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 try showing that $p^{6}-1$ is divisible by $2^{3} \cdot 3^{2} \cdot 7=504$ and this would be the largest positive integer that divides $p^{6}-1$ for all primes greater than 7 .
By Fermat's theorem, $7 \mid p^{6}-1$.
Next, since $p$ is odd, $8 \mid p^{2}-1=(p-1)(p+1)$, hence $8 \mid p^{6}-1$.
It remains to show that $9 \mid p^{6}-1$.
Any prime number $p, p>3$ is 1 or -1 modulo 3 .
In the first case both $p-1$ and $p^{2}+p+1$ are divisible by 3 , and in the second case, both $p+1$ and $p^{2}-p+1$ are divisible by 3 .
Consequently, the required number is indeed 504
| 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
$$
x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1} \equiv x_{1} \ldots x_{j-1} x_{j+1} \ldots x_{m} \quad(\bmod 11)
$$
which would lead to $x_{i} \equiv 0(\bmod 11)$ for some $i \neq j$, contradicting (1).
We now prove that 10 is indeed the required maximum.
Consider $x_{i}=i$, for all $i \in\{1,2, \ldots, 10\}$. The products $2 \cdot 3 \cdots \cdot 10,1 \cdot 3 \cdots \cdots 10, \ldots$, $1 \cdot 2 \cdots \cdot 9$ are all different $(\bmod 11)$, and so
$$
2 \cdot 3 \cdots \cdots 10+1 \cdot 3 \cdots \cdots 10+\cdots+1 \cdot 2 \cdots \cdot 9 \equiv 1+2+\cdots+10 \quad(\bmod 11)
$$
and condition b) is satisfied, since $1+2+\cdots+10=55=5 \cdot 11$.
| 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 $2^{4 \cdot 2+2}+1=1025=13 \cdot 78+11$. Hence 13 is a divisor of $2^{4 n+2}+1$ precisely when $n \equiv 1(\bmod 3)$. Hence, $A_{n}$ is an integer iff $n \equiv 1(\bmod 3)$.
b) Applying the identity $4 x^{4}+1=\left(2 x^{2}-2 x+1\right)\left(2 x^{2}+2 x+1\right)$, we have $2^{4 n+2}+1=$ $\left(2^{2 n+1}-2^{n+1}+1\right)\left(2^{2 n+1}+2^{n+1}+1\right)$. For $n=1, A_{1}=1$, which is not a prime. According to a), if $n \neq 1$, then $n \geq 4$. But then $2^{2 n+1}+2^{n+1}+1>2^{2 n+1}-2^{n+1}+1>65$, and $A_{n}$ has at least two factors. We conclude that $A_{n}$ can never be a prime.
| 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
$$
3 x y(x+y) \equiv 2 \quad(\bmod 7)
$$
or
$$
(x+y)^{3}-x^{3}-y^{3} \equiv 2 \quad(\bmod 7)
$$
Note that, by Fermat's Little Theorem, we have that for any integer $k$ the cubic residues are $k^{3} \equiv-1,0,1 \bmod 7$. It follows that in (4.1) some of $(x+y)^{3}, x^{3}$ and $y^{3}$ should be divisible by 7 , but in this case, $x y(x+y)$ is divisible by 7 and this is a contradiction. So, the only possibility is to have $z=0$ and consequently, $x y(x+y)+4=2$, or, equivalently, $x y(x+y)=-2$. The only solution of the latter is $(x, y)=(-1,-1)$, so the required triples are $(a, b, c)=(k+2, k+1, k), k \in \mathbb{Z}$, and all their cyclic permutations.
| (,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 $x, y \in\{a, b, c, d\}$. But then the left hand side in the above equality is congruent to $1(\bmod 3)$ and the right hand side congruent to $2(\bmod 3)$, contradiction.
Assume $a+b+c+d \equiv 1(\bmod 3)$. Then $x+y \equiv 2(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$, and $x \equiv 1(\bmod 3)$, for all $x, y \in\{a, b, c, d\}$. Hence, $a, b, c, d \in\{1,4,7\}$, and since $4 \mid \overline{a b c d}$, we have $c=d=4$. Therefore, $8 \mid \overline{a b 44}$, and since at least one more factor is even, it follows that $16 \overline{a b 44}$. Then $b \neq 4$, and the only possibilities are $b=1$, implying $a=4$, which is impossible because 4144 is not divisible by $5=1+4$, or $b=7$, implying $11 \mid \overline{a 744}$, hence $a=7$, which is also impossible because 7744 is not divisible by $14=7+7$.
We conclude that $3 \mid \overline{a b c d}$, hence also $3 \mid a+b+c+d$. Then at least one factor $x+y$ of $(a+b),(a+c),(a+d),(b+c),(b+d),(c+d)$ is a multiple of 3 , implying that also $3 \mid a+b+c+d-x-y$, so $9 \mid \overline{a b c d}$. Then $9 \mid a+b+c+d$, and $a+b+c+d \in\{9,18,27,36\}$. Using the inequality $x y \geq x+y-1$, valid for all $x, y \in \mathbb{N}^{*}$, if $a+b+c+d \in\{27,36\}$, then
$$
\overline{a b c d}=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \geq 26^{3}>10^{4}
$$
which is impossible.
Using the inequality $x y \geq 2(x+y)-4$ for all $x, y \geq 2$, if $a+b+c+d=18$ and all two-digit sums are greater than 1 , then $\overline{a b c d} \geq 32^{3}>10^{4}$. Hence, if $a+b+c+d=18$, some two-digit sum must be 1 , hence the complementary sum will be 17 , and the digits are $\{a, b, c, d\}=\{0,1,8,9\}$. But then $\overline{a b c d}=1 \cdot 17 \cdot 8 \cdot 9^{2} \cdot 10>10^{4}$.
We conclude that $a+b+c+d=9$. Then among $a, b, c, d$ there are either three odd or three even numbers, and $8 \mid \overline{a b c d}$.
If three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies that $d \in\{2,6\}$. If $d=6$, then $a=b=c=1$. But 1116 is not divisible by 7 , so this is not a solution. If $d=2$, then $a, b, c$ are either $1,1,5$ or $1,3,3$ in some order. In the first case $2 \cdot 6^{2} \cdot 3^{2} \cdot 7=4536 \neq \overline{a b c d}$. The second case cannot hold because the resulting number is not a multiple of 5 .
Hence, there has to be one odd and three even digits. At least one of the two-digits sums of even digits is a multiple of 4 , and since there cannot be two zero digits, we have either $x+y=4$ and $z+t=5$, or $x+y=8$ and $z+t=1$ for some ordering $x, y, z, t$ of $a, b, c, d$. In the first case we have $d=0$ and the digits are $0,1,4,4$, or $0,2,3,4$, or $0,2,2,5$. None of these is a solution because $1 \cdot 4^{2} \cdot 5^{2} \cdot 8=3200,2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7=5040$ and $2^{2} \cdot 5 \cdot 4 \cdot 7^{2}=3920$. In the second case two of the digits are 0 and 1 , and the other two have to be either 4 and 4 , or 2 and 6 . We already know that the first possibility fails. For the second, we get
$$
(0+1) \cdot(0+2) \cdot(0+6) \cdot(1+2) \cdot(1+6) \cdot(2+6)=2016
$$
and $\overline{a b c d}=2016$ is the only solution.
| 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
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}-\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) .
$$
The last one simplifies to
$$
a b+b c+c a=\frac{a+b+c}{a b c}
$$
First we show that $a+b+c$ and $a b+b c+c a$ are different from 0 . Suppose on contrary then from relation (1) or (2) we have $a+b+c=a b+b c+c a=0$. But then we would have
$$
a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)=0
$$
which means that $a=b=c=0$. This is not possible since $a, b, c$ should be different from 0 . Now multiplying (1) and (2) we have
$$
(a+b+c)(a b+b c+c a)=\frac{(a+b+c)(a b+b c+c a)}{(a b c)^{2}}
$$
Since $a+b+c$ and $a b+b c+c a$ are different from 0 , we get $(a b c)^{2}=1$ and using the fact that $a b c>0$ we obtain that $a b c=1$. So relations (1) and (2) transform to
$$
a+b+c=a b+b c+c a
$$
Therefore,
$$
(a-1)(b-1)(c-1)=a b c-a b-b c-c a+a+b+c-1=0 .
$$
This means that at least one of the numbers $a, b, c$ is equal to 1 . Suppose that $c=1$ then relations (1) and (2) transform to $a+b+1=a b+a+b \Rightarrow a b=1$. Taking $a=t$ then we have $b=\frac{1}{t}$. We can now verify that any triple $(a, b, c)=\left(t, \frac{1}{t}, 1\right)$ satisfies both conditions. $t \in \mathbb{R} \backslash\{0\}$. From the initial observation any triple $(a, b, c)=\left(t, \frac{1}{t},-1\right)$ satisfies both conditions. $t \in \mathbb{R} \backslash\{0\}$. So, all triples that satisfy both conditions are $(a, b, c)=\left(t, \frac{1}{t}, 1\right),\left(t, \frac{1}{t},-1\right)$ and all permutations for any $t \in \mathbb{R} \backslash\{0\}$.
Comment by PSC. After finding that $a b c=1$ and
$$
a+b+c=a b+b c+c a
$$
we can avoid the trick considering $(a-1)(b-1)(c-1)$ as follows. By the Vieta's relations we have that $a, b, c$ are roots of the polynomial
$$
P(x)=x^{3}-s x^{2}+s x-1
$$
which has one root equal to 1 . Then, we can conclude as in the above solution.
| (,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
$b_{n+1}=\frac{n+5}{n+3} b_{n}=\frac{(n+5)(n+4)}{(n+3)(n+2)} b_{n-1}=\frac{(n+5)(n+4)}{(n+2)(n+1)} b_{n-2}=\cdots=\frac{(n+5)(n+4)}{5 \cdot 4} b_{1}=(n+5)(n+4)$.
Therefore $b_{n}=(n+4)(n+3)=n^{2}+7 n+12$ and $a_{n}=n^{2}+7 n+1$.
Since $(n+1)^{2}=n^{2}+2 n+1<a_{n}<n^{2}+8 n+16=(n+4)^{2}$, if $a_{n}$ is a perfect square, then $a_{n}=(n+2)^{2}$ or $a_{n}=(n+3)^{2}$.
If $a_{n}=(n+2)^{2}$, then $n^{2}+4 n+4=n^{2}+7 n+1$ giving $n=1$. If $a_{n}=(n+3)^{2}$, then $n^{2}+6 n+9=$ $n^{2}+7 n+1$ giving $n=8$.
Comment. We provide some other methods to find $a_{n}$.
Method 1: Define $b_{n}=\frac{a_{n}+11}{n+3}$. Then $b_{1}=5$ and $a_{n}=(n+3) b_{n}-11$. So
$$
a_{n+1}=(n+4) b_{n+1}-11=\frac{(n+5) a_{n}+22}{n+3}=a_{n}+\frac{2\left(a_{n}+11\right)}{n+3}=(n+3) b_{n}-11+2 b_{n}
$$
giving $(n+4) b_{n+1}=(n+5) b_{n}$. Then
$$
b_{n+1}=\frac{n+5}{n+4} b_{n}=\frac{n+5}{n+3} b_{n-1}=\cdots=\frac{n+5}{5} b_{1}=n+5
$$
Then $b_{n}=n+4$, so $a_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.
Method 2: We have
$$
(n+3) a_{n+1}-(n+5) a_{n}=22
$$
and therefore
$$
\frac{a_{n+1}}{(n+5)(n+4)}-\frac{a_{n}}{(n+4)(n+3)}=\frac{22}{(n+3)(n+4)(n+5)}=11\left[\frac{1}{n+3}-\frac{2}{n+4}+\frac{1}{n+5}\right]
$$
Now define $b_{n}=\frac{a_{n}}{(n+4)(n+3)}$ to get
$$
b_{n+1}=b_{n}+11\left[\frac{1}{n+3}-\frac{2}{n+4}+\frac{1}{n+5}\right]
$$
which telescopically gives
$$
\begin{aligned}
b_{n+1} & =b_{1}+11\left[\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{2}{6}+\frac{1}{7}\right)+\cdots+\left(\frac{1}{n+3}-\frac{2}{n+4}+\frac{1}{n+5}\right)\right] \\
& =b_{1}+11\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{n+4}+\frac{1}{n+5}\right)=\frac{9}{20}+\frac{11}{20}-\frac{11}{(n+4)(n+5)}
\end{aligned}
$$
We get $b_{n+1}=(n+4)(n+5)-11$ from which it follows that $b_{n}=(n+3)(n+4)-11=n^{2}+7 n+1$.
| 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 the right-hand side we get
$$
M=p s+\frac{s^{2}}{p}+1+\frac{2 s}{p^{2}}+\frac{1}{p^{3}}
$$
Now by $s \geq 2 \sqrt{p}$ and setting $x=p \sqrt{p}>0$ we get
$$
M \geq 2 x+5+\frac{4}{x}+\frac{1}{x^{2}}
$$
We will now prove that
$$
2 x+5+\frac{4}{x}+\frac{1}{x^{2}} \geq \frac{11+5 \sqrt{5}}{2} \text {. }
$$
Indeed, the latter is equivalent to $4 x^{3}-(5 \sqrt{5}+1) x^{2}+8 x+2 \geq 0$, which can be rewritten as
$$
\left(x-\frac{1+\sqrt{5}}{2}\right)^{2}(4 x+3-\sqrt{5}) \geq 0
$$
which is true.
