that
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} \... |
Solution. Let $r$ be the radius of the inscribed circle of the triangle $A B C$. We have that
$$
P_{1}=\frac{r c}{2}, \quad P_{2}=\frac{r a}{2}, \quad P_{3}=\frac{r b}{2}
$$
It follows that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 246 |
G 5. Given a rectangle $A B C D$ such that $A B=b>2 a=B C$, let $E$ be the midpoint of $A D$. On a line parallel to $A B$ through point $E$, a point $G$ is chosen such that the area of $G C E$ is
$$
(G C E)=\frac{1}{2}\left(\frac{a^{3}}{b}+a b\right)
$$
Point $H$ is the foot of the perpendicular from $E$ to $G D$ an... |
Solution. Let $L$ be the foot of the perpendicular from $G$ to $E C$ and let $Q$ the point of intersection of the lines $E G$ and $B C$. Then,
$$
(G C E)=\frac{1}{2} E C \cdot G L=\frac{1}{2} \sqrt{a^{2}+b^{2}} \cdot G L
$$
So, $G L=\frac{a}{b} \sqrt{a^{2}+b^{2}}$.
$ which satisfy the equation
$$
x^{5}-y^{5}=16 x y
$$
|
Solution. If one of $x, y$ is 0 , the other has to be 0 too, and $(x, y)=(0,0)$ is one solution. If $x y \neq 0$, let $d=\operatorname{gcd}(x, y)$ and we write $x=d a, y=d b, a, b \in \mathbb{Z}$ with $(a, b)=1$. Then, the given equation is transformed into
$$
d^{3} a^{5}-d^{3} b^{5}=16 a b
$$
So, by the above equat... | (x,y)=(0,0)or(-2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 249 |
NT 2. Find all pairs $(m, n)$ of positive integers such that
$$
125 \cdot 2^{n}-3^{m}=271
$$
|
Solution. Considering the equation mod5 we get
$$
3^{m} \equiv-1 \quad(\bmod 5)
$$
so $m=4 k+2$ for some positive integer $k$. Then, considering the equation $\bmod 7$ we get
$$
\begin{aligned}
& -2^{n}-9^{2 k+1} \equiv 5 \quad(\bmod 7) \Rightarrow \\
& 2^{n}+2^{2 k+1} \equiv 2 \quad(\bmod 7)
\end{aligned}
$$
Sinc... | (,n)=(6,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 250 |
NT 4. Show that there exist infinitely many positive integers $n$ such that
$$
\frac{4^{n}+2^{n}+1}{n^{2}+n+1}
$$
is an integer.
|
Solution. Let $f(n)=n^{2}+n+1$. Note that
$$
f\left(n^{2}\right)=n^{4}+n^{2}+1=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)
$$
This means that $f(n) \mid f\left(n^{2}\right)$ for every positive integer $n$. By induction on $k$, one can easily see that $f(n) \mid f\left(n^{2^{k}}\right)$ for every positive integers $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 252 |
A1 Determine all integers $a, b, c$ satisfying the identities:
$$
\begin{gathered}
a+b+c=15 \\
(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540
\end{gathered}
$$
|
Solution I: We will use the following fact:
Lemma: If $x, y, z$ are integers such that
$$
x+y+z=0
$$
then
$$
x^{3}+y^{3}+z^{3}=3 x y z
$$
Proof: Let
$$
x+y+z=0 \text {. }
$$
Then we have
$$
x^{3}+y^{3}+z^{3}=x^{3}+y^{3}+(-x-y)^{3}=x^{3}+y^{3}-x^{3}-y^{3}-3 x y(x+y)=3 x y z
$$
Now, from
$$
a+b+c=15
$$
we obt... | (,b,)\in{(-1,0,16),(-2,1,16),(7,10,-2),(8,9,-2)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 253 |
A2 Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=4 \\
z^{2}+t^{2}=9 \\
x t+y z \geq 6
\end{array}\right.
$$
|
Solution I: From the conditions we have
$$
36=\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right)=(x t+y z)^{2}+(x z-y t)^{2} \geq 36+(x z-y t)^{2}
$$
and this implies $x z-y t=0$.
