di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Number Theory


Question: 
Find (in closed form) the difference between the number of positive integers at most $2^{2017}$ with even weight and the number of positive integers at most $2^{2017}$ with odd weight.

Solution:
For every integer $M \geq 0$, let $A_{M}=\sum_{n=-2^{M}+1}^{0}(-1)^{w(n)}$ and let $B_{M}=$ $\sum_{n=1}^{2^{M}}(-1)^{w(n)}$; thus, $B_{M}$ evaluates the difference of the number of even weight integers in the range 1 through $2^{M}$ and the number of odd weight integers in that range.



Notice that



$$

w(n)= \begin{cases}w\left(n+2^{M}\right)+1 & \text { if }-2^{M}+1 \leq n \leq-2^{M-1} \\ w\left(n-2^{M}\right) & \text { if } 2^{M-1}+1 \leq n \leq 2^{M}\end{cases}

$$







to get



$$

\begin{aligned}

& A_{M}=-\sum_{n=-2^{M}+1}^{-2^{M-1}}(-1)^{w\left(n+2^{M}\right)}+\sum_{n=-2^{M-1}+1}^{0}(-1)^{w(n)}=-B_{M-1}+A_{M-1}, \\

& B_{M}=\sum_{n=1}^{2^{M-1}}(-1)^{w(n)}+\sum_{n=2^{M-1}+1}^{2^{M}}(-1)^{w\left(n-2^{M}\right)}=B_{M-1}+A_{M-1} .

\end{aligned}

$$



Iteration yields



$$

\begin{aligned}

B_{M} & =A_{M-1}+B_{M-1}=\left(A_{M-2}-B_{M-2}\right)+\left(A_{M-2}+B_{M-2}\right)=2 A_{M-2} \\

& =2 A_{M-3}-2 B_{M-3}=2\left(A_{M-4}-B_{M-4}\right)-2\left(A_{M-4}+B_{M-4}\right)=-4 B_{M-4}

\end{aligned}

$$



Thus, $B_{2017}=(-4)^{504} B_{1}=2^{1008} B_{1}$; since $B_{1}=(-1)^{w(1)}+(-1)^{w(2)}=2$, it follows that $B_{2017}=$ $2^{1009}$

Final answer: $2^{1009}$
Subfield: Algebra


Question: Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \leq n$, and $k+1$ distinct integers $x_{1}, x_{2}, \ldots, x_{k+1}$ such that



$$

P\left(x_{1}\right)+P\left(x_{2}\right)+\cdots+P\left(x_{k}\right)=P\left(x_{k+1}\right) .

$$


Note. A polynomial is monic if the coefficient of the highest power is one.

Solution:
There is only one such integer, namely, $n=2$. In this case, if $P$ is a constant polynomial, the required condition is clearly satisfied; if $P=X+c$, then $P(c-1)+P(c+1)=$ $P(3 c)$; and if $P=X^{2}+q X+r$, then $P(X)=P(-X-q)$.



To rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of degree at most $n$ with integer coefficients, whose restriction to the integers is injective, and $P(x) \equiv 1$ $(\bmod\  n)$ for all integers $x$. This is easily seen by reading the relation in the statement modulo $n$, to deduce that $k \equiv 1(\bmod\  n)$, so $k=1$, since $1 \leq k \leq n$; hence $P\left(x_{1}\right)=P\left(x_{2}\right)$ for some distinct integers $x_{1}$ and $x_{2}$, which contradicts injectivity.



If $n=1$, let $P=X$, and if $n=4$, let $P=X^{4}+7 X^{2}+4 X+1$. In the latter case, clearly, $P(x) \equiv 1(\bmod\  4)$ for all integers $x$; and $P$ is injective on the integers, since $P(x)-P(y)=$ $(x-y)\left((x+y)\left(x^{2}+y^{2}+7\right)+4\right)$, and the absolute value of $(x+y)\left(x^{2}+y^{2}+7\right)$ is either 0 or at least 7 for integral $x$ and $y$.



Assume henceforth $n \geq 3, n \neq 4$, and let $f_{n}=(X-1)(X-2) \cdots(X-n)$. Clearly, $f_{n}(x) \equiv$ $0(\bmod n)$ for all integers $x$. If $n$ is odd, then $f_{n}$ is non-decreasing on the integers; and if, in addition, $n>3$, then $f_{n}(x) \equiv 0(\bmod n+1)$ for all integers $x$, since $f_{n}(0)=-n !=-1 \cdot 2 \cdot \cdots$. $\frac{n+1}{2} \cdot \cdots \cdot n \equiv 0(\bmod\  n+1)$.