Remark: Notice that the equality holds for $a=c=\sqrt{p}=\sqrt[3]{\frac{1+\sqrt{5}}{2}}$ and $b=\frac{1}{a c}$.
| ==\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 tried each ice-cream exactly once.
We will call a degustation schedule fair if the number of ice-creams that were tried by Viktor before Natalia is equal to the number of ice creams tried by Natalia before Viktor.
Prove that the number of fair schedules is strictly larger than $2020!\left(2^{1010}+(1010!)^{2}\right)$.
|
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!) $)^{2}$ distinct fair schedules by permuting the ice-creams within each group.
2) We divide all ice-creams into disjoint groups of 4 , and in each group we swap the first 2 ice-creams with the last 2 , which gives us $\left((2!)^{2}\right)^{504}=2^{1010}$ distinct schedules.
Now, to make the inequality strict, we consider 1 more schedule like 2 ), but with groups of 2 ice-creams instead of 4 .
| 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 numbers, and should differ by 1 from an already chosen number. (At the very first step Bob can choose any number he wants.) The game ends when all numbers from the set $A$ are chosen.
For example, if Alice picks $n=4$, then a valid game would be for Bob to choose 2, then Alice to choose 3, then Bob to choose 1, and then Alice to choose 4 .
Alice wins if the sum $S$ of all of the numbers that she has chosen is composite. Otherwise Bob wins. (In the above example $S=7$, so Bob wins.)
Decide which player has a winning strategy.
|
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 Alice chooses 3. Bob can now choose either 1 or 3 .
Case 2A: If Bob chooses 1, then Alice's numbers are $3,4,6,8$. So $S=21$ and Alice wins.
Case 2B: If Bob chooses 4, then Alice chooses 1 and ends with the numbers 1,3,6,8. So $S$ is even and alice wins.
Case 3: If Bob chooses 3, then Alice chooses 2. Bob can now choose either 1 or 4 .
Case 3A: If Bob chooses 1, then Alice's numbers are 2,4,6,8. So $S$ is even and Alice wins.
Case 3B: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 1 or 6 .
Case 3Bi: If Bob chooses 1, then Alice's numbers are 2,5,6,8. So $S=21$ and Alice wins.
Case 3Bii: If Bob chooses 6 , then Alice chooses 1. Then Alice's numbers are 2,5,1,8. So $S$ is even and Alice wins.
Case 4: If Bob chooses 4, then Alice chooses 5. Bob can now choose either 3 or 6 .
Case 4A: If Bob chooses 3, then Alice chooses 6. Bob can now choose either 2 or 7 .
Case 4Ai: If Bob chooses 2, then Alice chooses 1 and ends up with 5,6,1,8. So $S$ is even and Alice wins.
Case 4Aii: If Bob chooses 7, then Alice chooses 8 and ends up with $5,6,8,1$. So $S$ is even and Alice wins.
Case 4B: If Bob chooses 6, then Alice chooses 7. Bob can now choose either 3 or 8 .
Case 4Bi: If Bob chooses 3, then Alice chooses 2 and ends up with 5,7,2 and either 1 or 8 . So $S=15$ or $S=22$ and Alice wins.
Case 4Bii: If Bob chooses 8, then Alice's numbers are $5,7,3,1$. So $S$ is even and Alice wins.
Cases 5-8: If Bob chooses $k \in\{5,6,7,8\}$ then Alice follows the strategy in case $9-k$ but whenever she had to choose $\ell$, she instead chooses $9-\ell$. If at the end of that strategy she ended up with $S$, she will now end up with $S^{\prime}=4 \cdot 9-S=36-S$. Then $S^{\prime}$ is even or $S^{\prime}=15$ or $S^{\prime}=21$ so again she wins.
## GEOMETRY
| 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 of $A D$ and the line $C E$ intersects $\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\triangle B D X$ and $\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.
|
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:
$$
\angle T Y E=\angle T Y C=\angle T D C=90^{\circ}
$$
This means that $E Y$ is perpendicular to $T Y$, so $T Y$ must be tangent to $\omega$. We similarly show that $T X$ is tangent to $\omega$. Thus, $T X$ and $T Y$ are tangents from $T$ to $\omega$ which implies that $s$ is the perpendicular bisector of the segment $X Y$. Now denote by $\sigma$ the reflection of the plane with respect to $s$. Then the points $X$ and $Y$ are symmetric with respect to $s$, so $\sigma(X)=Y$. Also note that $\sigma(E)=E$, because $E$ is on $s$. Using the fact that $B C$ is perpendicular to $s$, we see that $B C$ is the reflection image of itself with respect to $s$. Now note that $B$ is the intersection point of the lines $E X$ and $B C$. This means that the image of $B$ is the intersection point of the lines $\sigma(E X)=E Y$ and $\sigma(B C)=B C$, which is $C$. From here we see that $\sigma(B)=C$, so $s$ is the perpendicular bisector of $B C$, which is what we needed to prove.
| 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 $\triangle B E Z$, respectively. Let ( $c_{2}$ ) be an arbitrary circle passing through the points $A$ and $E$. Suppose $\left(c_{1}\right)$ meets the line $C Z$ again at the point $F$, and meets $\left(c_{2}\right)$ again at the point $N$. If $P$ is the other point of intesection of $\left(c_{2}\right)$ with $A F$, prove that the points $N, B, P$ are collinear.
|
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,
then $\angle B E Z=\angle B F Z$. So by the similarity of triangles $\triangle Z B E$ and $\triangle C B Z$ we get
$$
\angle B F Z=\angle B E Z=\angle B Z C=\angle B Z F
$$
and therefore the triangle $\triangle B F Z$ is isosceles. Since $B F=B Z=A B$, then the triangle $\triangle A F Z$ is right-angled with $\angle A F Z=90^{\circ}$.
It now follows that the points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then
$$
\angle E N P=\angle E A P=\angle E A F=\angle E C F=\angle B C Z=\angle B Z E,
$$
where in the last equality we used again the similarity of the triangles $\triangle Z B E$ and $\triangle C B Z$. Since $N, B, E, Z$ are concyclic, then $\angle E N P=\angle B Z E=\angle E N B$, from which it follows that the points $N, B, P$ are collinear.
| 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 $(c)$ with $Z C$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $F E$ and $C Z$. If the tangent to $(c)$ at $Z$ meets $P A$ at $T$, prove that the points $T, E, B, Z$ are concyclic.
|
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 $\angle F E D=90^{\circ}$, then $\angle P E D=90^{\circ}$ and so
$$
\angle E P D=90^{\circ}-\angle E D C=90^{\circ}-\angle B C A=\angle E A C
$$
Therefore the points $E, A, C, P$ are concyclic. It follows that $\angle C P A=90^{\circ}$ and therefore the triangle $\angle P A Z$ is right-angled. Since also $B$ is the midpoint of $A Z$, then $P B=A B=B Z$ and so $\angle Z P B=$ $\angle P Z B$.
Furthermore, $\angle E P D=\angle E A C=\angle C B A=\angle E B A$ from which it follows that the points $P, E, B, Z$ are also concyclic.
Now observe that
$$
\angle P A E=\angle P C E=\angle Z P B-\angle P B E=\angle P Z B-\angle P Z E=\angle E Z B
$$
Therefore $P A$ is tangent to $(c)$ at $A$ as claimed.
It now follows that $T A=T Z$. Therefore
$$
\begin{aligned}
\angle P T Z & =180^{\circ}-2(\angle T A B)=180^{\circ}-2(\angle P A E+\angle E A B)=180^{\circ}-2(\angle E C P+\angle A C B) \\
& =180^{\circ}-2\left(90^{\circ}-\angle P Z B\right)=2(\angle P Z B)=\angle P Z B+\angle B P Z=\angle P B A .
\end{aligned}
$$
Thus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.
| 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 prime.
If $n \equiv 3 \bmod 4$, say $n=4 k+3$, then $m=8\left(64^{2 k+1}+1\right) \equiv 8\left((-1)^{2 k+1}+1\right) \equiv 0 \bmod 13$. Since also $m>13$, then $m$ is not prime.
| 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 cannot happen since for any integer $x$ we have that
$$
x^{2} \equiv 0,1,4 \quad(\bmod 8)
$$
Hence, $p$ must be an even prime number, which means that $p=2$. In this case, we obtain the equation
$$
9 a^{2}+17\left(b^{2}+c^{2}\right)=289
$$
It follows that $b^{2}+c^{2} \leq 17$. This is possible only for
$$
(b, c) \in\{(4,1),(3,2),(3,1),(2,2),(2,1),(1,1)\}
$$
It is easy to check that among these pairs only the $(4,1)$ gives an integer solution for $a$, namely $a=1$. Therefore, the given equation has only two solutions,
$$
(a, b, c, p) \in\{(1,1,4,2),(1,4,1,2)\}
$$
| (,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 difference of any two different chosen numbers is not divisible by at least one of the numbers 6,7 or 8 .
|
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 $r_{i}$ ( mod 168). The number $a_{i}$ is divisible by 6 ( 7 or 8 ) if and only if its remainder $r_{i}$ is divisible by 6 (respectively 7 or 8 ). The difference $\left|a_{i}-a_{j}\right|$ is divisible by 168 if and only if their remainders $r_{i}=r_{j}$.
Choosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as choosing $k$ their remainders ( mod 168) such that:
1. each chosen remainder is not divisible by 6,7 and 8 ;
2. all chosen remainders are different.
Suppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders are different and $k \leq 168$ (otherwise, there would be two equal remainders).
Denote by $B=\{0,1,2,3, \ldots, 167\}$ the set of all possible remainders ( $\bmod 168)$ and by $B_{m}$ the subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following subsets:
$$
\begin{gathered}
\left|B_{6}\right|=168: 6=28, \quad\left|B_{7}\right|=168: 7=24, \quad\left|B_{8}\right|=168: 8=21 \\
\left|B_{6} \cap B_{7}\right|=\left|B_{42}\right|=168: 42=4, \quad\left|B_{6} \cap B_{8}\right|=\left|B_{24}\right|=168: 24=7 \\
\left|B_{7} \cap B_{8}\right|=\left|B_{56}\right|=168: 56=3, \quad\left|B_{6} \cap B_{7} \cap B_{8}\right|=\left|B_{168}\right|=1
\end{gathered}
$$
Denote by $D=B_{6} \cup B_{7} \cup B_{8}$, the subset of all elements of $B$, which are divisible by at least one of the numbers 6,7 or 8 . By the Inclusion-Exclusion principle we got
$$
\begin{gathered}
|D|=\left|B_{6}\right|+\left|B_{7}\right|+\left|B_{8}\right|-\left(\left|B_{6} \cap B_{7}\right|+\left|B_{6} \cap B_{8}\right|+\left|B_{7} \cap B_{8}\right|\right)+\left|B_{6} \cap B_{7} \cap B_{8}\right|= \\
28+24+21-(4+7+3)+1=60 .
\end{gathered}
$$
Each chosen remainder belongs to the subset $B \backslash D$, since it is not divisible by 6,7 and 8 . Hence, $k \leq|B \backslash D|=168-60=108$.
Let us show that the greatest possible value is $k=108$. Consider $n=168$. Given any collection $A$ of 168 consecutive positive integers, replace each number with its remainder ( $\bmod 168$ ). Choose from these remainders 108 numbers, which constitute the set $B \backslash D$. Finally, take 108 numbers from the initial set $A$, having exactly these remainders. These $k=108$ numbers verify the required conditions.
| 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,
$$
3^{19} \equiv 3^{3} \cdot\left(3^{4}\right)^{4} \equiv 3^{3} \cdot 8^{4} \equiv 3^{3} \cdot(-9)^{2} \equiv 27 \cdot 81 \equiv 27 \cdot 8 \equiv 70 \quad(\bmod 73)
$$
and
$$
2^{19} \equiv 2 \cdot 64^{3} \equiv 2 \cdot(-9)^{3} \equiv-18 \cdot 81 \equiv-18 \cdot 8 \equiv-144 \equiv 2 \quad(\bmod 73)
$$
Thus $p$ can be only among $2,3,7,19$. We will prove all these work.
- For $p=19$ this follows by Fermat's Theorem as
$$
(x+y)^{19} \equiv x+y \quad(\bmod 19), \quad x^{19} \equiv x \quad(\bmod 19), \quad y^{19} \equiv y \quad(\bmod 19)
$$
- For $p=7$, we have that
$$
a^{19} \equiv a \quad(\bmod 7)
$$
for every integer $a$. Indeed, if $7 \mid a$, it is trivial, while if $7 \nmid a$, then by Fermat's Theorem we have
$$
7\left|a^{6}-1\right| a^{18}-1
$$
therefore
$$
7 \mid a\left(a^{18}-1\right)
$$
- For $p=3$, we will prove that
$$
b^{19} \equiv b \quad(\bmod 3)
$$
Indeed, if $3 \mid b$, it is trivial, while if $3 \nmid b$, then by Fermat's Theorem we have
$$
3\left|b^{2}-1\right| b^{18}-1
$$
therefore
$$
3 \mid b\left(b^{18}-1\right)
$$
- For $p=2$ it is true, since among $x+y, x$ and $y$ there are 0 or 2 odd numbers.
| 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 true that there are no positive integers $x$ and $d$ such that if $x, x+d$ and $x+2 d$ are divided by $3 k$ then the remainders belong to $A$ and those of $x$ and $x+d$ are different and are:
a) small?
b) medium?
c) large?