Now it is clear that
$$
x^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13
$$
and the maximum value of $z+x$ is $\sqrt{13}$. It is achieved ... | \sqrt{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 254 |
A3 Find all values of the real parameter $a$, for which the system
$$
\left\{\begin{array}{c}
(|x|+|y|-2)^{2}=1 \\
y=a x+5
\end{array}\right.
$$
has exactly three solutions.
|
Solution: The first equation is equivalent to
$$
|x|+|y|=1
$$
or
$$
|x|+|y|=3
$$
The graph of the first equation is symmetric with respect to both axes. In the first quadrant it is reduced to $x+y=1$, whose graph is segment connecting points $(1,0)$ and $(0,1)$. Thus, the graph of
$$
|x|+|y|=1
$$
is square with ... | =5or=-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 255 |
A4 Real numbers $x, y, z$ satisfy
$$
0<x, y, z<1
$$
and
$$
x y z=(1-x)(1-y)(1-z) .
$$
Show that
$$
\frac{1}{4} \leq \max \{(1-x) y,(1-y) z,(1-z) x\}
$$
|
Solution: It is clear that $a(1-a) \leq \frac{1}{4}$ for any real numbers $a$ (equivalent to $0\max \{(1-x) y,(1-y) x,(1-z) x\}
$$
Now
$$
(1-x) y\frac{1}{2}$.
Using same reasoning we conclude:
$$
z\frac{1}{2}
$$
Using these facts we derive:
$$
\frac{1}{8}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}>x y z=(1-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 256 |
A5 Let $x, y, z$ be positive real numbers. Prove that:
$$
\left(x^{2}+y+1\right)\left(x^{2}+z+1\right)\left(y^{2}+z+1\right)\left(y^{2}+x+1\right)\left(z^{2}+x+1\right)\left(z^{2}+y+1\right) \geq(x+y+z)^{6}
$$
|
Solution I: Applying Cauchy-Schwarz's inequality:
$$
\left(x^{2}+y+1\right)\left(z^{2}+y+1\right)=\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geq(x+y+z)^{2}
$$
Using the same reasoning we deduce:
$$
\left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq(x+y+z)^{2}
$$
and
$$
\left(y^{2}+x+1\right)\left(z^{2}+x+1\righ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 257 |
C2 Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament?
|
Solution: A given pair must play with three other pairs and these plays must be in different days, so at three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ an... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 258 |
C3 a) In how many ways can we read the word SARAJEVO from the table below, if it is allowed to jump from cell to an adjacent cell (by vertex or a side) cell?
b) After the letter in one cell... |
Solution: In the first of the tables below the number in each cell shows the number of ways to reach that cell from the start (which is the sum of the quantities in the cells, from which we can come), and in the second one are the number of ways to arrive from that cell to the end (which is the sum of the quantities i... | 750 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 259 |
G1 Let $A B C D$ be a parallelogram with $A C>B D$, and let $O$ be the point of intersection of $A C$ and $B D$. The circle with center at $O$ and radius $O A$ intersects the extensions of $A D$ and $A B$ at points $G$ and $L$, respectively. Let $Z$ be intersection point of lines $B D$ and $G L$. Prove that $\angle Z ... | ## Solution:
From the point $L$ we draw a parallel line to $B D$ that intersects lines $A C$ and $A G$ at points $N$ and $R$ respectively. Since $D O=O B$, we have that $N R=N L$, and point $N$ is the midpoint of segment $L R$.
Let $K$ be the midpoint of $G L$. Now, $N K \| R G$, and
$$
\angle A G L=\angle N K L=\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 261 |
G2 In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$.
|
Solution: Produce the legs of the trapezoid until they intersect at point $E$. The triangles $A B H$ and $E C D$ are congruent (ASA). The area of $A B C D$ is equal to area of triangle $E A H$ of hypotenuse
$$
A E=A D+D E=A D+A H=8
$$
Let $M$ be the midpoint of $A E$. Then
$$
M E=M A=M H=4
$$
and $\angle A M H=30^... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 262 |
G3 A parallelogram $A B C D$ with obtuse angle $\angle A B C$ is given. After rotating the triangle $A C D$ around the vertex $C$, we get a triangle $C D^{\prime} A^{\prime}$, such that points $B, C$ and $D^{\prime}$ are collinear. The extension of the median of triangle $C D^{\prime} A^{\prime}$ that passes through $... |
Solution: Let $A C \cap B D=\{X\}$ and $P D^{\prime} \cap C A^{\prime}=\{Y\}$. Because $A X=C X$ and $C Y=Y A^{\prime}$, we deduce:
$$
\triangle A B C \cong \triangle C D A \cong \triangle C D^{\prime} A^{\prime} \Rightarrow \triangle A B X \cong \triangle C D^{\prime} Y, \triangle B C X \cong \triangle D^{\prime} A^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 263 |
G4 Let $A B C D E$ be a convex pentagon such that $A B+C D=B C+D E$ and let $k$ be a semicircle with center on side $A E$ that touches the sides $A B, B C, C D$ and $D E$ of the pentagon, respectively, at points $P, Q, R$ and $S$ (different from the vertices of the pentagon). Prove that $P S \| A E$.
|
Solution: Let $O$ be center of $k$. We deduce that $B P=B Q, C Q=C R, D R=D S$, since those are tangents to the circle $k$. Using the condition $A B+C D=B C+D E$, we derive:
$$
A P+B P+C R+D R=B Q+C Q+D S+E S
$$
From here we have $A P=E S$.
