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


Question: Let $A_{1} A_{2} A_{3} A_{4}$ be a non-cyclic quadrilateral. Let $O_{1}$ and $r_{1}$ be the circumcenter and the circumradius of the triangle $A_{2} A_{3} A_{4}$. Define $O_{2}, O_{3}, O_{4}$ and $r_{2}, r_{3}, r_{4}$ in a similar way. Prove that

$$
\frac{1}{O_{1} A_{1}^{2}-r_{1}^{2}}+\frac{1}{O_{2} A_{2}^{2}-r_{2}^{2}}+\frac{1}{O_{3} A_{3}^{2}-r_{3}^{2}}+\frac{1}{O_{4} A_{4}^{2}-r_{4}^{2}}=0
$$

Solution:
Let $M$ be the point of intersection of the diagonals $A_{1} A_{3}$ and $A_{2} A_{4}$. On each diagonal choose a direction and let $x, y, z$, and $w$ be the signed distances from $M$ to the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$, respectively.

Let $\omega_{1}$ be the circumcircle of the triangle $A_{2} A_{3} A_{4}$ and let $B_{1}$ be the second intersection point of $\omega_{1}$ and $A_{1} A_{3}$ (thus, $B_{1}=A_{3}$ if and only if $A_{1} A_{3}$ is tangent to $\omega_{1}$ ). Since the expression $O_{1} A_{1}^{2}-r_{1}^{2}$ is the power of the point $A_{1}$ with respect to $\omega_{1}$, we get

$$
O_{1} A_{1}^{2}-r_{1}^{2}=A_{1} B_{1} \cdot A_{1} A_{3} .
$$

On the other hand, from the equality $M B_{1} \cdot M A_{3}=M A_{2} \cdot M A_{4}$ we obtain $M B_{1}=y w / z$. Hence, we have

$$
O_{1} A_{1}^{2}-r_{1}^{2}=\left(\frac{y w}{z}-x\right)(z-x)=\frac{z-x}{z}(y w-x z) .
$$

Substituting the analogous expressions into the sought sum we get

$$
\sum_{i=1}^{4} \frac{1}{O_{i} A_{i}^{2}-r_{i}^{2}}=\frac{1}{y w-x z}\left(\frac{z}{z-x}-\frac{w}{w-y}+\frac{x}{x-z}-\frac{y}{y-w}\right)=0
$$

as desired.
Introduce a Cartesian coordinate system in the plane. Every circle has an equation of the form $p(x, y)=x^{2}+y^{2}+l(x, y)=0$, where $l(x, y)$ is a polynomial of degree at most 1 . For any point $A=\left(x_{A}, y_{A}\right)$ we have $p\left(x_{A}, y_{A}\right)=d^{2}-r^{2}$, where $d$ is the distance from $A$ to the center of the circle and $r$ is the radius of the circle.

For each $i$ in $\{1,2,3,4\}$ let $p_{i}(x, y)=x^{2}+y^{2}+l_{i}(x, y)=0$ be the equation of the circle with center $O_{i}$ and radius $r_{i}$ and let $d_{i}$ be the distance from $A_{i}$ to $O_{i}$. Consider the equation

$$
\sum_{i=1}^{4} \frac{p_{i}(x, y)}{d_{i}^{2}-r_{i}^{2}}=1
\tag{1}
$$



Since the coordinates of the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$ satisfy (1) but these four points do not lie on a circle or on an line, equation (1) defines neither a circle, nor a line. Hence, the equation is an identity and the coefficient of the quadratic term $x^{2}+y^{2}$ also has to be zero, i.e.

$$
\sum_{i=1}^{4} \frac{1}{d_{i}^{2}-r_{i}^{2}}=0
$$
Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral whose sides $A D$ and $B C$ are not parallel. Suppose that the circles with diameters $A B$ and $C D$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_{E}$ be the circle through the feet of the perpendiculars from $E$ to the lines $A B, B C$, and $C D$. Let $\omega_{F}$ be the circle through the feet of the perpendiculars from $F$ to the lines $C D, D A$, and $A B$. Prove that the midpoint of the segment $E F$ lies on the line through the two intersection points of $\omega_{E}$ and $\omega_{F}$.

