di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $A C$ is tangent to the circle $A D Q$, and the line $B D$ is tangent to the circle $B C P$. Let $M$ and $N$ be the midpoints of $B C$ and $A D$, respectively. Prove that the following three lines are concurrent: line $C D$, the tangent of circle $A N Q$ at point $A$, and the tangent to circle $B M P$ at point $B$.

Solution:
We first prove that triangles $A D Q$ and $C D B$ are similar. Since $A B C D$ is cyclic, we have $\angle D A Q=\angle D C B$. By the tangency of $A C$ to the circle $A Q D$ we also have $\angle C B D=\angle C A D=\angle A Q D$. The claimed similarity is proven.

Let $R$ be the midpoint of $C D$. Points $N$ and $R$ correspond in the proven similarity, and so $\angle Q N A=\angle B R C$.

<img_3825>

Let $K$ be the second common point of line $C D$ with circle $A B R$ (i.e., if $C D$ intersects circle $A B R$, then $K \neq R$ is the other intersection; otherwise, if $C D$ is tangent to $C D$, then $K=R$ ). In both cases, we have $\angle B A K=\angle B R C=\angle Q N A$; that indicates that $A K$ is tangent to circle $A N Q$. It can be showed analogously that $B K$ is tangent to circle $B M P$.
We present a second solution, without using the condition that $A B C D$ is cyclic. Again, $M$ and $N$ can be any points on lines $B C$ and $A D$ such that $B M: M C=D N: N A$.

Let $A B$ and $C D$ meet at $T$ (if $A B \| C D$ then $T$ is their common ideal point). Let $C D$ meet the tangent to the circle $A N Q$ at $A$, and the tangent to the circle $B M P$ at $B$ at points $K_{1}$ and $K_{2}$, respectively.

<img_3226>

Let $I$ and $J$ be the ideal points of $A D$ and $B C$, respectively. Notice that the pencils $\left(A D, A C, A T, A K_{1}\right)$ and $(Q A, Q D, Q I, Q N)$ of lines are congruent, because $\angle K_{1} A D=\angle A Q N$, $\angle C A D=\angle A Q D$ and $\angle I A T=\angle I Q T$. Hence,

$$
\left(D, C ; T, K_{1}\right)=\left(A D, A C ; A T, A K_{1}\right)=(Q A, Q D ; Q I, Q N)=(A, D ; I, N)=\frac{D N}{N A}
$$

It can be obtained analogously that

$$
\left(D, C ; T, K_{2}\right)=\left(B D, B C ; B T, B K_{2}\right)=(P C, P B ; P J, P M)=(C, B ; I, N)=\frac{B M}{M C}
$$

From $B M: M C=D N: D A$ we get $\left(D, C ; T, K_{1}\right)=\left(D, C ; T, K_{2}\right)$ and hence $K_{1}=K_{2}$.
Subfield: Geometry


Question: Let $A B C$ be an acute-angled triangle with $A C>A B$, let $O$ be its circumcentre, and let $D$ be a point on the segment $B C$. The line through $D$ perpendicular to $B C$ intersects the lines $A O, A C$ and $A B$ at $W, X$ and $Y$, respectively. The circumcircles of triangles $A X Y$ and $A B C$ intersect again at $Z \neq A$.

Prove that if $O W=O D$, then $D Z$ is tangent to the circle $A X Y$.

Solution:
Let $A O$ intersect $B C$ at $E$. As $E D W$ is a right-angled triangle and $O$ is on $W E$, the condition $O W=O D$ means $O$ is the circumcentre of this triangle. So $O D=O E$ which establishes that $D, E$ are reflections in the perpendicular bisector of $B C$.

Now observe:

$$
180^{\circ}-\angle D X Z=\angle Z X Y=\angle Z A Y=\angle Z C D \text {, }
$$

which shows $C D X Z$ is cyclic.