Thus,
$$
\triangle A P O \cong \triangle E S O\left(A P=E S, \angle A P O=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 264 |
G5 Let $A, B, C$ and $O$ be four points in the plane, such that $\angle A B C>90^{\circ}$ and $O A=$ $O B=O C$. Define the point $D \in A B$ and the line $\ell$ such that $D \in \ell, A C \perp D C$ and $\ell \perp A O$. Line $\ell$ cuts $A C$ at $E$ and the circumcircle of $\triangle A B C$ at $F$. Prove that the cir... |
Solution: Let $\ell \cap A C=\{K\}$ and define $G$ to be the mirror image of the point $A$ with respect to $O$. Then $A G$ is a diameter of the circumcircle of the triangle $A B C$, therefore $A C \perp C G$. On the other hand we have $A C \perp D C$, and it implies that points $D, C, G$ are collinear.
Moreover, as $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 265 |
NT1 Determine all positive integer numbers $k$ for which the numbers $k+9$ are perfect squares and the only prime factors of $k$ are 2 and 3 .
|
Solution: We have an integer $x$ such that
$$
x^{2}=k+9
$$
$k=2^{a} 3^{b}, a, b \geq 0, a, b \in \mathbb{N}$.
Therefore,
$$
(x-3)(x+3)=k \text {. }
$$
If $b=0$ then we have $k=16$.
If $b>0$ then we have $3 \mid k+9$. Hence, $3 \mid x^{2}$ and $9 \mid k$.
Therefore, we have $b \geq 2$. Let $x=3 y$.
$$
(y-1)(y+1... | {16,27,72,216,432,2592} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 266 |
NT2 A group of $n>1$ pirates of different age owned total of 2009 coins. Initially each pirate (except for the youngest one) had one coin more than the next younger.
a) Find all possible values of $n$.
b) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the la... | ## Solution:
a) If $n$ is odd, then it is a divisor of $2009=7 \times 7 \times 41$. If $n>49$, then $n$ is at least $7 \times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7, n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pirate has ... | 1994 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 267 |
NT3 Find all pairs $(x, y)$ of integers which satisfy the equation
$$
(x+y)^{2}\left(x^{2}+y^{2}\right)=2009^{2}
$$
|
Solution: Let $x+y=s, x y=p$ with $s \in \mathbb{Z}^{*}$ and $p \in \mathbb{Z}$. The given equation can be written in the form
$$
s^{2}\left(s^{2}-2 p\right)=2009^{2}
$$
or
$$
s^{2}-2 p=\left(\frac{2009}{s}\right)^{2}
$$
So, $s$ divides $2009=7^{2} \times 41$ and it follows that $p \neq 0$.
If $p>0$, then $2009^{... | (40,9),(9,40),(-40,-9),(-9,-40) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 268 |
NT4 Determine all prime numbers $p_{1}, p_{2}, \ldots, p_{12}, p_{13}, p_{1} \leq p_{2} \leq \ldots \leq p_{12} \leq p_{13}$, such that
$$
p_{1}^{2}+p_{2}^{2}+\ldots+p_{12}^{2}=p_{13}^{2}
$$
and one of them is equal to $2 p_{1}+p_{9}$.
|
Solution: Obviously, $p_{13} \neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \geq 7$.
We have that $n^{2} \equiv 1(\bmod 8)$, when $n$ is odd. Let $k$ be the number of prime numbers equal to 2 . Looking at equation modulo 8 we g... | (2,2,2,2,2,2,2,3,3,5,7,7,13),(2,2,2,2,2,2,2,3,3,5,7,13,17),(2,2,2,2,2,2,2,3,3,5,7,29,31) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 269 |
C1 Inside of a square whose side length is 1 there are a few circles such that the sum of their circumferences is equal to 10 . Show that there exists a line that meets alt least four of these circles.