Finally, let $P=f_{n}+n X+1$ if $n$ is odd, and let $P=f_{n-1}+n X+1$ if $n$ is even. In either case, $P$ is strictly increasing, hence injective, on the integers, and $P(x) \equiv 1(\bmod n)$ for all integers $x$.

Final answer: 2
Subfield: Number Theory


Question: Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $x^{3}+y^{3}=$ $p(x y+p)$.

Solution:
Up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$. The verification is routine.



Set $s=x+y$. Rewrite the equation in the form $s\left(s^{2}-3 x y\right)=p(p+x y)$, and express $x y$ :



$$

x y=\frac{s^{3}-p^{2}}{3 s+p} \tag{*}

$$



In particular,



$$

s^{2} \geq 4 x y=\frac{4\left(s^{3}-p^{2}\right)}{3 s+p}

$$



or



$$

(s-2 p)\left(s^{2}+s p+2 p^{2}\right) \leq p^{2}-p^{3}<0

$$



so $s<2 p$.



If $p \mid s$, then $s=p$ and $x y=p(p-1) / 4$ which is impossible for $x+y=p$ (the equation $t^{2}-p t+p(p-1) / 4=0$ has no integer solutions).



If $p \nmid s$, rewrite $(*)$ in the form



$$

27 x y=\left(9 s^{2}-3 s p+p^{2}\right)-\frac{p^{2}(p+27)}{3 s+p} .

$$



Since $p \nmid s$, this could be integer only if $3 s+p \mid$ $p+27$, and hence $3 s+p \mid 27-s$.



If $s \neq 9$, then $|3 s-27| \geq 3 s+p$, so $27-3 s \geq$ $3 s+p$, or $27-p \geq 6 s$, whence $s \leq 4$. These cases are ruled out by hand.



If $s=x+y=9$, then $(*)$ yields $x y=27-p$. Up to a swap of $x$ and $y$, all such triples $(x, y, p)$ are $(1,8,19),(2,7,13)$, and $(4,5,7)$.
Set again $s=x+y$. It is readily checked that $s \leq 8$ provides no solutions, so assume $s \geq 9$. Notice that $x^{3}+y^{3}=s\left(x^{2}-x y+y^{2}\right) \geq$ $\frac{1}{4} s^{3}$ and $x y \leq \frac{1}{4} s^{2}$. The condition in the statement then implies $s^{2}(s-p) \leq 4 p^{2}$, so $s<p+4$.



Notice that $p$ divides one of $s$ and $x^{2}-x y+y^{2}$. The case $p \mid s$ is easily ruled out by the condition $s<p+4$ : The latter forces $s=p$, so $x^{2}-x y+y^{2}=x y+p$, i. e., $(x-y)^{2}=p$, which is impossible.



Hence $p \mid x^{2}-x y+y^{2}$, so $x^{2}-x y+y^{2}=k p$ and $x y+p=k s$ for some positive integer $k$, implying



$$

s^{2}+3 p=k(3 s+p) \tag{**}

$$



Recall that $p \nmid s$ to infer that $3 k \equiv s(\bmod p)$. We now present two approaches.



1st Approach. Write $3 k=s+m p$ for some integer $m$ and plug $k=\frac{1}{3}(s+m p)$ into $(* *)$ to get $s=(9-m p) /(3 m+1)$. The condition $s \geq 9$ then forces $m=0$, so $s=9$, in which case, up to a swap of the first two entries, the solutions turn out to be $(x, y, p)=(1,8,19),(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$.



2nd Approach. Notice that $k=\frac{s^{2}+3 p}{3 s+p}=3+$ $\frac{s(s-9)}{3 s+p} \leq 3+\frac{1}{3}(s-9)=\frac{1}{3} s \leq \frac{1}{3}(p+3)$, since $s<p+4$. Hence $3 k \leq p+3$, and the congruence $3 k \equiv s$ $(\bmod p)$ then forces either $3 k=s-p$ or $3 k=s$.



The case $3 k=s-p$ is easily ruled out: Otherwise, $(* *)$ boils down to $2 s+p+9=0$, which is clearly impossible.



Finally, if $3 k=s$, then $(* *)$ reduces to $s=9$. In this case, up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$.