<img_3510>

We next show that $A Z \| B C$. To do this, introduce point $Z^{\prime}$ on circle $A B C$ such that $A Z^{\prime} \| B C$. By the previous result, it suffices to prove that $C D X Z^{\prime}$ is cyclic. Notice that triangles $B A E$ and $C Z^{\prime} D$ are reflections in the perpendicular bisector of $B C$. Using this and that $A, O, E$ are collinear:

$$
\angle D Z^{\prime} C=\angle B A E=\angle B A O=90^{\circ}-\frac{1}{2} \angle A O B=90^{\circ}-\angle C=\angle D X C,
$$

so $D X Z^{\prime} C$ is cyclic, giving $Z \equiv Z^{\prime}$ as desired.

Using $A Z \| B C$ and $C D X Z$ cyclic we get:

$$
\angle A Z D=\angle C D Z=\angle C X Z=\angle A Y Z
$$

which by the converse of alternate segment theorem shows $D Z$ is tangent to circle $A X Y$.
Notice that point $Z$ is the Miquel-point of lines $A C, B C, B A$ and $D Y$; then $B, D, Z, Y$ and $C, D, X, Y$ are concyclic. Moreover, $Z$ is the centre of the spiral similarity that maps $B C$ to $Y X$.

By $B C \perp Y X$, the angle of that similarity is $90^{\circ}$; hence the circles $A B C Z$ and $A X Y Z$ are perpendicular, therefore the radius $O Z$ in circle $A B C Z$ is tangent to circle $A X Y Z$.

<img_3338>

By $O W=O D$, the triangle $O W D$ is isosceles, and

$$
\angle Z O A=2 \angle Z B A=2 \angle Z B Y=2 \angle Z D Y=\angle O D W+\angle D W O
$$

so $D$ lies on line $Z O$ that is tangent to circle $A X Y$.
Subfield: Number Theory


Question: A number is called Norwegian if it has three distinct positive divisors whose sum is equal to 2022. Determine the smallest Norwegian number.

(Note: The total number of positive divisors of a Norwegian number is allowed to be larger than 3.)

Solution:
Observe that 1344 is a Norwegian number as 6, 672 and 1344 are three distinct divisors of 1344 and $6+672+1344=2022$. It remains to show that this is the smallest such number.

Assume for contradiction that $N<1344$ is Norwegian and let $N / a, N / b$ and $N / c$ be the three distinct divisors of $N$, with $a<b<c$. Then

$$
2022=N\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)<1344\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$

and so

$$
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>\frac{2022}{1344}=\frac{337}{224}=\frac{3}{2}+\frac{1}{224} .
$$

If $a>1$ then

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}<\frac{3}{2}
$$

so it must be the case that $a=1$. Similarly, it must hold that $b<4$ since otherwise

$$
1+\frac{1}{b}+\frac{1}{c} \leqslant 1+\frac{1}{4}+\frac{1}{5}<\frac{3}{2}
$$

This leaves two cases to check, $b=2$ and $b=3$.

Case $b=3$. Then

$$
\frac{1}{c}>\frac{3}{2}+\frac{1}{224}-1-\frac{1}{3}>\frac{1}{6}
$$

so $c=4$ or $c=5$. If $c=4$ then

$$
2022=N\left(1+\frac{1}{3}+\frac{1}{4}\right)=\frac{19}{12} N
$$

but this is impossible as $19 \nmid 2022$. If $c=5$ then

$$
2022=N\left(1+\frac{1}{3}+\frac{1}{5}\right)=\frac{23}{15} N
$$

which again is impossible, as $23 \nmid 2022$.

Case $b=2$. Note that $c<224$ since

$$
\frac{1}{c}>\frac{3}{2}+\frac{1}{224}-1-\frac{1}{2}=\frac{1}{224}
$$

It holds that

$$
2022=N\left(1+\frac{1}{2}+\frac{1}{c}\right)=\frac{3 c+2}{2 c} N \Rightarrow(3 c+2) N=4044 c
$$

Since $(c, 3 c-2)=(c, 2) \in\{1,2\}$, then $3 c+2 \mid 8088=2^{3} \cdot 3 \cdot 337$ which implies that $3 c+2 \mid 2^{3} \cdot 337$. But since $3 c+2 \geqslant 3 \cdot 3+2>8=2^{3}$ and $3 c+2 \neq 337$, then it must hold that $3 c+2 \geqslant 2 \cdot 337$, contradicting $c<224$.