(In this problem we assume that if a multiple of $3 k$ is divided by $3 k$ then the remainder is $3 k$ rather than 0. )
|
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 $4 k2 k$. Therefore $6 k=4 k+2 kk$ so $d=(x+d)-x<k$. Hence $0 \leq x+2 d=(x+d)+d<3 k$. Thus the remainders $x, x+d$ and $x+2 d$ are in $A$ and
$$
2(x+d)=(x+2 d)+x
$$
a contradiction.
| 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 multiple of 17 then so is $m$ and hence 289 is a multiple of $17^{3}$, which is absurd. For $n$ even and not multiple of 17 , since
$$
\operatorname{gcd}\left(9 n^{2}+17, n^{2}+17\right) \mid 9\left(n^{2}+17\right)-\left(9 n^{2}+17\right)=2^{3} \cdot 17
$$
this gcd must be 1 . Therefore $n^{2}+17=a^{3}$ for an odd $a$, so
$$
n^{2}+25=(a+2)\left(a^{2}-2 a+4\right)
$$
For $a \equiv 1(\bmod 4)$ we have $a+2 \equiv 3(\bmod 4)$, while for $a \equiv 3(\bmod 4)$ we have $a^{2}-2 a+4 \equiv 3$ $(\bmod 4)$. Thus $(a+2)\left(a^{2}-2 a+4\right)$ has a prime divisor of type $4 \ell+3$. As it divides $n^{2}+25$, it has to divide $n$ and 5 , which is absurd.
| 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-Schwarz inequality for the left hand side we get
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{(a+b+c)^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}}}
$$
Using Cauchy-Schwarz inequality for three positive numbers $\alpha . \beta . \uparrow$, we have
$$
\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma} \leq \sqrt{3(\alpha+\beta+\gamma)}
$$
Using this result twice, we have
$$
\begin{aligned}
\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}} & \leq \sqrt{6(\sqrt{a}+\sqrt{b}+\sqrt{c})} \\
& \leq \sqrt{6 \sqrt{3(a+b+c)}}
\end{aligned}
$$
Combining (1) and (2) we get the desired result.
| 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
$$
(m+n)\left(m^{2}+n^{2}-m n-m-n\right) \geq 10
$$
If $m+n=2$ or $m+n=3$, then $(m, n)=(1,1),(1,2),(2,1)$ and we can check that none of them satisfies the condition $m^{3}+n^{3}>(m+n)^{2}$.
If $m+n=4$, then $(m, n)=(1,3),(2,2),(3,1)$. The pair $(m, n)=(2,2)$ doesn't satisfy the condition. The pairs $(m, n)=(1,3),(3,1)$ satisfy the condition and we can readily check that $m^{3}+n^{3} \geq(m+$ $n)^{2}+10$.
If $m+n \geq 5$ then we will show that
$$
m^{2}+n^{2}-m n-m-n \geq 2
$$
which is equivalent to
$$
(m-n)^{2}+(m-1)^{2}+(n-1)^{2} \geq 6
$$
If at least one of the numbers $m, n$ is greater or equal to 4 then $(m-1)^{2} \geq 9$ or $(n-1)^{2} \geq 9$ hence the desired result holds. As a result, it remains to check what happens if $m \leq 3$ and $n \leq 3$. Using the condition $m+n \geq 5$ we have that all such pairs are $(m, n)=(2,3),(3,2),(3,3)$.
All of them satisfy the condition and also the inequality $m^{2}+n^{2}-m n-m-n \geq 2$, thus we have the desired result.
| 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 as
$$
(m+n)\left(1+x^{3}\right)^{2} \geq 3 x^{3}\left(x^{3}+m+n+1\right)
$$
or
$$
(m+n)\left(x^{6}-x^{3}+1\right) \geq 3 x^{3}\left(x^{3}+1\right)
$$
By the AM-GM inequality we have $m \geq 3 x$ and $n \geq 3 x^{2}$, hence $m+n \geq 3 x(x+1)$. It is sufficient to prove that
$$
\begin{aligned}
x(x+1)\left(x^{6}-x^{3}+1\right) & \geq x^{3}(x+1)\left(x^{2}-x+1\right) \\
3\left(x^{6}-x^{3}+1\right) & \geq x^{2}\left(x^{2}-x+1\right) \\
\left(x^{2}-1\right)^{2} & \geq 0
\end{aligned}
$$
which is true.
| 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 $x_{1} x_{2} \ldots x_{n}$ as a function of $k$ and $n$
b) the least value of $k$, such that there exist $n, x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given conditions.
|
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
$$
x_{1}-x_{2}=k \frac{x_{2}-x_{3}}{x_{2} x_{3}}=k^{2} \frac{x_{3}-x_{4}}{\left(x_{2} x_{3}\right)\left(x_{3} x_{4}\right)}=\cdots=k^{n} \frac{x_{1}-x_{2}}{\left(x_{2} x_{3}\right)\left(x_{3} x_{4}\right) \ldots\left(x_{1} x_{2}\right)}
$$
Since $x_{1} \neq x_{2}$ we get
$$
x_{1} x_{2} \ldots x_{n}= \pm \sqrt{k^{n}}= \pm k^{\frac{n-1}{2}} \sqrt{k}
$$
If one among these two values, positive or negative, is obtained, then the other one will be also obtained by changing the sign of all $x_{i}$ since $n$ is odd.
b) From the above result, as $n$ is odd, we conclude that $k$ is a perfect square, so $k \geq 4$. For $k=4$ let $n=2019$ and $x_{3 j}=4, x_{3 j-1}=1, x_{3 j-2}=-2$ for $j=1,2, \ldots, 673$. So the required least value is $k=4$.
Comment by PSC. There are many ways to construct the example when $k=4$ and $n=2019$. Since $3 \mid 2019$, the idea is to find three numbers $x_{1}, x_{2}, x_{3}$ satisfying the given equations, not all equal, and repeat them as values for the rest of the $x_{i}$ 's. So, we want to find $x_{1}, x_{2}, x_{3}$ such that
$$
x_{1}+\frac{4}{x_{2}}=x_{2}+\frac{4}{x_{3}}=x_{3}+\frac{4}{x_{1}}
$$
As above, $x_{1} x_{2} x_{3}= \pm 8$. Suppose without loss of generality that $x_{1} x_{2} x_{3}=-8$. Then, solving the above system we see that if $x_{1} \neq 2$, then
$$
x_{2}=-\frac{4}{x_{1}-2} \text { and } x_{3}=2-\frac{4}{x_{1}}
$$
leading to infinitely many solutions. The example in the official solution is obtained by choosing $x_{1}=-2$.
Comment by PSC. An alternative formulation of the problem's statement could be the following: Let $k>1$ be a positive integer. Suppose that there exists an odd positive integer $n>2018$ and nonzero rational numbers $x_{1}, x_{2}, \ldots, x_{n}$, not all of them equal, that 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 the minimum value of $k$.
| 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
$$
E(a, b, c, d, x, y, z, t)=a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2}
$$
and assuming without loss of generality that $a \leq b \leq c \leq d$ and $x \leq y \leq z \leq t$, we have
$$
\begin{aligned}
E(a, b, c, d, x, y, z, t) & \leq E(0,0,0,0, a+x, b+y, c+z, d+t) \\
& \leq E(0,0,0,0,1, b+y, c+z, a+d+x+t-1) \\
& \leq E(0,0,0,0,1,1, c+z, a+b+d+x+y+t-2) \\
& \leq E(0,0,0,0,1,1,1,5)=28
\end{aligned}
$$
Note that if $(a, b, c, d, x, y, z, t) \neq(0,0,0,0,1,1,1,5)$, at least one of the above inequalities, obtained by the $\epsilon$ replacement mentioned above, should be a strict inequality. Thus, the maximum value of $E$ is 28 , and it is obtained only for $(a, b, c, d, x, y, z, t)=(0,0,0,0,1,1,1,5)$ and permutations of $a, b, c, d$ and of $x, y, z, t$.
| 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
$$
\frac{a}{\sqrt{a^{3}+5}} \leq \sqrt{\frac{2 a^{2}}{3(c+a)(a+b)}}
$$
Using again the AM-GM inequality we get
$$
\sqrt{\frac{2 a^{2}}{3(c+a)(a+b)}} \leq \sqrt{\frac{2}{3}}\left(\frac{\frac{a}{c+a}+\frac{a}{a+b}}{2}\right)=\frac{\sqrt{6}}{6}\left(\frac{a}{c+a}+\frac{a}{a+b}\right)
$$
From (1) and (2) we obtain
$$
\frac{a}{\sqrt{a^{3}+5}} \leq \frac{\sqrt{6}}{6}\left(\frac{a}{c+a}+\frac{a}{a+b}\right)
$$
Similar inequalities hold by cyclic permutations of the $a, b, c$ 's. Adding all these we get
$$
\sum_{\text {cyclic }} \frac{a}{\sqrt{a^{3}+5}} \leq \sum_{\text {cyc }} \frac{\sqrt{6}}{6}\left(\frac{a}{c+a}+\frac{a}{a+b}\right)=\frac{\sqrt{6}}{6} \cdot 3=\frac{\sqrt{6}}{2}
$$
which is the desired result.
| \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 $A_{k}$ can contain at most two elements of since if $n, m \in$ with $n>m$ then
$$
\sqrt{n}-\sqrt{m} \leqslant \sqrt{(k+1)^{2}-1}-\sqrt{k^{2}}<(k+1)-k=1
$$
In particular, since $\subseteq A_{1} \cup \cdots \cup A_{44}$, we have $|S| \leqslant 2 \cdot 44=88$.
On the other hand we claim that $A=\left\{m^{2}: 1 \leqslant m \leqslant 44\right\} \cup\left\{m^{2}+m: 1 \leqslant m \leqslant 44\right\}$ satisfies the properties and has $|A|=88$. We check property (b) as everything else is trivial.
So let $r, s, t$ be three elements of $A$ and assume $r<s<t$. There are two cases for $r$.
(i) If we have that $r=m^{2}$, then $t \geqslant(m+1)^{2}$ and so $\sqrt{t}-\sqrt{r} \geq 1$ verifying (b).
(ii) If we have that $r=m^{2}+m$, then $t \geqslant(m+1)^{2}+(m+1)$ and
$$
\begin{aligned}
\sqrt{t} \geqslant \sqrt{r}+1 & \Leftrightarrow \sqrt{(m+1)^{2}+(m+1)} \geqslant \sqrt{m^{2}+m}+1 \\
& \Leftrightarrow m^{2}+3 m+2 \geqslant m^{2}+m+1+2 \sqrt{m^{2}+m} \\
& \Leftrightarrow 2 m+1 \geqslant 2 \sqrt{m^{2}+m} \\
& \Leftrightarrow 4 m^{2}+4 m+1 \geqslant 4 m^{2}+4 m .
\end{aligned}
$$
So property (b) holds in this case as well.
## COMBINATORICS
| 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 neighbouring subsets for which $b-a=3$ is $n-3$. The number of the neighbouring subsets for which $b-a=4$ is $n-4$ and so on. It follows that the number of the neighbouring subsets of the set $\{1,2, \ldots, n\}$ is:
$$
(n-3)+(n-4)+\cdots+3+2+1=\frac{(n-3)(n-2)}{2}
$$
| \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 are different.