|
Solution
Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \pi$, it follows that the sum of the lengths of all me... | proof | Geometry | proof | Yes | Yes | olympiads | false | 270 |
C2 Can we divide an equilateral triangle $\triangle A B C$ into 2011 small triangles using 122 straight lines? (there should be 2011 triangles that are not themselves divided into smaller parts and there should be no polygons which are not triangles)
| ## Solution
Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2}=1444$ triangles. Then we erase 11 lines which are closest to the vertex $A$ and parallel to the side $B C$ and we draw 21 lines perpendicular to $B C$, the first starting from the vertex $A$ and... | 2011 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 271 |
C3 We can change a natural number $n$ in three ways:
a) If the number $n$ has at least two digits, we erase the last digit and we subtract that digit from the remaining number (for example, from 123 we get $12-3=9$ );
b) If the last digit is different from 0 , we can change the order of the digits in the opposite on... |
Solution
The answer is NO. We will prove that if the first number is divisible by 11, then all the numbers which we can get from $n$, are divisible by 11 . When we use $a$ ), from the number $10 a+b$, we will get the number $m=a-b=11 a-n$, so $11 \mid m$ since $11 \mid n$. It's well-known that a number is divisible b... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 272 |
C4 In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps, if: $... | ## Solution
We will denote the people by $A, B, C, \ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.
a) Five people form 10 pairs, so at least 10 swaps are necessary.
In fa... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 273 |
C5 A set $S$ of natural numbers is called good, if for each element $x \in S, x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\{1,2,3, \ldots, 63\}$.
|
Solution
Let set $B$ be the good subset of $A$ which have the maximum number of elements. We can easily see that the number 1 does not belong to $B$ since 1 divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides the sum ... | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 274 |
C7 Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area 3 , inside of the given rectangle, such that
the sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of the rect... | ## Solution
Let $A B C D$ be a rectangle with $A B=m$ and $A D=n$ where $m, n$ are natural numbers such that $m \geq n \geq 2$. Suppose that inside of the rectangle $A B C D$ is placed a rectangular lattice consisting of some identical squares whose areas are equals to 3 , where $k$ of them are placed along the side $... | (2,3);(3,4);(3,6);(3,8);(3,10);(3,12) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 275 |
C8 Determine the polygons with $n$ sides $(n \geq 4)$, not necessarily convex, which satisfy the property that the reflection of every vertex of polygon with respect to every diagonal of the polygon does not fall outside the polygon.
Note: Each segment joining two non-neighboring vertices of the polygon is a diagonal... | ## Solution
A polygon with this property has to be convex, otherwise we consider an edge of the convex hull of this set of vertices which is not an edge of this polygon. All the others vertices are situated in one of the half-planes determined by the support-line of this edge, therefore the reflections of the others v... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 276 |
C9 Decide if it is possible to consider 2011 points in a plane such that the distance between every two of these points is different from 1 and each unit circle centered at one of these points leaves exactly 1005 points outside the circle.
| ## Solution
NO. If such a configuration existed, the number of segments starting from each of the 2011 points towards the other one and having length less than 1 would be 1005 . Since each segment is counted twice, their total number would be $1005 \cdot 2011 / 2$ which is not an integer, contradiction!
| proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 277 |
A1. Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=a b+b c+c d+d e+e a$ and $B=a c+c e+e b+b d+d a$.
Show that
$$
2005 A+B \leq 0 \text { or } \quad A+2005 B \leq 0
$$
| ## Solution
We have
$$
0=(a+b+c+d+e)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+2 A+2 B
$$
This implies that
$$
A+B \leq 0 \text { or } 2006(\dot{A}+B)=(2005 A+B)+(A+2005 B) \leq 0
$$
This implies the conclusion.
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 278 |
A2. Find all positive integers $x, y$ satisfying the equation
$$
9\left(x^{2}+y^{2}+1\right)+2(3 x y+2)=2005
$$
| ## Solution
The given equation can be written into the form
$$
2(x+y)^{2}+(x-y)^{2}=664
$$
Therefore, both numbers $x+y$ and $x-y$ are even.
Let $x+y=2 m$ and $x-y=2 t, t \in \mathbb{Z}$.
Now from (1) we have that $t$ and $t^{2}$ are even and $m$ is odd.
So, if $t=2 k, k \in \mathbb{Z}$ and $m=2 n+1, n \in \mathb... | (x,y)=(11,7)or(x,y)=(7,11) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 279 |
A3. Find the maximum value of the area of a triangle having side lengths $a, b, c$ with
$$
a^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}
$$
| ## Solution
Without any loss of generality, we may assume that $a \leq b \leq c$.