Final answer: 1344
Subfield: Number Theory


Question: Find all positive integers $n>2$ such that

$$
n ! \mid \prod_{\substack{p<q \leqslant n \\ p, q \text { primes }}}(p+q)
$$

Solution:
Assume that $n$ satisfies $n ! \mid \prod_{p<q \leqslant n}(p+q)$ and let $2=p_{1}<p_{2}<\cdots<p_{m} \leqslant n$ be the primes in $\{1,2, \ldots, n\}$. Each such prime divides $n$ !. In particular, $p_{m} \mid p_{i}+p_{j}$ for some $p_{i}<p_{j} \leqslant n$. But

$$
0<\frac{p_{i}+p_{j}}{p_{m}}<\frac{p_{m}+p_{m}}{p_{m}}=2
$$

so $p_{m}=p_{i}+p_{j}$ which implies $m \geqslant 3, p_{i}=2$ and $p_{m}=2+p_{j}=2+p_{m-1}$.

Similarly, $p_{m-1} \mid p_{k}+p_{l}$ for some $p_{k}<p_{l} \leqslant n$. But

$$
0<\frac{p_{l}+p_{k}}{p_{m-1}} \leqslant \frac{p_{m}+p_{m-1}}{p_{m-1}}=\frac{2 p_{m-1}+2}{p_{m-1}}<3
$$

so either $p_{m-1}=p_{l}+p_{k}$ or $2 p_{m-1}=p_{l}+p_{k}$. As above, the former case gives $p_{m-1}=2+p_{m-2}$. If $2 p_{m-1}=p_{l}+p_{k}$, then $p_{m-1}<p_{k}$, so $k=m$ and

$$
2 p_{m-1}=p_{l}+p_{m-1}+2 \Rightarrow p_{m-1}=p_{l}+2=p_{m-2}+2
$$

Either way, $p_{m-2}>2$ and 3 divides one of $p_{m-2}, p_{m-1}=p_{m-2}+2$ and $p_{m}=p_{m-2}+4$. This implies $p_{m-2}=3$ and thus $p_{m}=7$, giving $7 \leqslant n<11$.

Finally, a quick computation shows that $7 ! \mid \prod_{p<q \leqslant 7}(p+q)$ but 8 ! $\nmid \prod_{p<q \leqslant 7}(p+q)$, so neither does 9 ! and 10 !.

Final answer: 7
Subfield: Number Theory


Question: Find all triples of positive integers $(a, b, p)$ with $p$ prime and

$$
a^{p}=b !+p
$$

Solution:
Clearly, $a>1$. We consider three cases.

Case 1: We have $a<p$. Then we either have $a \leqslant b$ which implies $a \mid a^{p}-b$ ! $=p$ leading to a contradiction, or $a>b$ which is also impossible since in this case we have $b ! \leqslant a !<a^{p}-p$, where the last inequality is true for any $p>a>1$.

Case 2: We have $a>p$. In this case $b !=a^{p}-p>p^{p}-p \geqslant p$ ! so $b>p$ which means that $a^{p}=b !+p$ is divisible by $p$. Hence, $a$ is divisible by $p$ and $b !=a^{p}-p$ is not divisible by $p^{2}$. This means that $b<2 p$. If $a<p^{2}$ then $a / p<p$ divides both $a^{p}$ and $b$ ! and hence it also divides $p=a^{p}-b$ ! which is impossible. On the other hand, the case $a \geqslant p^{2}$ is also impossible since then $a^{p} \geqslant\left(p^{2}\right)^{p}>(2 p-1) !+p \geqslant b !+p$.

Case 3: We have $a=p$. In this case $b !=p^{p}-p$. One can check that the values $p=2,3$ lead to the claimed solutions and $p=5$ does not lead to a solution. So we now assume that $p \geqslant 7$. We have $b !=p^{p}-p>p !$ and so $b \geqslant p+1$ which implies that

$$
v_{2}((p+1) !) \leqslant v_{2}(b !)=v_{2}\left(p^{p-1}-1\right) \stackrel{L T E}{=} 2 v_{2}(p-1)+v_{2}(p+1)-1=v_{2}\left(\frac{p-1}{2} \cdot(p-1) \cdot(p+1)\right)
$$

where in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of $(p+1)$ !. But, due to $p+1 \geqslant 8$, there are at least 4 even numbers among $1,2, \ldots, p+1$, so this case is not possible.
Clearly, $a>1$. We consider three cases.