Final answer: $(1,8,19), (2,7,13), (4,5,7)$
Subfield: Combinatorics


Question: Fix an integer $n \geqslant 3$. Let $\mathcal{S}$ be a set of $n$ points in the plane, no three of which are collinear. Given different points $A, B, C$ in $\mathcal{S}$, the triangle $A B C$ is nice for $A B$ if Area $(A B C) \leqslant \operatorname{Area}(A B X)$ for all $X$ in $\mathcal{S}$ different from $A$ and $B$. (Note that for a segment $A B$ there could be several nice triangles.) A triangle is beautiful if its vertices are all in $\mathcal{S}$ and it is nice for at least two of its sides.



Prove that there are at least $\frac{1}{2}(n-1)$ beautiful triangles.

Solution:
For convenience, a triangle whose vertices all lie in $\mathcal{S}$ will be referred to as a triangle in $\mathcal{S}$. The argument hinges on the following observation:



Given any partition of $\mathcal{S}$, amongst all triangles in $\mathcal{S}$ with at least one vertex in each part, those of minimal area are all adequate.



Indeed, amongst the triangles under consideration, one of minimal area is suitable for both sides with endpoints in different parts.



We now present two approaches for the lower bound.



1st Approach. By the above observation, the 3uniform hypergraph of adequate triangles is connected. It is a well-known fact that such a hypergraph has at least $\frac{1}{2}(n-1)$ hyperedges, whence the required lower bound



2nd Approach. For a partition $\mathcal{S}=\mathcal{A} \sqcup \mathcal{B}$, an area minimising triangle as above will be called $(\mathcal{A}, \mathcal{B})$ minimal. Thus, $(\mathcal{A}, \mathcal{B})$-minimal triangles are all adequate.



Consider now a partition of $\mathcal{S}=\mathcal{A} \sqcup \mathcal{B}$, where $|\mathcal{A}|=1$. Choose an $(\mathcal{A}, \mathcal{B})$-minimal triangle and add to $\mathcal{A}$ its vertices from $\mathcal{B}$ to obtain a new partition also written $\mathcal{S}=\mathcal{A} \sqcup \mathcal{B}$. Continuing, choose an $(\mathcal{A}, \mathcal{B})$-minimal triangle and add to $\mathcal{A}$ its vertices/vertex from $\mathcal{B}$ and so on and so forth all the way down for at least another $\frac{1}{2}(n-5)$ steps - this works at least as many times, since at each step, $\mathcal{B}$ loses at most two points. Clearly, each step provides a new adequate triangle, so the overall number of adequate triangles is at least $\frac{1}{2}(n-1)$, as required.
Subfield: Algebra


Question: Let $x$ and $y$ be positive real numbers such that $x+y^{2016} \geq 1$. Prove that $x^{2016}+y>$ $1-1 / 100$.

Solution:
If $x \geq 1-1 /(100 \cdot 2016)$, then



$$

x^{2016} \geq\left(1-\frac{1}{100 \cdot 2016}\right)^{2016}>1-2016 \cdot \frac{1}{100 \cdot 2016}=1-\frac{1}{100}

$$



by Bernoulli's inequality, whence the conclusion.



If $x<1-1 /(100 \cdot 2016)$, then $y \geq(1-x)^{1 / 2016}>(100 \cdot 2016)^{-1 / 2016}$, and it is sufficient to show that the latter is greater than $1-1 / 100=99 / 100$; alternatively, but equivalently, that



$$

\left(1+\frac{1}{99}\right)^{2016}>100 \cdot 2016

$$



To establish the latter, refer again to Bernoulli's inequality to write



$$

\left(1+\frac{1}{99}\right)^{2016}>\left(1+\frac{1}{99}\right)^{99 \cdot 20}>\left(1+99 \cdot \frac{1}{99}\right)^{20}=2^{20}>100 \cdot 2016 .

$$
Subfield: Geometry


Question: A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the $\operatorname{arc} A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$.
Prove that $R \geq r_{1}+r_{2}+r_{3}$.

Solution:
Let $\ell_{1}$ be the tangent to $\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\omega_{1}$. The tangents $\ell_{2}$ and $\ell_{3}$ are defined similarly. The lines $\ell_{1}$ and $\ell_{2}, \ell_{2}$ and $\ell_{3}, \ell_{3}$ and $\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.



Each of the triangles $\Delta_{1}=\triangle X S_{1} T_{1}, \Delta_{2}=\triangle T_{2} X S_{2}$, and $\Delta_{3}=\triangle S_{3} T_{3} X$ is similar to $\Delta=\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\Delta_{i}$ and $\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\rho_{i}$ is the inradius of $\Delta_{i}$, then $\rho_{1}+\rho_{2}+\rho_{3}=R$.