On the one hand, Tchebyshev's inequality gives
$$
(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \leq 3\left(a^{3}+b^{3}+c^{3}\right)
$$
Therefore using the given equation we get
$$
a+b+c \leq 3 \text { or } p \leq \frac{3}{2}
$$
where $p$ d... | S\leq\frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 280 |
A4. Find all the integer solutions of the equation
$$
9 x^{2} y^{2}+9 x y^{2}+6 x^{2} y+18 x y+x^{2}+2 y^{2}+5 x+7 y+6=0
$$
| ## Solution
The equation is equivalent to the following one
$$
\begin{aligned}
& \left(9 y^{2}+6 y+1\right) x^{2}+\left(9 y^{2}+18 y+5\right) x+2 y^{2}+7 y++6=0 \\
& \Leftrightarrow(3 y+1)^{2}\left(x^{2}+x\right)+4(3 y+1) x+2 y^{2}+7 y+6=0
\end{aligned}
$$
Therefore $3 y+1$ must divide $2 y^{2}+7 y+6$ and so it must... | (-2,0),(-3,0),(0,-2),(-1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 281 |
A5. Solve the equation
$$
8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\left(x^{2}+y^{2}+x y+1\right)
$$
in the set of integers.
| ## Solution
We transform the equation to the following one
$$
\left(x^{2}+y^{2}\right)(8 x+8 y-15)=15(x y+1)
$$
Since the right side is divisible by 3 , then $3 /\left(x^{2}+y^{2}\right)(8 x+8 y-15)$. But if $3 /\left(x^{2}+y^{2}\right)$, then $3 / x$ and $3 / y, 009$ will wive $15(x y+1)$ and $3 /(x y+1)$, which is... | (x,y)=(1,2)(x,y)=(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 282 |
G1. Let $A B C D$ be an isosceles trapezoid with $A B=A D=B C, A B / / D C, A B>D C$. Let $E$ be the point of intersection of the diagonals $A C$ and $B D$ and $N$ be the symmetric point of $\mathrm{B}$ with respect to the line $\mathrm{AC}$. Prove that quadrilateral $A N D E$ is cyclic.
| ## Solution
Let $\omega$ be a circle passing through the points $A, N, D$ and let $M$ the point where $\omega$ intersects $B D$ for the second time. The quadrilateral $A N D M$ is cyclic and it follows that
$$
\angle N D M+\angle N A M=\angle N D M+\angle B D C=180^{\circ}
$$
and
Figure 2
Assume that point $C$ lies on the line segment $B P$. By the Power of Point theorem we have $M A^{2}=M R \cdot M B$ and so $M P^{2}=M R \cdot M B$. The last equality i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 284 |
G3. Let $A B C D E F$ be a regular hexagon. The points $\mathrm{M}$ and $\mathrm{N}$ are internal points of the sides $\mathrm{DE}$ and $\mathrm{DC}$ respectively, such that $\angle A M N=90^{\circ}$ and $A N=\sqrt{2} \cdot C M$. Find the measure of the angle $\angle B A M$.
| ## Solution
Since $A C \perp C D$ and $A M \perp M N$ the quadrilateral $A M N C$ is inscribed. So, we have
$$
\angle M A N=\angle M C N
$$
Let $P$ be the projection of the point $M$ on the line $C D$. The triangles $A M N$ and $C P M$ are similar implying
$$
\frac{A M}{C P}=\frac{M N}{P M}=\frac{A N}{C M}=\sqrt{2}... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 285 |
G4. Let $\mathrm{ABC}$ be an isosceles triangle such that $A B=A C$ and $\angle \frac{A}{2}<\angle B$. On the extension of the altitude $\mathrm{AM}$ we get the points $\mathrm{D}$ and $\mathrm{Z}$ such that $\angle C B D=\angle A$ and $\angle Z B A=90^{\circ}$. $\mathrm{E}$ is the foot of the perpendicular from $\mat... | ## Solution
The points $A, B, K, Z$ and $C$ are co-cyclic.
Because ME//AC so we have
$$
\angle K E M=\angle E A C=\angle M B K
$$
Therefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have
$$
\begin{aligned}
& \angle A B F=\angle A B C-\angle F B C \\
& =\angle A K C-\angle E K M=\angle M K C
\end{aligned}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 286 |
G5. Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$, respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$, respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$, respectiv... | ## Solution
The points $B, P, C$ are collinear, and
$$
\angle A P C=\angle A P B=90^{\circ}
$$
Let $N$ be the midpoint of $D P$.
So we have:
$$
\begin{aligned}
& \angle N O P=\angle D A P \\
& =\angle E C P=\angle E C A+\angle A C P
\end{aligned}
$$
Since $O K / / B C$ and $O K$ is the bisector of $\angle A K P$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 287 |
G6. A point $O$ and the circles $k_{1}$ with center $O$ and radius $3, k_{2}$ with center $O$ and radius 5, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $A B C$ is equilateral triangle, find the maximum value of the distance $O C$.
| ## Solution
It is easy to see that the points $O$ and $C$ must be in different semi-planes with respect to the line $A B$.