Case 1: We have $a<p$. Then we either have $a \leqslant b$ which implies $a \mid a^{p}-b$ ! $=p$ leading to a contradiction, or $a>b$ which is also impossible since in this case we have $b ! \leqslant a !<a^{p}-p$, where the last inequality is true for any $p>a>1$.

Case 2: We have $a>p$. In this case $b !=a^{p}-p>p^{p}-p \geqslant p$ ! so $b>p$ which means that $a^{p}=b !+p$ is divisible by $p$. Hence, $a$ is divisible by $p$ and $b !=a^{p}-p$ is not divisible by $p^{2}$. This means that $b<2 p$. If $a<p^{2}$ then $a / p<p$ divides both $a^{p}$ and $b$ ! and hence it also divides $p=a^{p}-b$ ! which is impossible. On the other hand, the case $a \geqslant p^{2}$ is also impossible since then $a^{p} \geqslant\left(p^{2}\right)^{p}>(2 p-1) !+p \geqslant b !+p$.

Case 3: We have $a=p$. In this case $b !=p^{p}-p$. One can check that the values $p=2,3$ lead to the claimed solutions and $p=5$ does not lead to a solution. For $p \geqslant 5$ we have $b !=p\left(p^{p-1}-1\right)$. By Zsigmondy's Theorem there exists some prime $q$ that divides $p^{p-1}-1$ but does not divide $p^{k}-1$ for $k<p-1$. It follows that $\operatorname{ord}_{q}(p)=p-1$, and hence $q \equiv 1$ $\bmod (p-1)$. Note that $p \neq q$. But then we must have $q \geqslant 2 p-1$, giving

$b ! \geqslant(2 p-1) !=[1 \cdot(2 p-1)] \cdot[2 \cdot(2 p-2)] \cdots \cdots[(p-1) \cdot(p+1)] \cdot p>(2 p-1)^{p-1} p>p^{p}>p^{p}-p$, a contradiction.

Final answer: $(2,2,2),(3,4,3)$
Subfield: Number Theory


Question: For each $1 \leqslant i \leqslant 9$ and $T \in \mathbb{N}$, define $d_{i}(T)$ to be the total number of times the digit $i$ appears when all the multiples of 1829 between 1 and $T$ inclusive are written out in base 10 .

Show that there are infinitely many $T \in \mathbb{N}$ such that there are precisely two distinct values among $d_{1}(T), d_{2}(T), \ldots, d_{9}(T)$.

Solution:
Let $n:=1829$. First, we choose some $k$ such that $n \mid 10^{k}-1$. For instance, any multiple of $\varphi(n)$ would work since $n$ is coprime to 10 . We will show that either $T=10^{k}-1$ or $T=10^{k}-2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrary large.

For this it suffices to show that $\#\left\{d_{i}\left(10^{k}-1\right): 1 \leqslant i \leqslant 9\right\} \leqslant 2$. Indeed, if
$$
\#\left\{d_{i}\left(10^{k}-1\right): 1 \leqslant i \leqslant 9\right\}=1
$$
then, since $10^{k}-1$ which consists of all nines is a multiple of $n$, we have
$$
d_{i}\left(10^{k}-2\right)=d_{i}\left(10^{k}-1\right) \text { for } i \in\{1, \ldots, 8\} \text {, and } d_{9}\left(10^{k}-2\right)<d_{9}\left(10^{k}-1\right) \text {. }
$$
This means that $\#\left\{d_{i}\left(10^{k}-2\right): 1 \leqslant i \leqslant 9\right\}=2$.