Finally, notice that $\omega_{i}$ is interior to $\Delta_{i}$, so $r_{i} \leq \rho_{i}$, and the conclusion follows by the preceding.

<img_3506>



Fig. 1



<img_3720>



Fig. 2
Subfield: Geometry


Question: A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the $\operatorname{arc} A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$.
If $R=r_{1}+r_{2}+r_{3}$, prove that the six points where the circles $\omega_{i}$ touch the diagonals $A_{1} B_{2}, A_{2} B_{3}, A_{3} B_{1}$ are concyclic.

Solution:
Let $\ell_{1}$ be the tangent to $\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\omega_{1}$. The tangents $\ell_{2}$ and $\ell_{3}$ are defined similarly. The lines $\ell_{1}$ and $\ell_{2}, \ell_{2}$ and $\ell_{3}, \ell_{3}$ and $\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.



Each of the triangles $\Delta_{1}=\triangle X S_{1} T_{1}, \Delta_{2}=\triangle T_{2} X S_{2}$, and $\Delta_{3}=\triangle S_{3} T_{3} X$ is similar to $\Delta=\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\Delta_{i}$ and $\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\rho_{i}$ is the inradius of $\Delta_{i}$, then $\rho_{1}+\rho_{2}+\rho_{3}=R$.



Finally, notice that $\omega_{i}$ is interior to $\Delta_{i}$, so $r_{i} \leq \rho_{i}$, and the conclusion follows by the preceding.

<img_3506>



Fig. 1



<img_3720>



Fig. 2

By above, the equality $R=r_{1}+r_{2}+r_{3}$ holds if and only if $r_{i}=\rho_{i}$ for all $i$, which implies in turn that $\omega_{i}$ is the incircle of $\Delta_{i}$. Let $K_{i}, L_{i}, M_{i}$ be the points where $\omega_{i}$ touches the sides $X S_{i}, X T_{i}, S_{i} T_{i}$, respectively. We claim that the six points $K_{i}$ and $L_{i}(i=1,2,3)$ are equidistant from $X$.



Clearly, $X K_{i}=X L_{i}$, and we are to prove that $X K_{2}=X L_{1}$ and $X K_{3}=X L_{2}$. By similarity, $\angle T_{1} M_{1} L_{1}=\angle C_{3} M_{1} M_{2}$ and $\angle S_{2} M_{2} K_{2}=\angle C_{3} M_{2} M_{1}$, so the points $M_{1}, M_{2}, L_{1}, K_{2}$ are collinear. Consequently, $\angle X K_{2} L_{1}=\angle C_{3} M_{1} M_{2}=\angle C_{3} M_{2} M_{1}=\angle X L_{1} K_{2}$, so $X K_{2}=X L_{1}$. Similarly, $X K_{3}=X L_{2}$.
Subfield: Algebra


Question: Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$.
Prove that there are infinitely many positive integers $n$ such that $\lambda(n)=\lambda(n+1)=+1$

Solution:
Notice that we have $\Omega(m n)=\Omega(m)+\Omega(n)$ for all positive integers $m, n$ ( $\Omega$ is a completely additive arithmetic function), translating into $\lambda(m n)=\lambda(m) \cdot \lambda(n)$ (so $\lambda$ is a completely multiplicative arithmetic function), hence $\lambda(p)=-1$ for any prime $p$, and $\lambda\left(k^{2}\right)=\lambda(k)^{2}=+1$ for all positive integers $k$.



The start (first 100 terms) of the sequence $\mathfrak{S}=(\lambda(n))_{n \geq 1}$ is
+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1

The Pell equation $x^{2}-6 y^{2}=1$ has infinitely many solutions in positive integers; all solutions are given by $\left(x_{n}, y_{n}\right)$, where $x_{n}+y_{n} \sqrt{6}=(5+2 \sqrt{6})^{n}$. Since $\lambda\left(6 y^{2}\right)=1$ and also $\lambda\left(6 y^{2}+1\right)=\lambda\left(x^{2}\right)=1$, the thesis is proven.
Notice that we have $\Omega(m n)=\Omega(m)+\Omega(n)$ for all positive integers $m, n$ ( $\Omega$ is a completely additive arithmetic function), translating into $\lambda(m n)=\lambda(m) \cdot \lambda(n)$ (so $\lambda$ is a completely multiplicative arithmetic function), hence $\lambda(p)=-1$ for any prime $p$, and $\lambda\left(k^{2}\right)=\lambda(k)^{2}=+1$ for all positive integers $k$.