Find the largest possible value of $n$.
|
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 (which means go to the next digit). Then for example 324 can be obtained from 111 by the
string AAGAGAAA. There are in total
$$
\frac{8!}{6!\cdot 2!}=28
$$
such words, so $S$ contains 28 numbers. Now, from the conditions (3), (4), (5), if $\overline{a b c}$ is in $T$ then each of the other numbers of the form $\overline{* c}$ cannot be in $T$, neither $\overline{* * *}$ can be, nor $\overline{a * *}$. Since there are $a+b-2$ numbers of the first category, $a+c-2$ from the second and $b+c-2$ from the third one. In
these three categories there are
$$
(a+b-2)+(b+c-2)+(c+a-2)=2(a+b+c)-6=2 \cdot 9-6=12
$$
distinct numbers that cannot be in $T$ if $\overline{a b c}$ is in $T$. So, if $T$ has $n$ numbers, then $12 n$ are the forbidden ones that are in $S$, but each number from $S$ can be a forbidden number no more than three times, once for each of its digits, so
$$
n+\frac{12 n}{3} \leq 28 \Longleftrightarrow n \leq \frac{28}{5}
$$
and since $n$ is an integer, we get $n \leq 5$. A possible example for $n=5$ is
$$
T=\{144,252,315,423,531\}
$$
Comment by PSC. It is classical to compute the cardinality of $S$ and this can be done in many ways. In general, the number of solutions of the equation
$$
x_{1}+x_{2}+\cdots+x_{k}=n
$$
in positive integers, where the order of $x_{i}$ matters, is well known that equals to $\binom{n-1}{k-1}$. In our case,
we want to count the number of positive solutions to we want to count the number of positive solutions to $a+b+c=9$. By the above, this equals to $\binom{9-1}{3-1}=28$. Using the general result above, we can also find that there are $a+b-2$ numbers of the
form $\overline{* * c}$.
| 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 point $K$. The line parallel to $B D$ through $K$, intersects $A B$ at point $L$. If $M$ is the midpoint of $A B$ and $N$ is the foot of the perpendicular from $H$ to $C L$, prove that the line $M N$ bisects the segment $C H$.
|
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
$$
\angle A C K=\gamma+2 \alpha-180^{\circ}
$$
where $\gamma=\angle A C B$. From the relation $C K=C A$ we get
$$
\angle A L C=\angle A K C=180^{\circ}-\alpha-\frac{\gamma}{2}
$$
and thus from the triangle $A C L$ we obtain
$$
\angle A C L=180^{\circ}-\alpha-\angle A L C=\frac{\gamma}{2}
$$
which means that $C L$ is the angle bisector of $\angle A C B$, thus $\angle A C L=\angle B C L$. Moreover, from the fact that $C H \perp A B$ and the isosceles triangle $B O C$ has $\angle B O C=2 \alpha$, we get $\angle A C H=\angle B C O=90^{\circ}-\alpha$. It follows that,
$$
\angle N P H=2 \angle N C H=\angle O C H
$$
On the other hand, it is known that $2 C P=C H=2 O M$ and $C P \| O M$, so $C P M O$ is a parallelogram and
$$
\angle M P H=\angle O C H
$$
Now from (3) and (4) we obtain that
$$
\angle M P H=\angle N P H,
$$
which means that the points $M, N, P$ are collinear.
## Alternative formulation of the statement by PSC.
Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C>A C$, inscribed in a circle $\Gamma$. A point $D$ on $\Gamma$, which is on the arc $A B$ not containing $C$, is chosen such that $C B=C D$. A point $K$ is chosen on the segment $C D$ such that $C A=C K$. The line parallel to $B D$ through $K$, intersects $A B$ at point $L$. If $M$ is the midpoint of $A B$ and $N$ is the foot of the perpendicular from $H$ to $C L$, prove that the line $M N$ bisects the segment $C H$.
| 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 point $N$ Let $A^{\prime}$ be the symmetric of $A$ with respect to the line $E Z$ and $I, K$ the projections of $A^{\prime}$ onto $A B$ and $A C$ respectively. If $T$ is the point of intersection of the lines $I K$ and $D E$, prove that $\angle N A^{\prime} T=\angle A D T$.
|
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 $Z A L, A L E$ are similar, so $\angle Z A L=\angle A E Z$. By the similarity of the triangles $A B C, D A B$, we get $\angle A C B=\angle B A D$. We have also that $\angle A E Z=\angle B A D$, therefore
$$
\angle Z A L=\angle C A M=\angle A C B=\angle A C M
$$
Since $A F \perp C N$, we have that the right triangles $A F C$ and $C D A$ are equal. Thus the altitudes from the vertices $F, D$ of the triangles $A F C, C D A$ respectively are equal. It follows that $F D \| A C$ and since $D E \| A C$ we get that the points $E, D, F$ are collinear.
In the triangle $L F T$ we have, $A^{\prime} I \| F T$ and $\angle L A^{\prime} I=\angle L I A^{\prime}$, so $\angle L F T=\angle L T F$. Therefore the points $F, A^{\prime}, I, T$ belong to the same circle. Also, $\angle A^{\prime} I N=\angle A^{\prime} F N=90^{\circ}$ so the quadrilateral $I A^{\prime} F N$ is cyclic. Thus, the points $F, A^{\prime}, I, T, N$ all lie on a circle. From the above, we infer that
$$
\angle N A^{\prime} T=\angle T F N=\angle A C F=\angle F E Z=\angle A D T \text {. }
$$
| 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 similarly. Prove that the lines $A A_{1}, B B_{1}$, and $C C_{1}$ have a common point.
|
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 intersection of the perpendicular bisector of $A B$ with $A C$. It follows that $O$ is the orthocenter of triangle $A O_{1} O_{2}$. This means that $A O$ is perpendicular to $O_{1} O_{2}$. On the other hand, the segment $A A_{1}$ is the common chord of the two circles, thus it is perpendicular to $O_{1} O_{2}$. As a result, $A A_{1}$ passes through $O$. Similarly, $B B_{1}$ and $C C_{1}$ pass through $O$, so the three lines are concurrent at $O$.
Comment by PSC. We present here a different approach.
We first prove that $A_{1}, B$ and $C^{\prime}$ are collinear. Indeed, since $\angle B A B^{\prime}=\angle C A C^{\prime}=2 \angle B A C$, then from the circles $\left(A B B^{\prime}\right),\left(A C C^{\prime}\right)$ we get
$$
\angle A A_{1} B=90^{\circ}-\angle B A C=\angle A A_{1} C^{\prime}
$$
It follows that
$$
\angle A_{1} A C=\angle A_{1} C^{\prime} C=\angle B C^{\prime} C=90^{\circ}-\angle A B C
$$
On the other hand, if $O$ is the circumcenter of $A B C$, then
$$
\angle O A C=90^{\circ}-\angle A B C
$$
From (1) and (2) we conclude that $A_{1}, A$ and $O$ are collinear. Similarly, $B B_{1}$ and $C C_{1}$ pass through $O$, so the three lines are concurrent in $O$.
| 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
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} \geq 16
$$
|
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
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)
$$
From Leibniz's relation we have that if $H$ is the orthocenter, then
$$
O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2}
$$
It follows that
$$
9 R^{2} \geq a^{2}+b^{2}+c^{2}
$$
Therefore, using the AM-HM inequality and then (1), we get
$$
\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} \geq \frac{9}{a^{2}+b^{2}+c^{2}} \geq \frac{1}{R^{2}}
$$
Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{4}{r^{2} R^{2}} \geq \frac{16}{R^{4}}
$$
Comment by PSC. We can avoid using Leibniz's relation as follows: as in the above solution we have that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)
$$
Let $a+b+c=2 \tau, E=(A B C)$ and using the inequality $x^{2}+y^{2}+z^{2} \geq x y+y z+z x$ we get
$$
\begin{aligned}
\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} & \geq \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{2 \tau}{a b c} \\
& =\frac{\tau}{2 R E}=\frac{1}{2 R r}
\end{aligned}
$$
where we used the area formulas $E=\frac{a b c}{4 R}=\tau r$. Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{2}{r^{3} R} \geq \frac{16}{R^{4}}
$$
| 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 $G D$ and a point $I$ is taken on the diagonal $A C$ such that the triangles $A C E$ and $A E I$ are similar. The lines $B H$ and $I E$ intersect at $K$ and the lines $C A$ and $E H$ intersect at $J$. Prove that $K J \perp A B$.
|
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, $G L=\frac{a}{b} \sqrt{a^{2}+b^{2}}$.
Observing that the triangles $Q C E$ and $E L G$ are similar, we have $\frac{a}{b}=\frac{G L}{E L}$, which implies that $E L=\sqrt{a^{2}+b^{2}}$, or in other words $L \equiv C$.
Consider the circumcircle $\omega$ of the triangle $E B C$. Since
$$
\angle E B G=\angle E C G=\angle E H G=90^{\circ}
$$
the points $H$ and $G$ lie on $\omega$.
From the given similarity of the triangles $A C E$ and $A E I$, we have that
$$
\angle A I E=\angle A E C=90^{\circ}+\angle G E C=90^{\circ}+\angle G H C=\angle E H C
$$
therefore $E H C I$ is cyclic, thus $I$ lies on $\omega$.
Since $E B=E C$, we get that $\angle E I C=\angle E H B$, thus $\angle J I E=\angle E H K$. We conclude that $J I H K$ is cyclic, therefore
$$
\angle J K H=\angle H I C=\angle H B C
$$
It follows that $K J \| B C$, so $K J \perp A B$.
Comment. The proposer suggests a different way to finish the proof after proving that $I$ lies on $\omega$ : We apply Pascal's Theorem to the degenerated hexagon $E E H B C I$. Since $B C$ and $E E$ intersect at infinity, this implies that $K J$, which is the line through the intersections of the other two opposite pairs of sides of the hexagon, has to go through this point at infinity, thus it is parallel to $B C$, and so $K J \perp A B$.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 247 |
NT 1. Find all the integers pairs $(x, y)$ 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 equation, we conclude that $a \mid d^{3} b^{5}$ and thus $a \mid d^{3}$. Similarly $b \mid d^{3}$. Since $(a, b)=1$, we get that $a b \mid d^{3}$, so we can write $d^{3}=a b r$ with $r \in \mathbb{Z}$. Then, equation (1) becomes
$$
\begin{aligned}
a b r a^{5}-a b r b^{5} & =16 a b \Rightarrow \\
r\left(a^{5}-b^{5}\right) & =16
\end{aligned}
$$
Therefore, the difference $a^{5}-b^{5}$ must divide 16. Therefore, the difference $a^{5}-b^{5}$ must divide 16 . This means that
$$
a^{5}-b^{5}= \pm 1, \pm 2, \pm 4, \pm 8, \pm 16
$$
The smaller values of $\left|a^{5}-b^{5}\right|$ are 1 or 2 . Indeed, if $\left|a^{5}-b^{5}\right|=1$ then $a= \pm 1$ and $b=0$ or $a=0$ and $b= \pm 1$, a contradiction. If $\left|a^{5}-b^{5}\right|=2$, then $a=1$ and $b=-1$ or $a=-1$ and $b=1$. Then $r=-8$, and $d^{3}=-8$ or $d=-2$. Therefore, $(x, y)=(-2,2)$. If $\left|a^{5}-b^{5}\right|>2$ then, without loss of generality, let $a>b$ and $a \geq 2$. Putting $a=x+1$ with $x \geq 1$, we have
$$
\begin{aligned}
\left|a^{5}-b^{5}\right| & =\left|(x+1)^{5}-b^{5}\right| \\
& \geq\left|(x+1)^{5}-x^{5}\right| \\
& =\left|5 x^{4}+10 x^{3}+10 x^{2}+5 x+1\right| \geq 31
\end{aligned}
$$
which is impossible. Thus, the only solutions are $(x, y)=(0,0)$ or $(-2,2)$.
| (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
$$
\begin{aligned}
& -2^{n}-9^{2 k+1} \equiv 5 \quad(\bmod 7) \Rightarrow \\
& 2^{n}+2^{2 k+1} \equiv 2 \quad(\bmod 7)
\end{aligned}
$$
Since $2^{s} \equiv 1,2,4(\bmod 7)$ for $s \equiv 0,1,2(\bmod 3)$, respectively, the only possibility is $2^{n} \equiv 2^{2 k+1} \equiv$ $1(\bmod 7)$, so $3 \mid n$ and $3 \mid 2 k+1$. From the last one we get $3 \mid m$, so we can write $n=3 x$ and $m=3 y$. Therefore, the given equation takes the form
$$
5^{3} \cdot 2^{3 x}-3^{3 y}=271
$$
or
$$
\left(5 \cdot 2^{x}-3^{y}\right)\left(25 \cdot 2^{2 x}+5 \cdot 2^{x} \cdot 3^{y}+3^{2 y}\right)=271
$$
It follows that $25 \cdot 2^{2 x}+5 \cdot 2^{x} \cdot 3^{y}+3^{2 y} \leq 271$, and so $25 \cdot 2^{2 x} \leq 271$, or $x<2$. We conclude that $x=1$ and from (2) we get $y=2$. Thus, the only solution is $(m, n)=(6,3)$.
| (,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 $n$ and $k$. Note that the required condition is equivalent to $f(n) \mid f\left(2^{n}\right)$. From the discussion above, if there exists a positive integer $n$ so that $2^{n}$ can be written as $n^{2^{k}}$, for some positive integer $k$, then $f(n) \mid f\left(2^{n}\right)$. If we choose $n=2^{2^{m}}$ and $k=2^{m}-m$ for some positive integer $m$, then $2^{n}=n^{2^{k}}$ and since there are infinitely many positive integers of the form $n=2^{2^{m}}$, we have the desired result.
| 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 obtain:
$$
(a-3)+(b-5)+(c-7)=0
$$
Using the lemma and the given equations, we get:
$$
540=(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=3(a-3)(b-5)(c-7)
$$
Now,
$$
(a-3)(b-5)(c-7)=180=2 \times 2 \times 3 \times 3 \times 5
$$
Since
$$
(a-3)+(b-5)+(c-7)=0
$$
only possibility for the product $(a-3)(b-5)(c-7)$ is $(-4) \times(-5) \times 9$. Finally, we obtain the following systems of equations:
$$
\left\{\begin{array} { l }
{ a - 3 = - 4 } \\
{ b - 5 = - 5 } \\
{ c - 7 = 9 , }
\end{array} \quad \left\{\begin{array} { l }
{ a - 3 = - 5 } \\
{ b - 5 = - 4 } \\
{ c - 7 = 9 , }
\end{array} \quad \left\{\begin{array} { l }
{ a - 3 = - 4 } \\
{ b - 5 = - 5 } \\
{ c - 7 = 9 , }
\end{array} \quad \left\{\begin{array}{l}
a-3=-5 \\
b-5=-4 \\
c-7=9
\end{array}\right.\right.\right.\right.
$$
From here we get:
$$
(a, b, c) \in\{(-1,0,16),(-2,1,16),(7,10,-2),(8,9,-2)\}
$$
| (,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 for $x=\frac{4}{\sqrt{13}}, y=t=\frac{6}{\sqrt{13}}$ and $z=\frac{9}{\sqrt{13}}$.
| \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 vertices $(1,0),(0,1),(-1,0)$ and $(0,-1)$. Similarly, the graph of
$$
|x|+|y|=3
$$
is a square with vertices $(3,0),(0,3),(-3,0)$ and $(0,-3)$. The graph of the second equation of the system is a straight line with slope $a$ passing through (0,5). This line intersects the graph of the first equation in three points exactly, when passing through one of the points $(1,0)$ or $(-1,0)$. This happens if and only if $a=5$ or $a=-5$.
| =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:
$$
\frac{1}{8}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}>x y z=(1-x)(1-y)(1-z)>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}
$$
Contradiction!