Let $O P B$ be an equilateral triangle ( $P$ and $C$ on the same side of $O B$ ). Since $\angle P B C$ $=60^{\circ}-\angle A B P$ and $\angle O B A=60^{\circ}-\angle A B P$, then $\angle P B C=\a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 288 |
G7. Let $A B C D$ be a parallelogram, $\mathrm{P}$ a point on $C D$, and $Q$ a point on $A B$. Let also $M=A P \cap D Q, \quad N=B P \cap C Q, K=M N \cap A D$, and $L=M N \cap B C$. Show that $B L=D K$.
| ## Solution
Let $O$ be the intersection of the diagonals. Let $P_{1}$ be on $A B$ such that $P P_{1} / / A D$, and let $Q_{1}$ be on $C D$ such that $\mathrm{Q} Q_{1} / / A D$. Let $\sigma$ be the central symmetry with center $\mathrm{O}$. Let $\left.P^{\prime}=\sigma(P), Q^{\prime}=\sigma(Q), P_{1}^{\prime}=\sigma\le... | proof | Geometry | proof | Yes | Yes | olympiads | false | 289 |
NT1. Find all the natural numbers $m$ and $n$, such that the square of $m$ minus the product of $n$ with $k$, is 2 , where the number $k$ is obtained from $n$ by writing 1 on the left of the decimal notation of $n$.
| ## Solution
Let $t$ be the number of digits of $n$. Then $k=10^{t}+n$. So
$$
\mathrm{m}^{2}=n\left(10^{t}+\mathrm{n}\right)+2, \text { i.e. } \mathrm{m}^{2}-\mathrm{n}^{2}=10^{t} n+2
$$
This implies that $\mathrm{m}, \mathrm{n}$ are even and both $\mathrm{m}, \mathrm{n}$ are odd.
If $t=1$, then, 4 is divisor of $10... | =11,n=7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 290 |
NT2. Find all natural numbers $n$ such that $5^{n}+12^{n}$ is perfect square.
| ## Solution
By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.
If $n=2 k+1$ is odd, we consider the equation modulo 5 and we obtain
$$
\begin{aligned}
x^{2} & \equiv 5^{2 k+1}+12^{2 k+1}(\bmod 5) \equiv 2^{2 k} \cdot 2(\bmod 5) \\
& \equiv(-1)^{k} \cdot 2(\bmod 5) \equiv \pm 2(\bmod... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 291 |
NT3. Let $p$ be an odd prime. Prove that $p$ divides the integer
$$
\frac{2^{p!}-1}{2^{k}-1}
$$
for all integers $k=1,2, \ldots, p$.
| ## Solution
At first, note that $\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.
We start with the case $\mathrm{k}=\mathrm{p}$. Since $p \mid 2^{p}-2$, then $p / 22^{p}-1$ and so it suffices to prove that $p \mid 2^{(p)!}-1$. This is obvious as $p \mid 2^{p-1}-1$ and $\left(2^{p-1}-1\right) \mid 2^{(p)!}-1$.
If $\m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 292 |
NT4. Find all the three digit numbers $\overline{a b c}$ such that
$$
\overline{a b c}=a b c(a+b+c)
$$
| ## Solution
We will show that the only solutions are 135 and 144 .
We have $a>0, b>0, c>0$ and
$$
9(11 a+b)=(a+b+c)(a b c-1)
$$
- If $a+b+c \equiv 0(\bmod 3)$ and $a b c-1 \equiv 0(\bmod 3)$, then $a \equiv b \equiv c \equiv 1(\bmod 3)$ and $11 a+b \equiv 0(\bmod 3)$. It follows now that
$$
a+b+c \equiv 0(\bmod 9)... | 135144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 293 |
NT5. Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$.
| ## Solution
We start with a simple fact:
Lemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.
For a proof, just note that numbers $b, 2 b, \ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 294 |
C1. A triangle with area 2003 is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is 2005 . Show that at mest one of the smaller triangles has area less or equal to 1.
| ## Solution
Since all the vertices are 2005 , and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least 2003. So, it follows that at least one of the small triangles has area at most 1
| proof | Geometry | proof | Yes | Yes | olympiads | false | 295 |
C4. Let $p_{1}, p_{2}, \ldots, p_{2005}$ be different prime numbers. Let $\mathrm{S}$ be a set of natural numbers which elements have the property that their simple divisors are some of the numbers $p_{1}, p_{2}, \ldots, p_{2005}$ and product of any two elements from $\mathrm{S}$ is not perfect square.