The start (first 100 terms) of the sequence $\mathfrak{S}=(\lambda(n))_{n \geq 1}$ is
+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1

Take any existing pair with $\lambda(n)=$ $\lambda(n+1)=1$. Then $\lambda\left((2 n+1)^{2}-1\right)=\lambda\left(4 n^{2}+4 n\right)=\lambda(4) \cdot \lambda(n)$. $\lambda(n+1)=1$, and also $\lambda\left((2 n+1)^{2}\right)=\lambda(2 n+1)^{2}=1$, so we have built a larger $(1,1)$ pair.
Subfield: Combinatorics


Question: For every $n \geq 3$, determine all the configurations of $n$ distinct points $X_{1}, X_{2}, \ldots, X_{n}$ in the plane, with the property that for any pair of distinct points $X_{i}, X_{j}$ there exists a permutation $\sigma$ of the integers $\{1, \ldots, n\}$, such that $\mathrm{d}\left(X_{i}, X_{k}\right)=\mathrm{d}\left(X_{j}, X_{\sigma(k)}\right)$ for all $1 \leq k \leq n$.



(We write $\mathrm{d}(X, Y)$ to denote the distance between points $X$ and $Y$.)

Solution:
Let us first prove that the points must be concyclic. Assign to each point $X_{k}$ the vector $x_{k}$ in a system of orthogonal coordinates whose origin is the point of mass of the configuration, thus $\frac{1}{n} \sum_{k=1}^{n} x_{k}=0$.



Then $\mathrm{d}^{2}\left(X_{i}, X_{k}\right)=\left\|x_{i}-x_{k}\right\|^{2}=\left\langle x_{i}-x_{k}, x_{i}-x_{k}\right\rangle=$ $\left\|x_{i}\right\|^{2}-2\left\langle x_{i}, x_{k}\right\rangle+\left\|x_{k}\right\|^{2}$, hence $\sum_{k=1}^{n} \mathrm{~d}^{2}\left(X_{i}, X_{k}\right)=n\left\|x_{i}\right\|^{2}-2\left\langle x_{i}, \sum_{k=1}^{n} x_{k}\right\rangle+\sum_{k=1}^{n}\left\|x_{k}\right\|^{2}=n\left\|x_{i}\right\|^{2}+\sum_{k=1}^{n}\left\|x_{k}\right\|^{2}=n\left\|x_{j}\right\|^{2}+$ $\sum_{k=1}^{n}\left\|x_{\sigma(k)}\right\|^{2}=\sum_{k=1}^{n} \mathrm{~d}^{2}\left(X_{j}, X_{\sigma(k)}\right)$, therefore $\left\|x_{i}\right\|=\left\|x_{j}\right\|$ for all pairs $(i, j)$. The points are thus concyclic (lying on a circle centred at $O(0,0))$.



Let now $m$ be the least angular distance between any two points. Two points situated at angular distance $m$ must be adjacent on the circle. Let us connect each pair of such two points with an edge. The graph $G$ obtained must be regular, of degree $\operatorname{deg}(G)=1$ or 2 . If $n$ is odd, since $\sum_{k=1}^{n} \operatorname{deg}\left(X_{k}\right)=$ $n \operatorname{deg}(G)=2|E|$, we must have $\operatorname{deg}(G)=2$, hence the configuration is a regular $n$-gon.



If $n$ is even, we may have the configuration of a regular $n$-gon, but we also may have $\operatorname{deg}(G)=1$. In that case, let $M$ be the next least angular distance between any two points; such points must also be adjacent on the circle. Let us connect each pair of such two points with an edge, in order to get a graph $G^{\prime}$. A similar reasoning yields $\operatorname{deg}\left(G^{\prime}\right)=1$, thus the configuration is that of an equiangular $n$-gon (with alternating equal side-lengths).

Final answer: the configuration is that of an equiangular $n$-gon when $n$ is even and the configuration is a regular $n$-gon when $n$ is odd
Subfield: Combinatorics


Question: Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6 n$, and $60^{\circ}$ rotational symmetry (that is, there is a point $O$ such that a $60^{\circ}$ rotation about $O$ maps the polygon to itself). In light of the pandemic, the government of Zoomtopia would like to relocate its $3 n^{2}+3 n+1$ citizens at $3 n^{2}+3 n+1$ points in the kingdom so that every two citizens have a distance of at least 1 for proper social distancing. Prove that this is possible. (The kingdom is assumed to contain its boundary.)