Remark: The exercise along with its proof generalizes for any given (finite) number of numbers, and you can consider this new form in place of the proposed one:
Exercise: If for the real numbers $x_{1}, x_{2}, \ldots, x_{n}, 0<x_{i}<1$, for all indices $i$, and
$$
x_{1} x_{2} \ldots x_{n}=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{n}\right)
$$
show that
$$
\frac{1}{4} \leq \max _{1 \leq i \leq n}\left(1-x_{i}\right) x_{i+1}
$$
(where $x_{n+1}=x_{1}$ ).
Or you can consider the following variation:
Exercise: If for the real numbers $x_{1}, x_{2}, \ldots, x_{2009}, 0<x_{i}<1$, for all indices $i$, and
$$
x_{1} x_{2} \ldots x_{2009}=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{2009}\right)
$$
show that
$$
\frac{1}{4} \leq \max _{1 \leq i \leq 2009}\left(1-x_{i}\right) x_{i+1}
$$
(where $x_{2010}=x_{1}$ ).
| 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
$$
\left(y^{2}+x+1\right)\left(z^{2}+x+1\right) \geq(x+y+z)^{2}
$$
Multiplying these three inequalities we get the desired result.
| 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 $A C$ and $B C$ ) must be on day $x$. Similarly, one of the plays of $B E$ with $A C$ or $A D$ must be on day $x$. Thus, two of the plays in the chain $B C-D E-A C-B E-A D$ are on day $x$ (more than two among these cannot be on one day).
Consider the chain $A B-C D-E A-B D-C E-A B$. At least three days are needed for playing all the matches within it. For each of these days we conclude (as above) that there are exactly two of the plays in the chain $B C-D E-A C-B E-A D-B C$ on that day. This is impossible, as this chain consists of five plays.
It remains to show that four days will suffice:
Day 1: $A B-C D, A C-D E, A D-C E, A E-B C$
Day 2: $A B-D E, A C-B D, A D-B C, B E-C D$
Day 3: $A B-C E, A D-B E, A E-B D, B C-D E$
Day 4: $A C-B E, A E-C D, B D-C E$.
Remark: It is possible to have 5 games in one day (but not on each day).
| 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 was deleted, only 525 ways to read the word SARAJEVO remained. Find all possible positions of that 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 in the cells, to which we can go).
a) The answer is 750 , as seen from the second table.
b) If we delete the letter in a cell, the number of ways to read SARAJEVO will decrease by the product of the numbers in the corresponding cell in the two tables. As $750-525=225$, this product has to be 225. This happens only for two cells on the third row. Here is the table with the products:
| 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 $\angle Z C A=90^{\circ}$.
| ## 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
$$
\angle A G L=\angle N K L=\angle A C L
$$
Therefore, from the cyclic quadrilateral $N K C L$ we deduce:
$$
\angle K C N=\angle K L N
$$
Now, since $L R \| D Z$, we have
$$
\angle K L N=\angle K Z O
$$
It implies that quadrilateral $O K C Z$ is cyclic, and
$$
\angle O K Z=\angle O C Z
$$
Since $O K \perp G L$, we derive that $\angle Z C A=90^{\circ}$.
| 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 $\angle A M H=30^{\circ}$. Now, the altitude from $H$ to AM equals one half of $M H$, namely 2. Finally, the area is 8 .
| 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 $D^{\prime}$ intersects the straight line $B D$ at point $P$. Prove that $P C$ is the bisector of the angle $\angle B P D^{\prime}$.
|
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:
$$
\triangle A B C \cong \triangle C D A \cong \triangle C D^{\prime} A^{\prime} \Rightarrow \triangle A B X \cong \triangle C D^{\prime} Y, \triangle B C X \cong \triangle D^{\prime} A^{\prime} Y
$$
It follows that
$$
\angle A B X=\angle C D^{\prime} Y
$$
Let $M$ and $N$ be orthogonal projections of the point $C$ on the straight lines $P D^{\prime}$ and $B P$, respectively, and $Q$ is the orthogonal projection of the point $A$ on the straight line $B P$. Because $C D^{\prime}=A B$, we have that $\triangle A B Q \cong \triangle C D^{\prime} M$.
We conclude that $C M=A Q$. But, $A X=C X$ and $\triangle A Q X \cong \triangle C N X$. So, $C M=C N$ and $P C$ is the bisector of the angle $\angle B P D^{\prime}$.
Much shortened: $\triangle C D^{\prime} Y \equiv \triangle C D X$ means their altitudes from $C$ are also equal, i.e. $C M=C N$ and the conclusion.
| 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,
$$
\triangle A P O \cong \triangle E S O\left(A P=E S, \angle A P O=\angle E S O=90^{\circ}, P O=S O\right)
$$
This implies
$$
\angle O P S=\angle O S P
$$
Therefore,
$$
\angle A P S=\angle A P O+\angle O P S=90^{\circ}+\angle O P S=90^{\circ}+\angle O S P=\angle P S E
$$
Now, from quadrilateral $A P S E$ we deduce:
$$
2 \angle E A P+2 \angle A P S=\angle E A P+\angle A P S+\angle P S E+\angle S E A=360^{\circ}
$$
So,
$$
\angle E A P+\angle A P S=180^{\circ}
$$
and $A P S E$ is isosceles trapezoid. Therefore, $A E \| P S$.
| 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 circumcircles of triangles $B E F$ and $C F D$ are tangent at $F$.
|
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 $A E \perp D G$ and $D E \perp A G$, we obtain that $E$ is the orthocenter of triangle $A D G$ and $G E \perp A D$. As $A G$ is a diameter, we have $A B \perp B G$, and since $A D \perp G E$, the points $E, G$, and $B$ are collinear.
Notice that
$$
\angle C A G=90^{\circ}-\angle A G C=\angle K D C
$$
and
$$
\angle C A G=\angle G F C
$$
since both subtend the same arc.
Hence,
$$
\angle F D G=\angle G F C
$$
Therefore, $G F$ is tangent to the circumcircle of the triangle $C D F$ at point $F$.
We claim that line $G F$ is also tangent to the circumcircle of triangle $B E F$ at point $F$, which concludes the proof.
The claim is equivalent to $\angle G B F=\angle E F G$. Denote by $F^{\prime}$ the second intersection point - other than $F$ - of line $\ell$ with the circumcircle of triangle $A B C$. Observe that $\angle G B F=\angle G F^{\prime} F$, because both angles subtend the same arc, and $\angle F F^{\prime} G=\angle E F G$, since $A G$ is the perpendicular bisector of the chord $F F^{\prime}$, and we are done.
### 2.4 Number Theory
| 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 $x=3 y$.
$$
(y-1)(y+1)=2^{a} 3^{b-2}
$$
If $a=0$ then $b=3$ and we have $k=27$.
If $a \geq 1$, then the numbers $y-1$ and $y+1$ are even. Therefore, we have $a \geq 2$, and
$$
\frac{y-1}{2} \cdot \frac{y+1}{2}=2^{a-2} 3^{b-2}
$$
Since the numbers $\frac{y-1}{2}, \frac{y+1}{2}$ are consecutive numbers, these numbers have to be powers of 2 and 3 . Let $m=a-2, n=b-2$.
- If $2^{m}-3^{n}=1$ then we have $m \geq n$. For $n=0$ we have $m=1, a=3, b=2$ and $k=72$. For $n>0$ using $\bmod 3$ we have that $m$ is even number. Let $m=2 t$. Therefore,
$$
\left(2^{t}-1\right)\left(2^{t}+1\right)=3^{n}
$$
Hence, $t=1, m=2, n=1$ and $a=4, b=3, k=432$.
- If $3^{n}-2^{m}=1$, then $m>0$. For $m=1$ we have $n=1, a=3, b=3, k=216$. For $m>1$ using $\bmod 4$ we have that $n$ is even number. Let $n=2 t$.
$$
\left(3^{t}-1\right)\left(3^{t}+1\right)=2^{m}
$$
Therefore, $t=1, n=2, m=3, a=5, b=4, k=2592$.
Set of solutions: $\{16,27,72,216,432,2592\}$.
| {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 largest possible number of coins a pirate can have after several days.
| ## 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 41 coins, so the initial amounts are from $41-24=17$ to $41+24=65$ ).
If $n$ is even, then 2009 is multiple of the sum $S$ of the oldest and the youngest pirate. If $S<7 \times 41$, then $S$ is at most 39 and the pairs of pirates of sum $S$ is at least 41 , so we must have at least 82 pirates, a contradiction. So we can have just $S=7 \times 41=287$ and $S=49 \times 41=2009$; respectively, $n=2 \times 7=14$ or $n=2 \times 1=2$. Each of these cases is possible (e.g. if $n=14$, the initial amounts are from $144-7=137$ to $143+7=150$ ). In total, $n$ is one of the numbers $2,7,13,41$ and 49 .
b) If $n=7$, the average pirate has $7 \times 41=287$ coins, so the initial amounts are from 284 to 290; they have different residues modulo 7. The operation decreases one of the amounts by 6 and increases the other ones by 1 , so the residues will be different at all times. The largest possible amount in one pirate's possession will be achieved if all the others have as little as possible, namely $0,1,2,3,4$ and 5 coins (the residues modulo 7 have to be different). If this happens, the wealthiest pirate will have $2009-14=1994$ coins. Indeed, this can be achieved e.g. if every day (until that moment) the coins are given by the second wealthiest: while he has more than 5 coins, he can provide the 6 coins needed, and when he has no more than five, the coins at the poorest six pirates have to be $0,1,2,3,4,5$. Thus, $n=1994$ can be achieved.
| 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 $2009^{2}=s^{2}\left(s^{2}-2 p\right)=s^{4}-2 p s^{2}s^{4}$. We obtain that $s$ divides 2009 and $|s| \leq 41$. Thus, $s \in\{ \pm 1, \pm 7, \pm 41\}$. For these values of $s$ the equation has no integer solutions.
So, the given equation has only the solutions $(40,9),(9,40),(-40,-9),(-9,-40)$.
| (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 get:
$$
4 k+12-k \equiv 1 \quad(\bmod 8)
$$
So, $k \equiv 7(\bmod 8)$ and because $k \leq 12$ we get $k=7$. Therefore, $p_{1}=p_{2}=\ldots=p_{7}=2$. Furthermore, we are looking for solutions of equations:
$$
28+p_{8}^{2}+p_{9}^{2}+p_{10}^{2}+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}
$$
where $p_{8}, p_{9}, \ldots, p_{13}$ are odd prime numbers and one of them is equal to $p_{9}+4$.
Now, we know that when $n$ is not divisible by $3, n^{2} \equiv 1(\bmod 3)$. Let $s$ be number of prime numbers equal to 3 . Looking at equation modulo 3 we get:
$$
28+5-s \equiv 1 \quad(\bmod 3)
$$
Thus, $s \equiv 2(\bmod 3)$ and because $s \leq 5, s$ is either 2 or 5 . We will consider both cases. i. When $s=2$, we get $p_{8}=p_{9}=3$. Thus, we are looking for prime numbers $p_{10} \leq p_{11} \leq$ $p_{12} \leq p_{13}$ greater than 3 and at least one of them is 7 (certainly $p_{13} \neq 7$ ), that satisfy
$$
46+p_{10}^{2}+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}
$$
We know that $n^{2} \equiv 1(\bmod 5)$ or $n^{2}=4(\bmod 5)$ when $n$ is not divisible by 5 . It is not possible that $p_{10}=p_{11}=5$, because in that case $p_{12}$ must be equal to 7 and the left-hand side would be divisible by 5 , which contradicts the fact that $p_{13} \geq 7$. So, we proved that $p_{10}=5$ or $p_{10}=7$.