Solution:
Denote by $P, Q, R$, and $S$ the projections of $E$ on the lines $D A, A B, B C$, and $C D$ respectively. The points $P$ and $Q$ lie on the circle with diameter $A E$, so $\angle Q P E=\angle Q A E$; analogously, $\angle Q R E=\angle Q B E$. So $\angle Q P E+\angle Q R E=\angle Q A E+\angle Q B E=90^{\circ}$. By similar reasons, we have $\angle S P E+\angle S R E=90^{\circ}$, hence we get $\angle Q P S+\angle Q R S=90^{\circ}+90^{\circ}=180^{\circ}$, and the quadrilateral $P Q R S$ is inscribed in $\omega_{E}$. Analogously, all four projections of $F$ onto the sides of $A B C D$ lie on $\omega_{F}$.

Denote by $K$ the meeting point of the lines $A D$ and $B C$. Due to the arguments above, there is no loss of generality in assuming that $A$ lies on segment $D K$. Suppose that $\angle C K D \geq 90^{\circ}$; then the circle with diameter $C D$ covers the whole quadrilateral $A B C D$, so the points $E, F$ cannot lie inside this quadrilateral. Hence our assumption is wrong. Therefore, the lines $E P$ and $B C$ intersect at some point $P^{\prime}$, while the lines $E R$ and $A D$ intersect at some point $R^{\prime}$.

<img_3762>

Figure 1

We claim that the points $P^{\prime}$ and $R^{\prime}$ also belong to $\omega_{E}$. Since the points $R, E, Q, B$ are concyclic, $\angle Q R K=\angle Q E B=90^{\circ}-\angle Q B E=\angle Q A E=\angle Q P E$. So $\angle Q R K=\angle Q P P^{\prime}$, which means that the point $P^{\prime}$ lies on $\omega_{E}$. Analogously, $R^{\prime}$ also lies on $\omega_{E}$.

In the same manner, denote by $M$ and $N$ the projections of $F$ on the lines $A D$ and $B C$



respectively, and let $M^{\prime}=F M \cap B C, N^{\prime}=F N \cap A D$. By the same arguments, we obtain that the points $M^{\prime}$ and $N^{\prime}$ belong to $\omega_{F}$.

<img_3154>

Figure 2

Now we concentrate on Figure 2, where all unnecessary details are removed. Let $U=N N^{\prime} \cap$ $P P^{\prime}, V=M M^{\prime} \cap R R^{\prime}$. Due to the right angles at $N$ and $P$, the points $N, N^{\prime}, P, P^{\prime}$ are concyclic, so $U N \cdot U N^{\prime}=U P \cdot U P^{\prime}$ which means that $U$ belongs to the radical axis $g$ of the circles $\omega_{E}$ and $\omega_{F}$. Analogously, $V$ also belongs to $g$.

Finally, since $E U F V$ is a parallelogram, the radical axis $U V$ of $\omega_{E}$ and $\omega_{F}$ bisects $E F$.
Subfield: Geometry


Question: Let $A B C$ be an acute triangle with circumcircle $\Omega$. Let $B_{0}$ be the midpoint of $A C$ and let $C_{0}$ be the midpoint of $A B$. Let $D$ be the foot of the altitude from $A$, and let $G$ be the centroid of the triangle $A B C$. Let $\omega$ be a circle through $B_{0}$ and $C_{0}$ that is tangent to the circle $\Omega$ at a point $X \neq A$. Prove that the points $D, G$, and $X$ are collinear.

Solution:
If $A B=A C$, then the statement is trivial. So without loss of generality we may assume $A B<A C$. Denote the tangents to $\Omega$ at points $A$ and $X$ by $a$ and $x$, respectively.

Let $\Omega_{1}$ be the circumcircle of triangle $A B_{0} C_{0}$. The circles $\Omega$ and $\Omega_{1}$ are homothetic with center $A$, so they are tangent at $A$, and $a$ is their radical axis. Now, the lines $a, x$, and $B_{0} C_{0}$ are the three radical axes of the circles $\Omega, \Omega_{1}$, and $\omega$. Since $a \nmid B_{0} C_{0}$, these three lines are concurrent at some point $W$.