What is the ma... | ## Solution
Let $a, b$ be two arbitrary numbers from $\mathrm{S}$. They can be written as
$$
a=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{2005}^{a_{2005}} \text { and } b=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{2005}^{\beta_{2005}}
$$
In order for the product of the elements $a$ and $b$ to be a square all the sums ... | 2^{2005} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 297 |
A1 Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$\left(a^{5}+a^{4}+a^{3}+a^{2}+a+1\right)\left(b^{5}+b^{4}+b^{3}+b^{2}+b+1\right)\left(c^{5}+c^{4}+c^{3}+c^{2}+c+1\right) \geq 8\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
| ## Solution
We have $x^{5}+x^{4}+x^{3}+x^{2}+x+1=\left(x^{3}+1\right)\left(x^{2}+x+1\right)$ for all $x \in \mathbb{R}_{+}$.
Take $S=\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
The inequality becomes $S\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8 S$.
It remains to pro... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 298 |
A2 Let $x, y, z$ be positive real numbers. Prove that:
$$
\frac{x+2 y}{z+2 x+3 y}+\frac{y+2 z}{x+2 y+3 z}+\frac{z+2 x}{y+2 z+3 x} \leq \frac{3}{2}
$$
| ## Solution 1
Notice that $\sum_{c y c} \frac{x+2 y}{z+2 x+3 y}=\sum_{c y c}\left(1-\frac{x+y+z}{z+2 x+3 y}\right)=3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y}$.
We have to proof that $3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y} \leq \frac{3}{2}$ or $\frac{3}{2(x+y+z)} \leq \sum_{c y c} \frac{1}{z+2 x+3 y}$.
By Cauchy-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 299 |
A3 Let $a, b$ be positive real numbers. Prove that $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq a+b$.
|
Solution 1
Applying $x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}$ for $x=\sqrt{\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\sqrt{a b}$, we will obtain $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq \sqrt{\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \leq \sqrt{\frac{3\left(a^{2}+b^{2}+2 a b\right)}{3}}=a+b$.
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 300 |
A4 Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.
| ## Solution 1
We have $(x+y)\left(x^{2}+y^{2}\right) \geq(x+y)\left(x^{3}+y^{3}\right) \geq\left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq(1+1)\left(x^{2}+y^{2}\right) \geq(x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 301 |
A5 Determine the positive integers $a, b$ such that $a^{2} b^{2}+208=4\{l c m[a ; b]+g c d(a ; b)\}^{2}$.
| ## Solution
Let $d=\operatorname{gcd}(a, b)$ and $x, y \in \mathbb{Z}_{+}$such that $a=d x, b=d y$. Obviously, $(x, y)=1$. The equation is equivalent to $d^{4} x^{2} y^{2}+208=4 d^{2}(x y+1)^{2}$. Hence $d^{2} \mid 208$ or $d^{2} \mid 13 \cdot 4^{2}$, so $d \in\{1,2,4\}$. Take $t=x y$ with $t \in \mathbb{Z}_{+}$.
Cas... | (,b)\in{(2,12);(4,6);(6,4);(12;2)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 302 |
A6 Let $x_{i}>1$, for all $i \in\{1,2,3, \ldots, 2011\}$. Prove the inequality $\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?
| ## Solution 1
Realize that $\left(x_{i}-2\right)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4\left(x_{i}-1\right)$. So we get:
$\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. ... | 8044 | Inequalities | proof | Yes | Yes | olympiads | false | 303 |
A7 Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$
| ## Solution 1
By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{... | 3 | Inequalities | proof | Yes | Yes | olympiads | false | 304 |
A8 Decipher the equality $(\overline{L A R N}-\overline{A C A}):(\overline{C Y P}+\overline{R U S})=C^{Y^{P}} \cdot R^{U^{S}}$ where different symbols correspond to different digits and equal symbols correspond to equal digits. It is also supposed that all these digits are different from 0 .
| ## Solution
Denote $x=\overline{L A R N}-\overline{A C A}, y=\overline{C Y P}+\overline{R U S}$ and $z=C^{Y^{P}} \cdot R^{U^{S}}$. It is obvious that $1823-898 \leq x \leq 9187-121,135+246 \leq y \leq 975+864$, that is $925 \leq x \leq 9075$ and $381 \leq y \leq 1839$, whence it follows that $\frac{925}{1839} \leq \fr... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 305 |
A9 Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$.
| ## Solution 1
Case I. If $\min \left(x_{1}, x_{n}\right)=x_{1}$, we know that $x_{k} \geq \min \left(x_{k} ; x_{k+1}\right)$ for all $k \in\{1,2,3, \ldots, n-1\}$. So $x_{1}+x_{2}+\ldots+x_{n-1} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{1}$, hence $\sum_{k=2}^{n-1} x_{k}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 306 |
A1 If for the real numbers $x, y, z, k$ the following conditions are valid, $x \neq y \neq z \neq x$ and $x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right)=2008$, find the product $x y z$.