If $p_{10}=5$ then $p_{11}=7$ because $p_{11}$ is the least of remaining prime numbers. Thus, we are looking for solutions of equation
$$
120=p_{13}^{2}-p_{12}^{2}
$$
in prime numbers. Now, from
$$
2^{3} \cdot 3 \cdot 5=\left(p_{12}-p_{12}\right)\left(p_{13}+p_{12}\right)
$$
that desired solutions are $p_{12}=7, p_{13}=13 ; p_{12}=13, p_{13}=17 ; p_{12}=29, p_{13}=31$.
If $p_{10}=7$ we are solving equation:
$$
95+p_{11}^{2}+p_{12}^{2}=p_{13}^{2}
$$
in prime numbers greater than 5 . But left side can give residues 0 or 3 modulo 5 , while right side can give only 1 or 4 modulo 5 . So, in this case we do not have solution.
ii. When $s=5$ we get equation:
$$
28+45=73=p_{13}^{2}
$$
but 73 is not square or integer and we do not have solution in this case.
Finally, only solutions are:
$\{(2,2,2,2,2,2,2,3,3,5,7,7,13),(2,2,2,2,2,2,2,3,3,5,7,13,17),(2,2,2,2,2,2,2,3,3,5,7,29,31)\}$. NT5 Show that there are infinitely many positive integers $c$, such that the following equations both have solutions in positive integers:
$$
\left(x^{2}-c\right)\left(y^{2}-c\right)=z^{2}-c
$$
and
$$
\left(x^{2}+c\right)\left(y^{2}-c\right)=z^{2}-c
$$
| (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 mentioned projections is equal to $10 / \pi>3$. Because the side of the square is equal to 1 , we conclude that at least one point is covered with at least four of these projections. Hence, a perpendicular line to the projection side passing through this point meets at least four of the given circles, so this is a line with the desired property.
| 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 10 on each of the two sides, the lines which are closest to the vertex $A$, distributed symmetrically. In this way we get $26 \cdot 21+10=$ 556 new triangles. Therefore we obtain a total of 2000 triangles and we have used $37 \cdot 3-11+21=121$ lines. Let $D$ be the $12^{\text {th }}$ point on side $A B$, starting from $B$ (including it). The perpendicular to $B C$ passing through $D$ will be the last line we draw. In this way we obtain the required configuration.
| 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 one (for example, from 123 we get 321 );
c) We can multiply the number $n$ by a number from the set $\{1,2,3, \ldots, 2010\}$.
Can we get the number 21062011 from the number 1012011?
|
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 by 11 if and only if the difference between sum of digits on even places and sum of digits on odd places is divisible by 11 . Hence, when we use $b$ ), from a number which is divisible by 11, we will get a number which is also divisible by 11 . When we use $c$ ), the obtained number remains divisible by 11 . So, the answer is NO since 1012011 is divisible by 11 and 21062011 is not.
| 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: $a$ ) $n=5$; b) $n=6$.
| ## 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 fact, 10 swaps are sufficient:
Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e$.
Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d$.
Swap $B E$, then $C D$; the state is now $A a, B d, C e, D b, E c$.
Swap $B D$, then $C E$; the state is now $A a, B b, C c, D d, E e$.
All requirements are fulfilled now, so the answer is 10 .
b) Six people form 15 pairs, so at least 15 swaps are necessary. We will prove that the final number of swaps must be even. Call a pair formed by a ball and a person inverted if letter of the ball lies after letter of the person in the alphabet. Let $T$ be the number of inverted pairs; at the start we have $T=0$. Each swap changes $T$ by 1 , so it changes the parity of $T$. Since in the end $T=0$, the total number of swaps must be even. Hence, at least 16 swaps are necessary. In fact 16 swaps are sufficient:
Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e, F f$. Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d, F f$. Swap $F B$, then $B E$, then $E F$; the state is now $A a, B d, C b, D e, E c, F f$. Swap $F C$, then $C D$, then $D F$; the state is now $A a, B d, C e, D b, E c, F f$. Swap $B D$, then $C E$, then twice $A F$, the state is now $A a, B b, C c, D d, E e, F f$. All requirements are fulfilled now, so the answer is 16 .
| 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 of all numbers in the set $B$. If $B$ has exactly 62 elements, than $B=\{2,3,4, \ldots, 62\}$, but this set can't be good since the sum of its elements is 2015 which is divisible by 5 . Therefore $B$ has at most 61 elements. Now we are looking for the set, whose elements does not divide their sum, so the best way to do that is making a sum of elements be a prime number. $2+3+4+\ldots+63=2015$ and if we remove the number 4, we will obtain the prime number 2011. Hence the set $B=\{2,3,5,6,7, \ldots, 63\}$ is a good one. We conclude that our number is 61 .
| 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 rectangle. Determine the dimensions of all rectangles with this property.
| ## 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 $A B$ and $l$ of them along the side $A D$.
The sum of areas of all of this squares is equal to $3 k l$. Besides of the obvious conditions $k \sqrt{3}m$ and $(l+1) \sqrt{3}>n$ (2).
The proposed problem is to determine all pairs $(m, n) \in \mathbb{N} \times \mathbb{N}$ with $m \geq n \geq 2$, for which the ratio $R_{m, n}=\frac{3 k l}{m n}$ is equal to 0,5 where $k, l$ are natural numbers determined by the conditions (1) and (2).
Observe that for $n \geq 6$, using (2), we get $R_{m, n}=\frac{k \sqrt{3} \cdot l \sqrt{3}}{m n}>\frac{(m-\sqrt{3})(n-\sqrt{3})}{m n}=$ $\left(1-\frac{\sqrt{3}}{m}\right)\left(1-\frac{\sqrt{3}}{n}\right) \geq\left(1-\frac{\sqrt{3}}{6}\right)^{2}=\frac{1}{2}+\frac{7}{12}-\frac{\sqrt{3}}{3}>\frac{1}{2}+\frac{\sqrt{48}}{12}-\frac{\sqrt{3}}{3}=0,5$
So, the condition $R_{m, n}=0,5$ yields $n \leq 5$ or $n \in\{2,3,4,5\}$. We have 4 possible cases:
Case 1: $n=2$. Then $l=1$ and thus as above we get $R_{m, 2}=\frac{3 k}{2 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{2 m}=$ $\frac{\sqrt{3}}{2} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0,5 for each $m>\frac{\sqrt{27}+3}{2}>\frac{5+3}{2}=4$, hence $m \in\{2,3,4\}$. Direct calculations give $R_{2,2}=R_{2,4}=0,75$ and $R_{2,3}=0,5$.
Case 2: $n=3$. Then $l=1$ and thus as above we get $R_{m, 3}=\frac{3 k}{3 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{3 m}=$ $\frac{\sqrt{3}}{3} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0,5 for each $m>4 \sqrt{3}+6>12$, hence $m \in$ $\{3,4, \ldots, 12\}$. Direct calculations give $R_{3,3}=0,(3), R_{3,5}=0,4, R_{3,7}=4 / 7, R_{3,9}=$ $5 / 9, R_{3,11}=6 / 11$ and $R_{3,4}=R_{3,6}=R_{3,8}=R_{3,10}=R_{3,12}=0,5$.
Case 3: $n=4$. Then $l=2$ and thus as above we get $R_{m, 4}=\frac{6 k}{4 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{2 m}=$ $\frac{\sqrt{3}}{2} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0,5 for each $m>\frac{\sqrt{27}+3}{2}>\frac{5+3}{2}=4$.
Hence $m=4$ and a calculation gives $R_{4,4}=0,75$.
Case 4: $n=5$. Then $l=2$ and thus as above we get $R_{m, 5}=\frac{6 k}{5 m}>\frac{2 \sqrt{3} \cdot(m-\sqrt{3})}{5 m}=$
$\frac{2 \sqrt{3}}{5} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0,5 for each $m>\frac{12(4 \sqrt{3}+5}{23}>\frac{12 \cdot 11}{23}>6$, hence $m \in\{5,6\}$. Direct calculations give $R_{5,5}=0,48$ and $R_{5,6}=0,6$.
We conclude that: $R_{i, j}=0,5$ for $(i, j) \in\{(2,3) ;(3,4) ;(, 3,6) ;(3,8) ;(3,10) ;(3,12)\}$.
These pairs are the dimensions of all rectangles with desired property.
| (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. The reflection is considered with respect to the support line of the 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 vertices falls outside the polygon.
Now we choose a diagonal. It divides the polygon into two parts, $P 1$ and $P 2$. The reflection of $P 1$ falls into the interior of $P 2$ and viceversa. As a consequence, the diagonal is a symmetry axis for the polygon. Then every diagonal of the polygon bisects the angles of the polygon and this means that there are 4 vertices and the polygon is a rhombus. Each rhombus satisfies the desired condition.
| 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 $m=2 n+1, n \in \mathbb{N}$, then from (1) we get
$$
k^{2}=41-2 n(n+1)
$$
Thus $41-2 n(n+1) \geq 0$ or $2 n^{2}+2 n-41 \leq 0$. The last inequality is satisfied for the positive integers $n=1,2,3,4$ and for $n=0$.
However, only for $n=4$, equation (2) gives a perfect square $k^{2}=1 \Leftrightarrow k= \pm 1$. Therefore the solutions are $(x, y)=(11,7)$ or $(x, y)=(7,11)$.
| (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 $p$ denotes the semi perimeter of the triangle.
On the other hand,
$$
p=(p-a)+(p-b)+(p-c) \geq 3 \sqrt[3]{(p-a)(p-b)(p-c)}
$$
Hence
$$
\begin{aligned}
p^{3} \geq 27(p-a)(p-b)(p-c) & \Leftrightarrow p^{4} \geq 27 p(p-a)(p-b)(p-c) \\
& \Leftrightarrow p^{2} \geq 3 \sqrt{3} \cdot S
\end{aligned}
$$
where $S$ is the area of the triangle.
Thus $S \leq \frac{\sqrt{3}}{4}$ and equality holds whenever when $a=b=c=1$.
## Comment
Cauchy's inequality implies the following two inequalities are true:
$$
\frac{a+b+c}{3} \leq \frac{a^{2}+b^{2}+c^{2}}{a+b+c} \leq \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}
$$
Now note that
$$
\frac{a+b+c}{3} \leq \frac{a^{2}+b^{2}+c^{2}}{a+b+c}
$$
gives
$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$
whereas $\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \leq \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}$, because of our assumptions, becomes $\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \leq 1$, and so,
$$
a^{2}+b^{2}+c^{2} \leq a+b+c
$$
Combining (1) and (2) we get
$(a+b+c)^{2} \leq 3(a+b+c)$ and then $a+b+c \leq 3$.
| 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 also divide
$$
9\left(2 y^{2}+7 y+6\right)=18 y^{2}+63 y+54=2(3 y+1)^{2}+17(3 y+1)+35
$$
from which it follows that it must divide 35 as well. Since $3 y+1 \in \mathbb{Z}$ we conclude that $y \in\{0,-2,2,-12\}$ and it is easy now to get all the solutions $(-2,0),(-3,0),(0,-2),(-1,2)$.
| (-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 impossible. Hence $3 /(x+y)$ and 3 does not divide $x$ or $y$. Without loss of generality we can assume that $x=3 a+1$ and $y=3 b+2$. Substituting in the equation, we obtain
$$
\left(x^{2}+y^{2}\right)(8(a+b)+3)=5(x y+1)
$$
Since $x y+1 \equiv 0(\bmod 3)$, we conclude that $3 /(a+b)$.
Now we distinguish the following cases:
- If $a+b=0$, then $x=3 a+1$ and $y=-3 a+2$ from which we get
$$
\left(9 a^{2}+6 a+1+9 a^{2}-12 a+4\right) \cdot 3=5\left(-9 a^{2}+3 a+3\right) \text { or } 3 a^{2}-a=0
$$
But $a=\frac{1}{3}$ is not an integer, so $a=0$ and $x=1, y=2$. Thus, by symmetry, we have two solutions $(x, y)=(1,2)$ and $(x, y)=(2,1)$.
- If $a+b \neq 0$, then $|8(a+b)+3| \geq 21$. So we obtain
$$
\left|\left(x^{2}+y^{2}\right)(8(a+b)+3)\right| \geq 21 x^{2}+21 y^{2} \geq|5 x y+5|
$$
which means that the equation has no other solutions.
## Geometry
| (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 1
$$
\angle N A M=\angle B D C
$$
Now we have
$$
\angle B D C=\angle A C D=\angle N A C
$$
and
$$
\angle N A M=\angle N A C
$$
So the points $A, M, C$ are collinear and $M \equiv E$.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 283 |
G2. Let $A B C$ be a triangle inscribed in a circle $K$. The tangent from $A$ to the circle meets the line $B C$ at point $P$. Let $M$ be the midpoint of the line segment $A P$ and let $R$ be the intersection point of the circle $K$ with the line $B M$. The line $P R$ meets again the circle $K$ at the point $S$. Prove that the lines $A P$ and $C S$ are parallel.
| ## Solution
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 implies that the triangles $M R$ and $M P B$ are similar. Hence $\angle M P R=\angle M B P$ and since $\angle P S C=\angle M B P$, the claim is proved.
Slight changes are to be made if the point $B$ lies on the line segment $P C$.