To prove that $\#\left\{d_{i}\left(10^{k}-1\right)\right\} \leqslant 2$ we need an observation. Let $\overline{a_{k-1} a_{k-2} \ldots a_{0}} \in\left\{1, \ldots, 10^{k}-1\right\}$ be the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\overline{a_{k-1} a_{k-2} \ldots a_{0}}$ is divisible by $n$ if and only if $\overline{a_{k-2} \ldots a_{0} a_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that
$$
10 \cdot \overline{a_{k-1} a_{k-2} \ldots a_{0}}-\overline{a_{k-2} \ldots a_{0} a_{k-1}}=\left(10^{k}-1\right) \cdot a_{k-1}
$$
is divisible by $n$.

This observation shows that the set of multiples of $n$ between 1 and $10^{k}-1$ is invariant under simultaneous cyclic permutation of digits when numbers are written with leading zeroes. Hence, for each $i \in\{1, \ldots, 9\}$ the number $d_{i}\left(10^{k}-1\right)$ is $k$ times larger than the number of $k$ digit numbers which start from the digit $i$ and are divisible by $n$. Since the latter number is either $\left\lfloor 10^{k-1} / n\right\rfloor$ or $1+\left\lfloor 10^{k-1} / n\right\rfloor$, we conclude that $\#\left\{d_{i}\left(10^{k}-1\right)\right\} \leqslant 2$.
Subfield: Number Theory


Question: Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$ consider its prime factorisation; define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$. A positive integer $n$ is called special if $p(n)+p(n+1)$ and $q(n)+q(n+1)$ are both even integers. Prove that there is a constant $c>0$ independent of the set $Q$ such that for any positive integer $N>100$, the number of special integers in $[1, N]$ is at least $c N$.

(For example, if $Q=\{3,7\}$, then $p(42)=3, q(42)=2, p(63)=3, q(63)=3, p(2022)=3$, $q(2022)=1$.)

Solution:
Let us call two positive integers $m, n$ friends if $p(m)+p(n)$ and $q(m)+q(n)$ are both even integers. We start by noting that the pairs $(p(k), q(k))$ modulo 2 can take at most 4 different values; thus, among any five different positive integers there are two which are friends.

In addition, both functions $p$ and $q$ satisfy $f(a b)=f(a)+f(b)$ for any $a, b$. Therefore, if $m$ and $n$ are divisible by $d$, then both $p$ and $q$ satisfy the equality $f(m)+f(n)=f(m / d)+$ $f(n / d)+2 f(d)$. This implies that $m, n$ are friends if and only if $m / d, n / d$ are friends.

Let us call a set of integers $\left\{n_{1}, n_{2}, \ldots, n_{5}\right\}$ an interesting set if for any indexes $i, j$, the difference $d_{i j}=\left|n_{i}-n_{j}\right|$ divides both $n_{i}$ and $n_{j}$. We claim that if elements of an interesting set are all positive, then we can obtain a special integer. Indeed, if we were able to construct such a set, then there would be a pair of integers $\left\{n_{i}, n_{j}\right\}$ which are friends, according to the first observation. Additionally, the second observation yields that the quotients $n_{i} / d_{i j}, n_{j} / d_{i j}$ form a pair of friends, which happen to be consecutive integers, thus giving a special integer as desired.

In order to construct a family of interesting sets, we can start by observing that the set $\{0,6,8,9,12\}$ is an interesting set. Using that $72=2^{3} \cdot 3^{2}$ is the least common multiple of all pairwise differences in this set, we obtain a family of interesting sets by considering
$$
\{72 k, 72 k+6,72 k+8,72 k+9,72 k+12\}
$$
for any $k \geqslant 1$. If we consider the quotients (of these numbers by the appropriate differences), then we obtain that the set
$$
S_{k}=\{6 k, 8 k, 9 k, 12 k, 12 k+1,18 k+2,24 k+2,24 k+3,36 k+3,72 k+8\}
$$
has at least one special integer. In particular, the interval $[1,100 k]$ contains the sets $S_{1}, S_{2}, \ldots, S_{k}$, each of which has a special number. Any special number can be contained in at most ten sets $S_{k}$, from where we conclude that the number of special integers in $[1,100 k]$ is at least $k / 10$.