|
Solution
$x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right) \Rightarrow x^{2}+x z+z^{2}=-k(x+z):(1)$ and $y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right) \Rightarrow y^{2}+y x+x^{2}=-k(y+x):(2)$
- From (1) $-(2) \Rightarrow x+y+z=-k:(*)$
- If $x+z=0$, then from $... | xy1004 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 307 |
A2 Find all real numbers $a, b, c, d$ such that $a+b+c+d=20$ and $a b+a c+a d+b c+b d+c d=$ 150 .
| ## Solution
$400=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2 \cdot 150$, so $a^{2}+b^{2}+c^{2}+d^{2}=100$. Now $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}=3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-2(a b+$ $a c+a d+b c+b d+c d)=300-300=0$. Thus $a=b=c=d=5$.
| =b===5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 308 |
A3 Let the real parameter $p$ be such that the system
$$
\left\{\begin{array}{l}
p\left(x^{2}-y^{2}\right)=\left(p^{2}-1\right) x y \\
|x-1|+|y|=1
\end{array}\right.
$$
has at least three different real solutions. Find $p$ and solve the system for that $p$.
| ## Solution
The second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it that s... | p=1orp=-1,0\leqx\leq1,withy=\x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 309 |
A4 Find all triples $(x, y, z)$ of real numbers that satisfy the system
$$
\left\{\begin{array}{l}
x+y+z=2008 \\
x^{2}+y^{2}+z^{2}=6024^{2} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2008}
\end{array}\right.
$$
| ## Solution
The last equation implies $x y z=2008(x y+y z+z x)$, therefore $x y z-2008(x y+y z+z x)+$ $2008^{2}(x+y+z)-2008^{3}=0$.
$(x-2008)(y-2008)(z-2008)=0$.
Thus one of the variable equals 2008. Let this be $x$. Then the first equation implies $y=-z$. From the second one it now follows that $2 y^{2}=6024^{2}-20... | (2008,4016,-4016) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 310 |
A5 Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=3 \\
x+y+z \leq 12
\end{array}\right.
$$
| ## Solution
If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have
$$
\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right) \leq 22
$$
From AM-GM we have
$$
\frac{4 x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9 x}{z} \geq 6,... | (2,4,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 311 |
A7 Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq(1+2 a)(1+2 b)(1+2 c)
$$
| ## Solution 1
By Cauchy-Schwarz inequality and $a b c=1$ we get
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\s... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 313 |
A8 Show that
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 4\left(\frac{x}{x y+1}+\frac{y}{y z+1}+\frac{z}{z x+1}\right)^{2}
$$
for any real positive numbers $x, y$ and $z$.
| ## Solution
The idea is to split the inequality in two, showing that
$$
\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
can be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\fr... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 314 |
A9 Consider an integer $n \geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequenc... | ## Solution
After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.
Notice that the 256 numbers left after the first operation are $3,7, \ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 number... | 683 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 315 |
C1 On a $5 \times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an emp... |
Solution
a) Position 20 white markers on the board such that the left-most column is empty. This
positioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on positi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 316 |
C2 Kostas and Helene have the following dialogue:
Kostas: I have in my mind three positive real numbers with product 1 and sum equal to the sum of all their pairwise products.
Helene: I think that I know the numbers you have in mind. They are all equal to 1.
Kostas: In fact, the numbers you mentioned satisfy my con... | ## Solution
Kostas is right according to the following analysis:
If $x, y, z$ are the three positive real numbers Kostas thought about, then they satisfy the following equations:
$$
\begin{gathered}
x y+y z+z x=x+y+z \\
x y z=1
\end{gathered}
$$
Subtracting (1) from (2) by parts we obtain
$$
\begin{gathered}
x y z... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 317 |
C3 Integers $1,2, \ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.
| ## Solution
Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.
For our assumption,
$$
S \equiv 0+1+\ldots+2 n-1=\frac{(2 n-1) 2 n}{2}=(2 n-1) n \equiv n \quad(\bmod 2 n)
$$
But, if we sum, breaking all sums into its components, we derive
$$
S \equiv 2(1+\l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 318 |
G1 Two perpendicular chords of a circle, $A M, B N$, which intersect at point $K$, define on the circle four arcs with pairwise different length, with $A B$ being the smallest of them.
We draw the chords $A D, B C$ with $A D \| B C$ and $C, D$ different from $N, M$. If $L$ is the point of intersection of $D N, M C$ a... | ## Solution
First we prove that $N L \perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.
From the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:
$$
\varangle D C L=\varangle D A M \text { and } \varangle C D L=\varangle C B N \text {. }
$$
So we obtain
$$
\varangl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 320 |
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