The points $A$ and $D$ are symmetric with respect to the line $B_{0} C_{0}$; hence $W X=W A=W D$. This means that $W$ is the center of the circumcircle $\gamma$ of triangle $A D X$. Moreover, we have $\angle W A O=\angle W X O=90^{\circ}$, where $O$ denotes the center of $\Omega$. Hence $\angle A W X+\angle A O X=180^{\circ}$.

<img_3871>

Denote by $T$ the second intersection point of $\Omega$ and the line $D X$. Note that $O$ belongs to $\Omega_{1}$. Using the circles $\gamma$ and $\Omega$, we find $\angle D A T=\angle A D X-\angle A T D=\frac{1}{2}\left(360^{\circ}-\angle A W X\right)-\frac{1}{2} \angle A O X=$ $180^{\circ}-\frac{1}{2}(\angle A W X+\angle A O X)=90^{\circ}$. So, $A D \perp A T$, and hence $A T \| B C$. Thus, $A T C B$ is an isosceles trapezoid inscribed in $\Omega$.

Denote by $A_{0}$ the midpoint of $B C$, and consider the image of $A T C B$ under the homothety $h$ with center $G$ and factor $-\frac{1}{2}$. We have $h(A)=A_{0}, h(B)=B_{0}$, and $h(C)=C_{0}$. From the



symmetry about $B_{0} C_{0}$, we have $\angle T C B=\angle C B A=\angle B_{0} C_{0} A=\angle D C_{0} B_{0}$. Using $A T \| D A_{0}$, we conclude $h(T)=D$. Hence the points $D, G$, and $T$ are collinear, and $X$ lies on the same line.
We define the points $A_{0}, O$, and $W$ as in the previous solution and we concentrate on the case $A B<A C$. Let $Q$ be the perpendicular projection of $A_{0}$ on $B_{0} C_{0}$.

Since $\angle W A O=\angle W Q O=\angle O X W=90^{\circ}$, the five points $A, W, X, O$, and $Q$ lie on a common circle. Furthermore, the reflections with respect to $B_{0} C_{0}$ and $O W$ map $A$ to $D$ and $X$, respectively. For these reasons, we have

$$
\angle W Q D=\angle A Q W=\angle A X W=\angle W A X=\angle W Q X
$$

Thus the three points $Q, D$, and $X$ lie on a common line, say $\ell$.

<img_3834>

To complete the argument, we note that the homothety centered at $G$ sending the triangle $A B C$ to the triangle $A_{0} B_{0} C_{0}$ maps the altitude $A D$ to the altitude $A_{0} Q$. Therefore it maps $D$ to $Q$, so the points $D, G$, and $Q$ are collinear. Hence $G$ lies on $\ell$ as well.
Subfield: Number Theory


Question: For any integer $d>0$, let $f(d)$ be the smallest positive integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16$, and $f(6)=12$ ). Prove that for every integer $k \geq 0$ the number $f\left(2^{k}\right)$ divides $f\left(2^{k+1}\right)$.

Solution:
For any positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. Let $n=\prod_{p} p^{a(p)}$ be the prime factorization of $n$ where $p$ ranges over the prime numbers, the integers $a(p)$ are nonnegative and all but finitely many $a(p)$ are zero. Then we have $d(n)=\prod_{p}(a(p)+1)$. Thus, $d(n)$ is a power of 2 if and only if for every prime $p$ there is a nonnegative integer $b(p)$ with $a(p)=2^{b(p)}-1=1+2+2^{2}+\cdots+2^{b(p)-1}$. We then have

$$
n=\prod_{p} \prod_{i=0}^{b(p)-1} p^{2^{i}}, \quad \text { and } \quad d(n)=2^{k} \quad \text { with } \quad k=\sum_{p} b(p)
$$

Let $\mathcal{S}$ be the set of all numbers of the form $p^{2^{r}}$ with $p$ prime and $r$ a nonnegative integer. Then we deduce that $d(n)$ is a power of 2 if and only if $n$ is the product of the elements of some finite subset $\mathcal{T}$ of $\mathcal{S}$ that satisfies the following condition: for all $t \in \mathcal{T}$ and $s \in \mathcal{S}$ with $s \mid t$ we have $s \in \mathcal{T}$. Moreover, if $d(n)=2^{k}$ then the corresponding set $\mathcal{T}$ has $k$ elements.