Figure 3
| 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
$$
\frac{A M}{C P}=\frac{M N}{P M}=\frac{A N}{C M}=\sqrt{2}
$$
So, we have
$$
\frac{M P}{M N}=\frac{1}{\sqrt{2}} \Rightarrow \angle M N P=45^{\circ}
$$
Figure 4
Hence we have
$$
\angle C A M=\angle M N P=45^{\circ}
$$
and finally, we obtain
$$
\angle B A M=\angle B A C+\angle C A M=75^{\circ}
$$
| 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 $\mathrm{M}$ to the altitude $\mathrm{BF}$ and $\mathrm{K}$ is the foot of the perpendicular from $\mathrm{Z}$ to $\mathrm{AE}$. Prove that $\angle K D Z=\angle K B D=\angle K Z B$.
| ## 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
$$
\begin{aligned}
& \angle A B F=\angle A B C-\angle F B C \\
& =\angle A K C-\angle E K M=\angle M K C
\end{aligned}
$$
Also, we have
$$
\begin{aligned}
& \angle A B F=90^{\circ}-\angle B A F=90^{\circ}-\angle M B D \\
& =\angle B D M=\angle M D C
\end{aligned}
$$
From (1) and (2) we get $\angle M K C=\angle M D C$ and so the points $M, K, D$ and $C$ are co-cyclic.
Consequently,
$$
\angle K D M=\angle K C M=\angle B A K=\angle B Z K \text {, }
$$
and because the line $\mathrm{BD}$ is tangent to the circumcircle of triangle $A B C$, we have
$$
\angle K B D=\angle B A K
$$
Figure 5
Finally, we have
$$
\angle K D Z=\angle K B D=\angle K Z B
$$
| 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 $E$, respectively. Prove that the centre of the circumcircle of the triangle $D E P$ lies on the circumcircle $O K P$.
| ## 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 $\angle A K P$ we get
Figure 6
$$
\angle A C P=O K P
$$
Also, since $A P \perp O K$ and $M K \perp P E$ we have that
$$
\angle A P E=\angle M K O
$$
The points $A, E, C, P$ are co-cyclic, and so $\angle E C A=\angle A P E$.
Therefore, from (1), (2) and (3) we have that $\angle N O P=\angle M K P$.
Thus $O, M, K$ and $P$ are co-cyclic.
## Comment
Points B and C may not be included in the statement of the problem
| 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 $\angle P B C=\angle O B A$. Hence the triangles $A O B$ and $C P B$ are equal and $P C=O A$. From the triangle $O P C$ we have
$$
O C \leq O P+P C=O B+O A=8
$$
Hence, the maximum yalue of the distance $O C$ is 8 (when the point $P$ lies on $O C$ )
Figure 8
Figure 9
## Complement to the solution
Indeed there exists a triangle $\mathrm{OAB}$ with $\mathrm{OA}=3, \mathrm{OB}=5$ and $\mathrm{OC}=8$.
To construct such a triangle, let's first consider a point $M$ on the minor arc $\widehat{A_{0} B_{0}}$ of the circumference $\left(c_{0}\right)$ of an arbitrary equilateral triangle $A_{0} B_{0} C_{0}$. As $\mathrm{M}$ moves along $\widehat{\mathrm{A}_{0} \mathrm{~B}_{0}}$ from the midpoint position $\mathrm{M}_{0}$ towards $\mathrm{A}_{0}$, the ratio $\frac{\mathrm{MA}_{0}}{\mathrm{MB}_{0}}$ takes on all the decreasing values from 1 to 0 . Thus there exists a position of $\mathrm{M}$ such that $\frac{\mathrm{MA}_{0}}{\mathrm{MB}_{0}}=\frac{3}{5}$. Now a homothesy centered at the center of $\left(\mathrm{c}_{0}\right)$ can take $\mathrm{A}_{0}, \mathrm{~B}_{0}, \mathrm{C}_{0}, \mathrm{M}$ to the new positions $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{O}$ so that $\mathrm{OA}=3$ and $\mathrm{OB}=5$. Then, since $\mathrm{C}$ lies on the minor arc $\overparen{\mathrm{AB}}$ of the circumference (c) of the equilateral triangle $\mathrm{ABC}$ we get $\mathrm{OC}=\mathrm{OA}+\mathrm{OB}=3+5=8$ as wanted, (figure 9).
| 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 $\left.P^{\prime}=\sigma(P), Q^{\prime}=\sigma(Q), P_{1}^{\prime}=\sigma\left(P_{1}\right)\right)$ and, (figure 1).
Let $M_{1}=A Q_{1} \cap D P_{1}, N_{1}=B Q_{1} \cap C P_{1}, N^{\prime}=A Q^{\prime} \cap D P^{\prime}$ and $M^{\prime}=B Q^{\prime} \cap C P^{\prime}$.
Then: $M^{\prime}=\sigma(M), N^{\prime}=\sigma(N), M_{1}^{\prime}=\sigma\left(M_{1}\right)$ and $N_{1}^{\prime}=\sigma\left(N_{1}\right)$.
Since $A P$ and $D P_{1}$ are the diagonals of the parallelogram $A P_{1} P D, C P_{1}$ and $B P$ are the diagonals of the parallelogram $P_{1} B C P$, and $A Q_{1}$ and $D O$ are the diagonals of the parallelogram $A Q Q_{1} D$, it follows that the points $U, V, W$ (figure 2) are collinear and they lie on the line passing through the midpoints $R$ of $A D$ and $Z$ of $B C$. The diagonals AM and $D M_{1}$ the quadrilateral $A M_{1} M D$ intersect at $U$ and the diagonals $A M_{1}$ and - $D M$ intersect at $W$. Since the midpoint of $A D$ is on the line $U W$, it follows that the quadrilateral $A M_{1} M D$ is a trapezoid. Hence, $M M_{1}$ is parallel to $A D$ and the midpoint $S$ of $M M_{1}$ lies on the line $U W$, (figure 2).
Figure 11
Similarly $M^{\prime} M_{1}^{\prime}$ is parallel to $A D$ and its midpoint lies on $U W$. So $M_{1} M^{\prime} M_{1}^{\prime} M$ is a parallelogram whose diagonals intersect at $\mathrm{O}$.
Similarly, $N_{1}^{\prime} N N_{1} N^{\prime}$ is a parallelogram whose diagonals intersect at $O$.
All these imply that $M, N, M^{\prime}, N^{\prime}$ and $O$ are collinear, i.e. $O$ lies on the line $K L$. This implies that $K=\sigma(L)$, and since $D=\sigma(B)$, the conclusion follows.
Figure 12
| 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 $10 n+2$, so, $n$ is odd. We check that the only solution in this case is $\mathrm{m}=11$ and $\mathrm{n}=7$.
If $t>1$, then 4 is divisor of $\mathrm{m}^{2}-\mathrm{n}^{2}$, but 4 is not divisor of $10^{t}+2$.
Hence the only solution is $\mathrm{m}=11$ and $\mathrm{n}=7$.
| =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
$$
\begin{aligned}
x^{2} & \equiv 5^{2 k+1}+12^{2 k+1}(\bmod 5) \equiv 2^{2 k} \cdot 2(\bmod 5) \\
& \equiv(-1)^{k} \cdot 2(\bmod 5) \equiv \pm 2(\bmod 5)
\end{aligned}
$$
This is not possible, because the square residue of any natural number module 5 is 0,1 or 4. Therefore $n$ is even and $x^{2}=5^{2 k}+12^{2 k}$. Rearrange this equation in the form
$$
5^{2 k}=\left(x-12^{k}\right)\left(x+12^{k}\right)
$$
If 5 divides both factors on the right, it must also divide their difference, that is
$$
5 \mid\left(x+12^{k}\right)-\left(x-12^{k}\right)=2 \cdot 12^{k}
$$
which is not possible. Therefore we must have
$$
x-12^{k}=1 \text { and } x+12^{k}=5^{2 k}
$$
By adding the above equalities we get
$$
5^{2 k}-1=2 \cdot 12^{k}
$$
For $k \geq 2$, we have the inequality
$$
25^{k}-1>24^{k}=2^{k} \cdot 12^{k}>2 \cdot 12^{k}
$$
Thus we conclude that there exists a unique solution to our problem, namely $n=2$.
| 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 $\mathrm{k}=1,2, \ldots, \mathrm{p}-1$, let $m=\frac{(p-1)!}{k} \in \mathbb{N}$ and observe that $p!=k m p$. Consider $a \in \mathbb{N}$ so that $p^{a} \mid 2^{k}-1$ and observe that it suffices to prove $p^{a+1} \mid 2^{p!}-1$. The case $a=0$ is solved as the case $k=p$. If else, write $2^{k}=1+p^{a} \cdot l, l \in \mathbb{N}$ and rising at the power mp gives
$$
2^{p!}=\left(1+p^{a} \cdot l\right)^{m p}=1+m p \cdot p^{a} \cdot l+M p^{2 a}
$$
where $M n$ stands for a multiply of $\mathrm{n}$. Now it is clear that $p^{a+1} \mid 2^{p!}-1$, as claimed.
Comment. The case $\mathrm{k}=\mathrm{p}$ can be included in the case $a=0$.
| 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
$$
a+b+c \equiv 0(\bmod 9) ; \text { or } a b c-1 \equiv 0(\bmod 9)
$$
- If . $a b c-1 \equiv 0(\bmod 9)$
we have $11 a+b=(a+b+c) k$, where $k$ is an integer
and is easy to see that we must have $19$.
Now we will deal with the case when $a+b+c \equiv 0(\bmod 9)$ or $a+b+c=9 l$, where $l$ is an integer.
- If $l \geq 2$ we have $a+b+c \geq 18, \max \{a, b, c\} \geq 6$ and it is easy to see that $a b c \geq 72$ and $a b c(a+b+c)>1000$,so the case $l \geq 2$ is impossible.
- If $l=1$ we have
$$
11 a+b=a b c-1 \text { or } 11 a+b+1=a b c \leq\left(\frac{a+b+c}{3}\right)^{3}=27
$$
So we have only two cases: $a=1$ or $a=2$.
- If $a=1$, we have $b+c=8$ and $11+b=b c-1$ or $b+(c-1)=7$ and $b(c-1)=12$ and the solutions are $(a, b, c)=(1,3,5)$ and $(a, b, c)=(1,4,4)$, and the answer is 135 and 144.
- If $a=2$ we have $b(2 c-1)=23$ and there is no solution for the problem.
| 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 equal to 1 .
We prove that if $x, y=0,1,2, \ldots, p-1$ and $\mathrm{p}$ divides $x^{5}-y^{5}$, then $x=y$.
Indeed, assume that $x \neq y$. If $x=0$, then $p \mid y^{5}$ and so $y=0$, a contradiction.
To this point we have $x, y \neq 0$. Since
$$
p \mid(x-y)\left(x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4}\right) \text { and } p /(x-y)
$$
we have
$$
\begin{aligned}
& p l\left(x^{2}+y^{2}\right)^{2}+x y\left(x^{2}+y^{2}\right)-x^{2} y^{2} \text {, and so } \\
& p \|\left(2\left(x^{2}+y^{2}\right)+x y\right)^{2}-5 x^{2} y^{2}
\end{aligned}
$$
As $p / x y$, from the lemma we find an integer $s$ so that $s x y=k p+1, k \in \mathbb{N}$. Then
$$
p \mid\left[s\left(2 x^{2}+2 y^{2}+x y\right)\right]^{2}-5\left(k^{2} p^{2}+2 k p+1\right)
$$
and so $p \mid z^{2}-5$, where $z=s\left(2 x^{2}+2 y^{2}+x y\right)$, a contradiction.
Consequeatly $r=y$.
Since we have proved that numbers $0^{5}, 1^{5}, \ldots,(p-1)^{5}$ have distinct remainders when divided by $p$, the same goes for the numbers $0^{5}+a, 1^{5}+a, \ldots,(p-1)^{5}+a$ and the conclusion can be reached easily.
## Comments
1. For beauty we may choose $a=-2$ or any other value.
2. Moreover, we may ask only for one value of $m$, instead of "infinitely many".
3. A simple version will be to ask for a proof that the numbers $0^{5}, 1^{5}, \ldots,(p-1)^{5}$ have distinct remainders when divided by $p$.
## Combinatorics
| 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 maximum number of elements in $\mathrm{S}$ ?
| ## 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 of the corresponding exponents need to be even from where we can conclude that for every $i, a_{i}$ and $\beta_{i}$ have the same parity. If we replace all exponents of $\mathrm{a}$ and $b$ by their remainders modulo 2 , then we get two numbers $a^{\prime}, b^{\prime}$ whose product is a perfect square if and only if ab is a perfect square.
In order for the product $a^{\prime} b^{\prime}$ not to be a perfect square, at least one pair of the corresponding exponents modulo 2, need to be of opposite parity.
Since we form 2005 such pairs modulo 2, and each number in these pairs is 1 or 2 , we conclude that we can obtain $2^{2005}$ distinct products none of which is a perfect square.
Now if we are given $2^{2005}+1$ numbers, thanks to Dirichlet's principle, there are at least two with the same sequence of modulo 2 exponents, thus giving a product equal to a square.
So, the maximal number of the elements of $S$ is $2^{2005}$.
| 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 prove that $\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8$.