Finally, let $N=100 k+r$, with $k \geqslant 1$ and $0 \leqslant r<100$, so that we have $N<100(k+1) \leqslant$ $200 k$. Then the number of special integers in $[1, N]$ is at least $k / 10>N / 2000$, as we wanted to prove.
Subfield: Algebra


Question: Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Q}_{>0}$ satisfying

$$
f\left(x^{2} f(y)^{2}\right)=f(x)^{2} f(y)
\tag{*}
$$

for all $x, y \in \mathbb{Q}_{>0}$.

Solution:
Take any $a, b \in \mathbb{Q}_{>0}$. By substituting $x=f(a), y=b$ and $x=f(b), y=a$ into $(*)$ we get

$$
f(f(a))^{2} f(b)=f\left(f(a)^{2} f(b)^{2}\right)=f(f(b))^{2} f(a)
$$

which yields

$$
\frac{f(f(a))^{2}}{f(a)}=\frac{f(f(b))^{2}}{f(b)} \quad \text { for all } a, b \in \mathbb{Q}>0
$$

In other words, this shows that there exists a constant $C \in \mathbb{Q}_{>0}$ such that $f(f(a))^{2}=C f(a)$, or

$$
\left(\frac{f(f(a))}{C}\right)^{2}=\frac{f(a)}{C} \quad \text { for all } a \in \mathbb{Q}_{>0}
\tag{1}
$$

Denote by $f^{n}(x)=\underbrace{f(f(\ldots(f}_{n}(x)) \ldots))$ the $n^{\text {th }}$ iteration of $f$. Equality (1) yields

$$
\frac{f(a)}{C}=\left(\frac{f^{2}(a)}{C}\right)^{2}=\left(\frac{f^{3}(a)}{C}\right)^{4}=\cdots=\left(\frac{f^{n+1}(a)}{C}\right)^{2^{n}}
$$

for all positive integer $n$. So, $f(a) / C$ is the $2^{n}$-th power of a rational number for all positive integer $n$. This is impossible unless $f(a) / C=1$, since otherwise the exponent of some prime in the prime decomposition of $f(a) / C$ is not divisible by sufficiently large powers of 2 . Therefore, $f(a)=C$ for all $a \in \mathbb{Q}_{>0}$.

Finally, after substituting $f \equiv C$ into $(*)$ we get $C=C^{3}$, whence $C=1$. So $f(x) \equiv 1$ is the unique function satisfying $(*)$.

Final answer: $f(x)=1$
Subfield: Algebra


Question: Find the maximal value of

$$
S=\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}}+\sqrt[3]{\frac{d}{a+7}}
$$

where $a, b, c, d$ are nonnegative real numbers which satisfy $a+b+c+d=100$.

Solution:
Since the value $8 / \sqrt[3]{7}$ is reached, it suffices to prove that $S \leqslant 8 / \sqrt[3]{7}$.

Assume that $x, y, z, t$ is a permutation of the variables, with $x \leqslant y \leqslant z \leqslant t$. Then, by the rearrangement inequality,

$$
S \leqslant\left(\sqrt[3]{\frac{x}{t+7}}+\sqrt[3]{\frac{t}{x+7}}\right)+\left(\sqrt[3]{\frac{y}{z+7}}+\sqrt[3]{\frac{z}{y+7}}\right)
$$

Claim. The first bracket above does not exceed $\sqrt[3]{\frac{x+t+14}{7}}$.

Proof. Since

$$
X^{3}+Y^{3}+3 X Y Z-Z^{3}=\frac{1}{2}(X+Y-Z)\left((X-Y)^{2}+(X+Z)^{2}+(Y+Z)^{2}\right)
$$

the inequality $X+Y \leqslant Z$ is equivalent (when $X, Y, Z \geqslant 0$ ) to $X^{3}+Y^{3}+3 X Y Z \leqslant Z^{3}$. Therefore, the claim is equivalent to

$$
\frac{x}{t+7}+\frac{t}{x+7}+3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}} \leqslant \frac{x+t+14}{7}
$$

Notice that

$$
\begin{array}{r}
3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}}=3 \sqrt[3]{\frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)}} \\
\leqslant \frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)}
\end{array}
$$

by the AM-GM inequality, so it suffices to prove

$$
\frac{x}{t+7}+\frac{t}{x+7}+\frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \leqslant \frac{x+t+14}{7}
$$

A straightforward check verifies that the last inequality is in fact an equality.