Note that the $\operatorname{set} \mathcal{T}_{k}$ consisting of the smallest $k$ elements from $\mathcal{S}$ obviously satisfies the condition above. Thus, given $k$, the smallest $n$ with $d(n)=2^{k}$ is the product of the elements of $\mathcal{T}_{k}$. This $n$ is $f\left(2^{k}\right)$. Since obviously $\mathcal{T}_{k} \subset \mathcal{T}_{k+1}$, it follows that $f\left(2^{k}\right) \mid f\left(2^{k+1}\right)$.
Subfield: Number Theory


Question: Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^{n}-y^{n}$.

Solution:
Obviously, all functions in the answer satisfy the condition of the problem. We will show that there are no other functions satisfying that condition.

Let $f$ be a function satisfying the given condition. For each integer $n$, the function $g$ defined by $g(x)=f(x)+n$ also satisfies the same condition. Therefore, by subtracting $f(0)$ from $f(x)$ we may assume that $f(0)=0$.

For any prime $p$, the condition on $f$ with $(x, y)=(p, 0)$ states that $f(p)$ divides $p^{n}$. Since the set of primes is infinite, there exist integers $d$ and $\varepsilon$ with $0 \leq d \leq n$ and $\varepsilon \in\{1,-1\}$ such that for infinitely many primes $p$ we have $f(p)=\varepsilon p^{d}$. Denote the set of these primes by $P$. Since a function $g$ satisfies the given condition if and only if $-g$ satisfies the same condition, we may suppose $\varepsilon=1$.

The case $d=0$ is easily ruled out, because 0 does not divide any nonzero integer. Suppose $d \geq 1$ and write $n$ as $m d+r$, where $m$ and $r$ are integers such that $m \geq 1$ and $0 \leq r \leq d-1$. Let $x$ be an arbitrary integer. For each prime $p$ in $P$, the difference $f(p)-f(x)$ divides $p^{n}-x^{n}$. Using the equality $f(p)=p^{d}$, we get

$$
p^{n}-x^{n}=p^{r}\left(p^{d}\right)^{m}-x^{n} \equiv p^{r} f(x)^{m}-x^{n} \equiv 0 \quad\left(\bmod p^{d}-f(x)\right)
$$

Since we have $r<d$, for large enough primes $p \in P$ we obtain

$$
\left|p^{r} f(x)^{m}-x^{n}\right|<p^{d}-f(x)
$$

Hence $p^{r} f(x)^{m}-x^{n}$ has to be zero. This implies $r=0$ and $x^{n}=\left(x^{d}\right)^{m}=f(x)^{m}$. Since $m$ is odd, we obtain $f(x)=x^{d}$.

Final answer: All functions $f$ of the form $f(x)=\varepsilon x^{d}+c$, where $\varepsilon$ is in $\{1,-1\}$, the integer $d$ is a positive divisor of $n$, and $c$ is an integer
Subfield: Number Theory


Question: For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences

$$
t(n+a)-t(n), \quad t(n+a+1)-t(n+1), \quad \ldots, \quad t(n+2 a-1)-t(n+a-1)
$$

are divisible by 4 .

Solution:
A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs.

Now suppose that $a$ is a positive integer not equal to 1,3 , and 5 . We will show that there are no winning pairs $(a, n)$ by distinguishing three cases.