Solution:
Let $P$ denote the given polygon, i.e., the kingdom of Zoomtopia. Throughout the solution, we interpret polygons with integer sides and perimeter $6 k$ as $6 k$-gons with unit sides (some of their angles may equal $180^{\circ}$ ). The argument hinges on the claim below:



Claim. Let $P$ be a convex polygon satisfying the problem conditions - i.e., it has integer sides, perimeter $6 n$, and $60^{\circ}$ rotational symmetry. Then $P$ can be tiled with unit equilateral triangles and unit lozenges with angles at least $60^{\circ}$, with tiles meeting completely along edges, so that the tile configuration has a total of exactly $3 n^{2}+3 n+1$ distinct vertices.



Proof. Induct on $n$. The base case, $n=1$, is clear.



Now take a polygon $P$ of perimeter $6 n \geq 12$. Place six equilateral triangles inwards on six edges corresponding to each other upon rotation at $60^{\circ}$. It is possible to stick a lozenge to each other edge, as shown in the Figure below.



We show that all angles of the lozenges are at least $60^{\circ}$. Let an edge $X Y$ of the polygon bearing some lozenge lie along a boundary segment between edges $A B$ and $C D$ bearing equilateral triangles $A B P$ and $C D Q$. Then the angle formed by $\overrightarrow{X Y}$ and $\overrightarrow{B P}$ is between those formed by $\overrightarrow{A B}, \overrightarrow{B P}$ and $\overrightarrow{C D}, \overrightarrow{C Q}$, i.e., between $60^{\circ}$ and $120^{\circ}$, as desired.



Removing all obtained tiles, we get a $60^{\circ}$-symmetric convex $6(n-1)$-gon with unit sides which can be tiled by the inductive hypothesis. Finally, the number of vertices in the tiling of $P$ is $6 n+3(n-1)^{2}+3(n-1)+1=3 n^{2}+3 n+1$, as desired.



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Using the Claim above, we now show that the citizens may be placed at the $3 n^{2}+3 n+1$ tile vertices.



Consider any tile $T_{1}$; its vertices are at least 1 apart from each other. Moreover, let $B A C$ be a part of the boundary of some tile $T$, and let $X$ be any point of the boundary of $T$, lying outside the half-open intervals $[A, B)$ and $[A, C)$ (in this case, we say that $X$ is not adjacent to $A$ ). Then $A X \geq \sqrt{3} / 2$.



Now consider any two tile vertices $A$ and $B$. If they are vertices of the same tile we already know $A B \geq 1$; otherwise, the segment $A B$ crosses the boundaries of some tiles containing $A$ and $B$ at some points $X$ and $Y$ not adjacent to $A$ and $B$, respectively. Hence $A B \geq A X+Y B \geq \sqrt{3}>1$.
Subfield: Algebra


Question: For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime.

Solution:
The largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \ldots, y_{k}$ are primes for some $k \geq 1$ then $a=x_{1}$ is also prime.



Now we claim that for every odd prime $a$ at least one of the numbers $y_{1}, y_{2}, y_{3}$ is composite (and thus $k<3$ ). Assume, to the contrary, that $y_{1}, y_{2}$, and $y_{3}$ are primes; then $x_{1}, x_{2}, x_{3}$ are primes as well. Since $x_{1} \geq 3$ is odd, we have $x_{2}>3$ and $x_{2} \equiv 3(\bmod 4)$; consequently, $x_{3} \equiv 7$ $(\bmod 8)$. This implies that 2 is a quadratic residue modulo $p=x_{3}$, so $2 \equiv s^{2}(\bmod p)$ for some integer $s$, and hence $2^{x_{2}}=2^{(p-1) / 2} \equiv s^{p-1} \equiv 1(\bmod p)$. This means that $p \mid y_{2}$, thus $2^{x_{2}}-1=x_{3}=2 x_{2}+1$. But it is easy to show that $2^{t}-1>2 t+1$ for all integer $t>3$. A contradiction.



Finally, if $a=2$, then the numbers $y_{1}=3$ and $y_{2}=31$ are primes, while $y_{3}=2^{11}-1$ is divisible by 23 ; in this case we may choose $k=2$ but not $k=3$.

Final answer: 2