The claim leads now to

$$
S \leqslant \sqrt[3]{\frac{x+t+14}{7}}+\sqrt[3]{\frac{y+z+14}{7}} \leqslant 2 \sqrt[3]{\frac{x+y+z+t+28}{14}}=\frac{8}{\sqrt[3]{7}}
$$

the last inequality being due to the AM-CM inequality (or to the fact that $\sqrt[3]{ }$ is concave on $[0, \infty))$.
We present a different proof for the estimate $S \leqslant 8 / \sqrt[3]{7}$.

Start by using Hölder's inequality:

$$
S^{3}=\left(\sum_{\mathrm{cyc}} \frac{\sqrt[6]{a} \cdot \sqrt[6]{a}}{\sqrt[3]{b+7}}\right)^{3} \leqslant \sum_{\mathrm{cyc}}(\sqrt[6]{a})^{3} \cdot \sum_{\mathrm{cyc}}(\sqrt[6]{a})^{3} \cdot \sum_{\mathrm{cyc}}\left(\frac{1}{\sqrt[3]{b+7}}\right)^{3}=\left(\sum_{\text {cyc }} \sqrt{a}\right)^{2} \sum_{\mathrm{cyc}} \frac{1}{b+7}
$$

Notice that

$$
\frac{(x-1)^{2}(x-7)^{2}}{x^{2}+7} \geqslant 0 \Longleftrightarrow x^{2}-16 x+71 \geqslant \frac{448}{x^{2}+7}
$$

yields

$$
\sum \frac{1}{b+7} \leqslant \frac{1}{448} \sum(b-16 \sqrt{b}+71)=\frac{1}{448}\left(384-16 \sum \sqrt{b}\right)=\frac{48-2 \sum \sqrt{b}}{56} .
$$

Finally,

$$
S^{3} \leqslant \frac{1}{56}\left(\sum \sqrt{a}\right)^{2}\left(48-2 \sum \sqrt{a}\right) \leqslant \frac{1}{56}\left(\frac{\sum \sqrt{a}+\sum \sqrt{a}+\left(48-2 \sum \sqrt{a}\right)}{3}\right)^{3}=\frac{512}{7}
$$

by the AM-GM inequality. The conclusion follows.

Final answer: $\frac{8}{\sqrt[3]{7}}$
Subfield: Combinatorics


Question: Let $n \geqslant 3$ be an integer. Prove that there exists a set $S$ of $2 n$ positive integers satisfying the following property: For every $m=2,3, \ldots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$.

Solution:
We show that one of possible examples is the set

$$
S=\left\{1 \cdot 3^{k}, 2 \cdot 3^{k}: k=1,2, \ldots, n-1\right\} \cup\left\{1, \frac{3^{n}+9}{2}-1\right\}
$$

It is readily verified that all the numbers listed above are distinct (notice that the last two are not divisible by 3 ).

The sum of elements in $S$ is

$$
\Sigma=1+\left(\frac{3^{n}+9}{2}-1\right)+\sum_{k=1}^{n-1}\left(1 \cdot 3^{k}+2 \cdot 3^{k}\right)=\frac{3^{n}+9}{2}+\sum_{k=1}^{n-1} 3^{k+1}=\frac{3^{n}+9}{2}+\frac{3^{n+1}-9}{2}=2 \cdot 3^{n}
$$

Hence, in order to show that this set satisfies the problem requirements, it suffices to present, for every $m=2,3, \ldots, n$, an $m$-element subset $A_{m} \subset S$ whose sum of elements equals $3^{n}$.

Such a subset is

$$
A_{m}=\left\{2 \cdot 3^{k}: k=n-m+1, n-m+2, \ldots, n-1\right\} \cup\left\{1 \cdot 3^{n-m+1}\right\} .
$$

Clearly, $\left|A_{m}\right|=m$. The sum of elements in $A_{m}$ is

$$
3^{n-m+1}+\sum_{k=n-m+1}^{n-1} 2 \cdot 3^{k}=3^{n-m+1}+\frac{2 \cdot 3^{n}-2 \cdot 3^{n-m+1}}{2}=3^{n}
$$

as required.