Case 1: $a$ is even. In this case we have $a=2^{\alpha} d$ for some positive integer $\alpha$ and some odd $d$. Since $a \geq 2^{\alpha}$, for each positive integer $n$ there exists an $i \in\{0,1, \ldots, a-1\}$ such that $n+i=2^{\alpha-1} e$, where $e$ is some odd integer. Then we have $t(n+i)=t\left(2^{\alpha-1} e\right)=e$ and

$$
t(n+a+i)=t\left(2^{\alpha} d+2^{\alpha-1} e\right)=2 d+e \equiv e+2 \quad(\bmod 4) .
$$

So we get $t(n+i)-t(n+a+i) \equiv 2(\bmod 4)$, and $(a, n)$ is not a winning pair.

Case 2: $a$ is odd and $a>8$. For each positive integer $n$, there exists an $i \in\{0,1, \ldots, a-5\}$ such that $n+i=2 d$ for some odd $d$. We get

$$
t(n+i)=d \not \equiv d+2=t(n+i+4) \quad(\bmod 4)
$$

and

$$
t(n+a+i)=n+a+i \equiv n+a+i+4=t(n+a+i+4) \quad(\bmod 4)
$$

Therefore, the integers $t(n+a+i)-t(n+i)$ and $t(n+a+i+4)-t(n+i+4)$ cannot be both divisible by 4 , and therefore there are no winning pairs in this case.

Case 3: $a=7$. For each positive integer $n$, there exists an $i \in\{0,1, \ldots, 6\}$ such that $n+i$ is either of the form $8 k+3$ or of the form $8 k+6$, where $k$ is a nonnegative integer. But we have

$$
t(8 k+3) \equiv 3 \not \equiv 1 \equiv 4 k+5=t(8 k+3+7) \quad(\bmod 4)
$$

and

$$
t(8 k+6)=4 k+3 \equiv 3 \not \equiv 1 \equiv t(8 k+6+7) \quad(\bmod 4)
$$

Hence, there are no winning pairs of the form $(7, n)$.

Final answer: 1,3,5
Subfield: Number Theory


Question: Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients such that no nonconstant polynomial with rational coefficients divides both $P(x)$ and $Q(x)$. Suppose that for every positive integer $n$ the integers $P(n)$ and $Q(n)$ are positive, and $2^{Q(n)}-1$ divides $3^{P(n)}-1$. Prove that $Q(x)$ is a constant polynomial.

Solution:
First we show that there exists an integer $d$ such that for all positive integers $n$ we have $\operatorname{gcd}(P(n), Q(n)) \leq d$.

Since $P(x)$ and $Q(x)$ are coprime (over the polynomials with rational coefficients), EuCLID's algorithm provides some polynomials $R_{0}(x), S_{0}(x)$ with rational coefficients such that $P(x) R_{0}(x)-$ $Q(x) S_{0}(x)=1$. Multiplying by a suitable positive integer $d$, we obtain polynomials $R(x)=$ $d \cdot R_{0}(x)$ and $S(x)=d \cdot S_{0}(x)$ with integer coefficients for which $P(x) R(x)-Q(x) S(x)=d$. Then we have $\operatorname{gcd}(P(n), Q(n)) \leq d$ for any integer $n$.

To prove the problem statement, suppose that $Q(x)$ is not constant. Then the sequence $Q(n)$ is not bounded and we can choose a positive integer $m$ for which

$$
M=2^{Q(m)}-1 \geq 3^{\max \{P(1), P(2), \ldots, P(d)\}}
\tag{1}
$$

Since $M=2^{Q(n)}-1 \mid 3^{P(n)}-1$, we have $2,3 \not \backslash M$. Let $a$ and $b$ be the multiplicative orders of 2 and 3 modulo $M$, respectively. Obviously, $a=Q(m)$ since the lower powers of 2 do not reach $M$. Since $M$ divides $3^{P(m)}-1$, we have $b \mid P(m)$. Then $\operatorname{gcd}(a, b) \leq \operatorname{gcd}(P(m), Q(m)) \leq d$. Since the expression $a x-b y$ attains all integer values divisible by $\operatorname{gcd}(a, b)$ when $x$ and $y$ run over all nonnegative integer values, there exist some nonnegative integers $x, y$ such that $1 \leq m+a x-b y \leq d$.