By $A M-G M$ we have $x^{3}+1 \geq 2 \sqrt{x^{3}}$ for all $x \in \mathbb{R}_{+}$.
So $\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 2^{3} \cdot \sqrt{a^{3} b^{3} c^{3}}=8$ and we are done.
Equality holds when $a=b=c=1$.
| 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-Schwarz we obtain $\sum_{\text {cyc }} \frac{1}{z+2 x+3 y} \geq \frac{(1+1+1)^{2}}{\sum_{\text {cyc }}(z+2 x+3 y)}=\frac{3}{2(x+y+z)}$.
| 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 $x=y=1$.
So the greatest possible value of the product $x y$ is 1 .
| 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 $t \in \mathbb{Z}_{+}$.
Case I. If $d=1$, then $(x y)^{2}+208=4(x y+1)^{2}$ or $3 t^{2}+8 t-204=0$, without solutions.
Case II. If $d=2$, then $16 x^{2} y^{2}+208=16(x y+1)^{2}$ or $t^{2}+13=t^{2}+2 t+1 \Rightarrow t=6$, so $(x, y) \in\{(1,6) ;(2,3) ;(3,2) ;(6,1)\} \Rightarrow(a, b) \in\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\}$.
Case III. If $d=4$, then $16^{2} x^{2} y^{2}+208=4 \cdot 16(x y+1)^{2}$ or $16 t^{2}+13=4(t+1)^{2}$ and if $t \in \mathbb{Z}$, then 13 must be even, contradiction!
Finally, the solutions are $(a, b) \in\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\}$.
| (,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:
$\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. By $A M-G M$ :
$\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1} \geq 2011 \cdot \sqrt[2011]{\frac{x_{1}-1}{x_{2}-1} \cdot \frac{x_{2}-1}{x_{3}-1} \cdot \ldots \cdot \frac{x_{2011}-1}{x_{1}-1}}=2011$
Finally, we obtain that $\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 8044$.
Equality holds when $\left(x_{i}-2\right)^{2}=0,(\forall) i=\overline{1,2011}$, or $x_{1}=x_{2}=\ldots=x_{2011}=2$.
| 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:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a}$.
By $A M-G M$ we have $a b+b c+c a \geq 3$ and $a+b+c \geq 3$. But $3\left(a^{2}+b^{2}+c^{2}\right) \geq(a+b+c)^{2} \geq$ $3(a+b+c)$. So $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 3+a+b+c+a b+b c+c a$. Hence $\sum_{c y c} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a} \geq \frac{3(a+b+c)^{2}}{(a+b+c)^{2}}=3$.
| 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 $\frac{925}{1839} \leq \frac{x}{y} \leq \frac{9075}{381}$, or $0,502 \ldots \leq \frac{x}{y} \leq 23,81 \ldots$ Since $\frac{x}{y}=z$ is an integer, it follows that $1 \leq \frac{x}{y} \leq 23$, hence $1 \leq C^{Y^{P}} \cdot R^{U^{S}} \leq 23$. So both values $C^{Y^{P}}$ and $R^{U^{S}}$ are $\leq 23$. From this and the fact that $2^{2^{3}}>23$ it follows that at least one of the symbols in the expression $C^{Y^{P}}$ and at least one of the symbols in the expression $R^{U^{S}}$ correspond to the digit 1. This is impossible because of the assumption that all the symbols in the set $\{C, Y, P, R, U, S\}$ correspond to different digits.
| 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 $\sum_{k=2}^{n-1} x_{k} \geq 0$.
Case II. If $\min \left(x_{1}, x_{n}\right)=x_{n}$, we know that $x_{k} \geq \min \left(x_{k-1} ; x_{k}\right)$ for all $k \in\{2,3,4, \ldots, n\}$. So $x_{2}+x_{3}+\ldots+x_{n} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{n}$, hence $\sum_{k=2}^{n-1} x_{k} \geq 0$.
| 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 $(1) \Rightarrow x^{2}+x z+z^{2}=0 \Rightarrow(x+z)^{2}=x z \Rightarrow x z=0$
So $x=z=0$, contradiction since $x \neq z$ and therefore $(1) \Rightarrow-k=\frac{x^{2}+x z+z^{2}}{x+z}$
Similarly we have: $-k=\frac{y^{2}+y x+x^{2}}{y+x}$.
So $\frac{x^{2}+x z+z^{2}}{x+z}=\frac{y^{2}+x y+x^{2}}{x+y}$ from which $x y+y z+z x=0:(* *)$.
We substitute $k$ in $x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=2008$ from the relation $(*)$ and using the $(* *)$, we finally obtain that $2 x y z=2008$ and therefore $x y z=1004$.
Remark: $x, y, z$ must be the distinct real solutions of the equation $t^{3}+k t^{2}-1004=0$. Such solutions exist if (and only if) $k>3 \sqrt[3]{251}$.
| 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 satisfy the inequalities is a segment with endpoints $(1,1)$ and $(2,0)$. Now taking into account the invariance under the mentioned replacements, we conclude that the set of points satisfying the second equation is the square $\diamond$ with vertices $(1,1),(2,0),(1,-1)$ and $(0,0)$.
The first equation is equivalent to
$p x^{2}-p^{2} x y+x y-p y^{2}=0$
$p x(x-p y)+y(x-p y)=0$
$(p x+y)(x-p y)=0$.
Thus $y=-p x$ or $x=p y$. These are equations of two perpendicular lines passing through the origin, which is also a vertex of $\diamond$. If one of them passes through an interior point of the square, the other cannot have any common points with $\diamond$ other than $(0,0)$, so the system has two solutions. Since we have at least three different real solutions, the lines must contain some sides of $\diamond$, i.e. the slopes of the lines have to be 1 and -1 . This happens if $p=1$ or $p=-1$. In either case $x^{2}=y^{2},|x|=|y|$, so the second equation becomes $|1-x|+|x|=1$. It is true exactly when $0 \leq x \leq 1$ and $y= \pm x$.
| 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 $2 y^{2}=6024^{2}-2008^{2}=2008^{2}(9-1)=$ $2 \cdot 4016^{2}$. Thus $(x, y, z)$ is the triple $(2008,4016,-4016)$ or any of its rearrangements.
| (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
$$
\frac{4 x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9 x}{z} \geq 6, \quad \frac{4 z}{y}+\frac{9 y}{z} \geq 12
$$
Therefore
$$
22 \leq\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right)
$$
Now from (1) and (3) we get
$$
\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right)=22
$$
which means that in (2), everywhere equality holds i.e. we have equality between means, also $x+y+z=12$.
Therefore $\frac{4 x}{y}=\frac{y}{x}, \frac{z}{x}=\frac{9 x}{z}$ and, as $x>0, y>0, z>0$, we get $y=2 x, z=3 x$. Finally if we substitute for $y$ and $z$, in $x+y+z=12$, we get $x=2$, therefore $y=2 \cdot 2=4$ and $z=3 \cdot 2=6$.
Thus the unique solution is $(x, y, z)=(2,4,6)$.
| (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
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\sqrt{\frac{1}{c a}} \cdot \sqrt{c a}\right)=(2+b c)=(2 a b c+b c)=b c(1+2 a)
\end{gathered}
$$
Analogously we get $\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right)} \geq c a(1+2 b)$ and
$\sqrt{\left(c a+a b+\frac{1}{b c}\right)\left(a b+b c+\frac{1}{c a}\right)} \geq a b(1+2 a)$.
Multiplying these three inequalities we get:
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq a^{2} b^{2} c^{2}(1+2 a)(1+2 b)(1+2 c)=
$$
$(1+2 a)(1+2 b)(1+2 c)$ because $a b c=1$.
Equality holds if and only if $a=b=c=1$.
| 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
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
On the other hand, as
$$
\sqrt{\frac{x}{y}} \geq \frac{2 x}{x y+1} \Leftrightarrow(\sqrt{x y}-1)^{2} \geq 0
$$
by summation one has
$$
\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}} \geq \frac{2 x}{x y+1}+\frac{2 y}{y z+1}+\frac{2 z}{z x+1}
$$
The rest is obvious.
| 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 sequence 9,1 . Upon the sequence $1,2,3, \ldots, 1024$ the operation is performed successively for 5 times. Show that at the end only 1 number remains and find this number.
| ## 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 numbers left after the second operation are in arithmetical progression of ratio 16 and so on.
Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the first term in the 5 sequences obtained after each of the 5 operations. Thus $a_{1}=3$ and $a_{5}$ is the requested number. The sequence before the fifth operation has 4 numbers, namely
$$
a_{4}, a_{4}+256, a_{4}+512, a_{4}+768
$$
and $a_{5}=a_{4}+512$. Similarly, $a_{4}=a_{3}+128, a_{3}=a_{2}+32, a_{2}=a_{1}+8$.
Summing up yields $a_{5}=a_{1}+8+32+128+512=3+680=683$.
### 2.2 Combinatorics
| 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 empty square such that none of the neighboring squares contains a white marker (two squares are called neighboring if they contain a common side). If it is possible for the child to succeed in coloring all the markers black, we say that the initial positioning of the markers was good.
a) Prove that if $n=20$, then a good initial positioning exists.
b) Prove that if $n=21$, then a good initial positioning does not exist.
|
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 position $(i, j-1)$.
b) Suppose there exists a good positioning with 21 white markers on the board i.e. there exists a re-coloring of them all, one by one. In any moment when there are 21 markers on the board, there must be at least one column completely filled with markers, and there must be at least one row completely filled with markers. So, there exists a "cross" of markers on the board. At the initial position, each such cross is completely white, at the final position each such cross is completely black, and at every moment when there are 21 markers on the board, each such cross is monochromatic. But this cannot be, since every two crosses have at least two common squares and therefore it is not possible for a white cross to vanish and for a black cross to appear by re-coloring of only one marker. Contradiction!
| 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 conditions, but I did not think of these numbers. The numbers you mentioned have the minimal sum between all possible solutions of the problem.
Can you decide if Kostas is right? (Explain your answer).
| ## 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
$$
\begin{gathered}
x y z-(x y+y z+z x)=1-(x+y+z) \\
\Leftrightarrow x y z-x y-y z-z x+x+y+z-1=0 \\
\Leftrightarrow x y(z-1)-x(z-1)-y(z-1)+(z-1)=0 \\
\Leftrightarrow(z-1)(x y-x-y+1)=0 \\
(z-1)(x-1)(y-1)=0 \\
\Leftrightarrow x=1 \text { or } y=1 \text { or } z=1 .
\end{gathered}
$$
For $x=1$, from (1) and (2) we have the equation $y z=1$, which has the solutions
$$
(y, z)=\left(a, \frac{1}{a}\right), a>0
$$
And therefore the solutions of the problem are the triples
$$
(x, y, z)=\left(1, a, \frac{1}{a}\right), a>0
$$
Similarly, considering $y=1$ or $z=1$ we get the solutions
$$
(x, y, z)=\left(a, 1, \frac{1}{a}\right) \text { or }(x, y, z)=\left(a, \frac{1}{a}, 1\right), a>0
$$
Since for each $a>0$ we have
$$
x+y+z=1+a+\frac{1}{a} \geq 1+2=3
$$
and equality is valid only for $a=1$, we conclude that among the solutions of the problem, the triple $(x, y, z)=(1,1,1)$ is the one whose sum $x+y+z$ is minimal.
| 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
$$
S \equiv 2(1+\ldots+2 n)=2 \cdot \frac{2 n(2 n+1)}{2}=2 n(2 n+1) \equiv 0 \quad(\bmod 2 n)
$$
From the last two conclusions we derive $n \equiv 0(\bmod 2 n)$. Contradiction.
Therefore, there are two sums with the same remainder modulo $2 n$.
Remark: The result is no longer true if one replaces $2 n$ by $2 n+1$. Indeed, one could assign the number $k$ to the box labeled $k$, thus obtaining the sums $2 k, k=\overline{1,2 n+1}$. Two such numbers give different remainders when divided by $2 n+1$.
| 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 $D N, M C$ and $T$ the point of intersection of $D C, K L$, prove that $\angle K T C=\angle K N L$.
| ## 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
$$
\varangle D C L+\varangle C D L=\varangle D A M+\varangle C B N .
$$
And because $A D \| B C$, if $Z$ the point of intersection of $A M, B C$ then $\varangle D A M=\varangle B Z A$, and we have
$$
\varangle D C L+\varangle C D L=\varangle B Z A+\varangle C B N=90^{\circ}
$$
Let $P$ the point of intersection of $K L, A C$, then $N P \perp A C$, because the line $K P L$ is a Simson line of the point $N$ with respect to the triangle $A C M$.
From the cyclic quadrilaterals $N P C L$ and $A N D C$ we obtain:
$$
\varangle C P L=\varangle C N L \text { and } \varangle C N L=\varangle C A D \text {, }
$$
so $\varangle C P L=\varangle C A D$, that is $K L\|A D\| B C$ therefore $\varangle K T C=\varangle A D C$ (1).
But $\varangle A D C=\varangle A N C=\varangle A N K+\varangle K N C=\varangle C N L+\varangle K N C$, so
$$
\varangle A D C=\varangle K N L
$$
From (1) and (2) we obtain the result.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 320 |
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