By $Q(m+a x) \equiv Q(m)(\bmod a)$ we have

$$
2^{Q(m+a x)} \equiv 2^{Q(m)} \equiv 1 \quad(\bmod M)
$$

and therefore

$$
M\left|2^{Q(m+a x)}-1\right| 3^{P(m+a x)}-1 .
$$

Then, by $P(m+a x-b y) \equiv P(m+a x)(\bmod b)$ we have

$$
3^{P(m+a x-b y)} \equiv 3^{P(m+a x)} \equiv 1 \quad(\bmod M)
$$

Since $P(m+a x-b y)>0$ this implies $M \leq 3^{P(m+a x-b y)}-1$. But $P(m+a x-b y)$ is listed among $P(1), P(2), \ldots, P(d)$, so

$$
M<3^{P(m+a x-b y)} \leq 3^{\max \{P(1), P(2), \ldots, P(d)\}}
$$



which contradicts (1).
Subfield: Algebra


Question: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality

$$
f([x] y)=f(x)[f(y)]
\tag{1}
$$

holds for all $x, y \in \mathbb{R}$. Here, by $[x]$ we denote the greatest integer not exceeding $x$.

Solution:
First, setting $x=0$ in (1) we get

$$
f(0)=f(0)[f(y)] \tag{2}
$$

for all $y \in \mathbb{R}$. Now, two cases are possible.

Case 1. Assume that $f(0) \neq 0$. Then from $(2)$ we conclude that $[f(y)]=1$ for all $y \in \mathbb{R}$. Therefore, equation (1) becomes $f([x] y)=f(x)$, and substituting $y=0$ we have $f(x)=f(0)=C \neq 0$. Finally, from $[f(y)]=1=[C]$ we obtain that $1 \leq C<2$.

Case 2. Now we have $f(0)=0$. Here we consider two subcases.

Subcase 2a. Suppose that there exists $0<\alpha<1$ such that $f(\alpha) \neq 0$. Then setting $x=\alpha$ in (1) we obtain $0=f(0)=f(\alpha)[f(y)]$ for all $y \in \mathbb{R}$. Hence, $[f(y)]=0$ for all $y \in \mathbb{R}$. Finally, substituting $x=1$ in (1) provides $f(y)=0$ for all $y \in \mathbb{R}$, thus contradicting the condition $f(\alpha) \neq 0$.

Subcase 2b. Conversely, we have $f(\alpha)=0$ for all $0 \leq \alpha<1$. Consider any real $z$; there exists an integer $N$ such that $\alpha=\frac{z}{N} \in[0,1)$ (one may set $N=[z]+1$ if $z \geq 0$ and $N=[z]-1$ otherwise). Now, from (1) we get $f(z)=f([N] \alpha)=f(N)[f(\alpha)]=0$ for all $z \in \mathbb{R}$.

Finally, a straightforward check shows that all the obtained functions satisfy (1).
Assume that $[f(y)]=0$ for some $y$; then the substitution $x=1$ provides $f(y)=f(1)[f(y)]=0$. Hence, if $[f(y)]=0$ for all $y$, then $f(y)=0$ for all $y$. This function obviously satisfies the problem conditions.

So we are left to consider the case when $[f(a)] \neq 0$ for some $a$. Then we have

$$
f([x] a)=f(x)[f(a)], \quad \text { or } \quad f(x)=\frac{f([x] a)}{[f(a)]}\tag{3}
$$

This means that $f\left(x_{1}\right)=f\left(x_{2}\right)$ whenever $\left[x_{1}\right]=\left[x_{2}\right]$, hence $f(x)=f([x])$, and we may assume that $a$ is an integer.

Now we have

$$
f(a)=f\left(2 a \cdot \frac{1}{2}\right)=f(2 a)\left[f\left(\frac{1}{2}\right)\right]=f(2 a)[f(0)] ;
$$

this implies $[f(0)] \neq 0$, so we may even assume that $a=0$. Therefore equation (3) provides

$$
f(x)=\frac{f(0)}{[f(0)]}=C \neq 0
$$

for each $x$. Now, condition (1) becomes equivalent to the equation $C=C[C]$ which holds exactly when $[C]=1$.

Final answer: $f(x)=$ const $=C$, where $C=0$ or $1 \leq C<2$
Subfield: Algebra


Question: Let $x_{1}, \ldots, x_{100}$ be nonnegative real numbers such that $x_{i}+x_{i+1}+x_{i+2} \leq 1$ for all $i=1, \ldots, 100$ (we put $x_{101}=x_{1}, x_{102}=x_{2}$ ). Find the maximal possible value of the sum

$$
S=\sum_{i=1}^{100} x_{i} x_{i+2}
$$

Solution:
Let $x_{2 i}=0, x_{2 i-1}=\frac{1}{2}$ for all $i=1, \ldots, 50$. Then we have $S=50 \cdot\left(\frac{1}{2}\right)^{2}=\frac{25}{2}$. So, we are left to show that $S \leq \frac{25}{2}$ for all values of $x_{i}$ 's satisfying the problem conditions.

Consider any $1 \leq i \leq 50$. By the problem condition, we get $x_{2 i-1} \leq 1-x_{2 i}-x_{2 i+1}$ and $x_{2 i+2} \leq 1-x_{2 i}-x_{2 i+1}$. Hence by the AM-GM inequality we get

$$
\begin{aligned}
x_{2 i-1} x_{2 i+1} & +x_{2 i} x_{2 i+2} \leq\left(1-x_{2 i}-x_{2 i+1}\right) x_{2 i+1}+x_{2 i}\left(1-x_{2 i}-x_{2 i+1}\right) \\
& =\left(x_{2 i}+x_{2 i+1}\right)\left(1-x_{2 i}-x_{2 i+1}\right) \leq\left(\frac{\left(x_{2 i}+x_{2 i+1}\right)+\left(1-x_{2 i}-x_{2 i+1}\right)}{2}\right)^{2}=\frac{1}{4} .
\end{aligned}
$$

Summing up these inequalities for $i=1,2, \ldots, 50$, we get the desired inequality

$$
\sum_{i=1}^{50}\left(x_{2 i-1} x_{2 i+1}+x_{2 i} x_{2 i+2}\right) \leq 50 \cdot \frac{1}{4}=\frac{25}{2}
$$
We present another proof of the estimate. From the problem condition, we get

$$
\begin{aligned}
S=\sum_{i=1}^{100} x_{i} x_{i+2} \leq \sum_{i=1}^{100} x_{i}\left(1-x_{i}-x_{i+1}\right) & =\sum_{i=1}^{100} x_{i}-\sum_{i=1}^{100} x_{i}^{2}-\sum_{i=1}^{100} x_{i} x_{i+1} \\
& =\sum_{i=1}^{100} x_{i}-\frac{1}{2} \sum_{i=1}^{100}\left(x_{i}+x_{i+1}\right)^{2}
\end{aligned}
$$

By the AM-QM inequality, we have $\sum\left(x_{i}+x_{i+1}\right)^{2} \geq \frac{1}{100}\left(\sum\left(x_{i}+x_{i+1}\right)\right)^{2}$, so

$$
\begin{aligned}
S \leq \sum_{i=1}^{100} x_{i}-\frac{1}{200}\left(\sum_{i=1}^{100}\left(x_{i}+x_{i+1}\right)\right)^{2} & =\sum_{i=1}^{100} x_{i}-\frac{2}{100}\left(\sum_{i=1}^{100} x_{i}\right)^{2} \\
& =\frac{2}{100}\left(\sum_{i=1}^{100} x_{i}\right)\left(\frac{100}{2}-\sum_{i=1}^{100} x_{i}\right) .
\end{aligned}
$$

And finally, by the AM-GM inequality

$$
S \leq \frac{2}{100} \cdot\left(\frac{1}{2}\left(\sum_{i=1}^{100} x_{i}+\frac{100}{2}-\sum_{i=1}^{100} x_{i}\right)\right)^{2}=\frac{2}{100} \cdot\left(\frac{100}{4}\right)^{2}=\frac{25}{2}
$$

Final answer: $\frac{25}{2}$