problem: Let $A B C$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $G B$ and $G C$, respectively, such that  $$ \angle A B S=\angle A C R=180^{\circ}-\angle B G C . $$  Prove that $\angle R A S+\angle B A C=\angle B G C$.

solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-04.jpg?height=718&width=1192&top_left_y=1069&top_left_x=432) 【 Solution 1 using power of a point. From the given condition that $\measuredangle A C R=\measuredangle C G M$, we get that  $$ M A^{2}=M C^{2}=M G \cdot M R \Longrightarrow \measuredangle R A C=\measuredangle M G A $$  Analogously,  $$ \measuredangle B A S=\measuredangle A G N $$  Hence,  $$ \measuredangle R A S+\measuredangle B A C=\measuredangle R A C+\measuredangle B A S=\measuredangle M G A+\measuredangle A G N=\measuredangle M G N=\measuredangle B G C . $$


solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-04.jpg?height=718&width=1192&top_left_y=1069&top_left_x=432) ब Solution 2 using similar triangles. As before, $\triangle M G C \sim \triangle M C R$ and $\triangle N G B \sim$ $\triangle N B S$. We obtain  $$ \frac{|A C|}{|C R|}=\frac{2|M C|}{|C R|}=\frac{2|M G|}{|G C|}=\frac{|G B|}{2|N G|}=\frac{|B S|}{2|B N|}=\frac{|B S|}{|A B|} $$  which together with $\angle A C R=\angle A B S$ yields  $$ \triangle A C R \sim \triangle S B A \Longrightarrow \measuredangle B A S=\measuredangle C R A $$  Hence  $$ \measuredangle R A S+\measuredangle B A C=\measuredangle R A C+\measuredangle B A S=\measuredangle R A C+\measuredangle C R A=-\measuredangle A C R=\measuredangle B G C, $$  which proves the statement.


solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-04.jpg?height=718&width=1192&top_left_y=1069&top_left_x=432) I Solution 3 using parallelograms. Let $M$ and $N$ be defined as above. Let $P$ be the reflection of $G$ in $M$ and let $Q$ the reflection of $G$ in $N$. Then $A G C P$ and $A G B Q$ are parallelograms. ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-05.jpg?height=694&width=1194&top_left_y=681&top_left_x=434)  Claim - Quadrilaterals $A P C R$ and $A Q B S$ are concyclic.  Thus from $\overline{P C} \| \overline{G A}$ we get  $$ \measuredangle R A C=\measuredangle R P C=\measuredangle G P C=\measuredangle P G A $$  and similarly  $$ \measuredangle B A S=\measuredangle B Q S=\measuredangle B Q G=\measuredangle A G Q $$  We conclude that  $$ \measuredangle R A S+\measuredangle B A C=\measuredangle R A C+\measuredangle B A S=\measuredangle P G A+\measuredangle A G Q=\measuredangle P G Q=\measuredangle B G C $$


solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-04.jpg?height=718&width=1192&top_left_y=1069&top_left_x=432) ब Solution 4 also using parallelograms, by Ankan Bhattacharya. Construct parallelograms $A R C K$ and $A S B L$. Since  $$ \measuredangle C A K=\measuredangle A C R=\measuredangle C G B=\measuredangle C G K, $$  it follows that $A G C K$ is cyclic. Similarly, $A G B L$ is also cyclic. ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-06.jpg?height=660&width=1198&top_left_y=241&top_left_x=429)  Finally, observe that  $$ \begin{aligned} \angle R A S+\angle B A C & =\measuredangle B A S+\measuredangle R A C \\ & =\measuredangle A B L+\measuredangle K C A \\ & =\measuredangle A G L+\measuredangle K G A \\ & =\measuredangle K G L \\ & =\angle B G C \end{aligned} $$  as requested.


solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-04.jpg?height=718&width=1192&top_left_y=1069&top_left_x=432) 『 Solution 5 using complex numbers, by Milan Haiman. Note that $\angle R A S+\angle B A C=$ $\angle B A S+\angle R A C$. We compute $\angle B A S$ in complex numbers; then $\angle R A C$ will then be known by symmetry.  Let $a, b, c$ be points on the unit circle representing $A, B, C$ respectively. Let $g=\frac{1}{3}(a+b+c)$ represent the centroid $G$, and let $s$ represent $S$.  ## Claim - We have  $$ \frac{s-a}{b-a}=\frac{a b-2 b c+c a}{2 a b-b c-c a} . $$   $$ s=\frac{a+b}{2}+t(c-g) $$  for some $t \in \mathbb{R}$. By the given angle condition, we have that  $$ \frac{(s-b) /(b-a)}{(c-g) /(g-b)} \in \mathbb{R} $$  Note that  $$ \frac{s-b}{b-a}=t \frac{c-g}{b-a}-\frac{1}{2} $$  So,  $$ t \frac{g-b}{b-a}-\frac{g-b}{2(c-g)} \in \mathbb{R} $$  Thus  $$ t=\frac{\operatorname{Im}\left(\frac{g-b}{2(c-g)}\right)}{\operatorname{Im}\left(\frac{g-b}{b-a}\right)}=\frac{1}{2} \cdot \frac{\left(\frac{g-b}{c-g}\right)-\overline{\left(\frac{g-b}{c-g}\right)}}{\left(\frac{g-b}{b-a}\right)-\overline{\left(\frac{g-b}{b-a}\right)}} $$  Let $N$ and $D$ be the numerator and denominator of the second factor above. We want to compute  $$ \frac{s-a}{b-a}=\frac{1}{2}+t \frac{c-g}{b-a}=\frac{(b-a)+2 t(c-g)}{2(b-a)}=\frac{(b-a) D+(c-g) N}{2(b-a) D} $$  We have  $$ \begin{aligned} (c-g) N & =g-b-(c-g) \overline{\left(\frac{g-b}{c-g}\right)} \\ & =\frac{a+b+c}{3}-b-\left(c-\frac{a+b+c}{3}\right) \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{b}}{\frac{3}{c}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}} \\ & =\frac{(a+c-2 b)(2 a b-b c-c a)-(2 c-a-b)(a b+b c-2 c a)}{3(2 a b-b c-c a)} \\ & =\frac{3\left(a^{2} b+b^{2} c+c^{2} a-a b^{2}-b c^{2}-c a^{2}\right)}{3(2 a b-b c-c a)} \\ & =\frac{(a-b)(b-c)(a-c)}{2 a b-b c-c a} \end{aligned} $$  We also compute  $$ \begin{aligned} (b-a) D & =g-b-(b-a) \overline{\left(\frac{g-b}{b-a}\right)} \\ & =\frac{a+b+c}{3}-b-(b-a) \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{b}}{\frac{3}{b}-\frac{3}{a}} \\ & =\frac{(a+c-2 b) c+(a b+b c-2 c a)}{3 c} \\ & =\frac{a b-b c-c a+c^{2}}{3 c} \\ & =\frac{(a-c)(b-c)}{3 c} \end{aligned} $$  So, we obtain  $$ \frac{s-a}{b-a}=\frac{\frac{1}{3 c}+\frac{a-b}{2 a b-b c-c a}}{\frac{2}{3 c}}=\frac{2 a b-b c-c a+3 c(a-b)}{2(2 a b-b c-c a)}=\frac{a b-2 b c+c a}{2 a b-b c-c a} $$  By symmetry,  $$ \frac{r-a}{c-a}=\frac{a b-2 b c+c a}{2 c a-a b-b c} $$  Hence their ratio  $$ \frac{s-a}{b-a} \div \frac{r-a}{c-a}=\frac{2 a b-b c-c a}{2 c a-a b-b c} $$  has argument $\angle R A C+\angle B A S$.  We also have that $\angle B G C$ is the argument of  $$ \frac{b-g}{c-g}=\frac{2 b-a-c}{2 c-a-b} $$  Note that these two complex numbers are inverse-conjugates, and thus have the same argument. So we're done.


problem: Let $n \geq m \geq 1$ be integers. Prove that  $$ \sum_{k=m}^{n}\left(\frac{1}{k^{2}}+\frac{1}{k^{3}}\right) \geq m \cdot\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2} $$

solution: 【 First solution (authors). By Cauchy-Schwarz, we have  $$ \begin{aligned} \sum_{k=m}^{n} \frac{k+1}{k^{3}} & =\sum_{k=m}^{n} \frac{\left(\frac{1}{k^{2}}\right)^{2}}{\frac{1}{k(k+1)}} \\ & \geq \frac{\left(\frac{1}{m^{2}}+\frac{1}{(m+1)^{2}}+\cdots+\frac{1}{n^{2}}\right)^{2}}{\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}+\cdots+\frac{1}{n(n+1)}} \\ & =\frac{\left(\frac{1}{m^{2}}+\frac{1}{(m+1)^{2}}+\cdots+\frac{1}{n^{2}}\right)^{2}}{\frac{1}{m}-\frac{1}{n+1}} \\ & >\frac{\left(\sum_{k=m}^{n} \frac{1}{k^{2}}\right)^{2}}{\frac{1}{m}} \end{aligned} $$  as desired. Remark (Bound on error). Let $A=\sum_{k=m}^{n} k^{-2}$ and $B=\sum_{k=m}^{n} k^{-3}$. The inequality above becomes tighter for large $m$ and $n \gg m$. If we use Lagrange's identity in place of Cauchy-Schwarz, we get  $$ A+B-m A^{2}=m \cdot \sum_{m \leq a<b} \frac{(a-b)^{2}}{a^{3} b^{3}(a+1)(b+1)} $$  We can upper bound this error by  $$ \leq m \cdot \sum_{m \leq a<b} \frac{1}{a^{3}(a+1) b(b+1)}=m \cdot \sum_{m \leq a} \frac{1}{a^{3}(a+1)^{2}} \approx m \cdot \frac{1}{m^{4}}=\frac{1}{m^{3}}, $$  which is still generous as $(a-b)^{2} \ll b^{2}$ for $b$ not much larger than $a$, so the real error is probably around $\frac{1}{10 \mathrm{~m}^{3}}$. This exhibits the tightness of the inequality since it implies  $$ m A^{2}+O(B / m)>A+B $$  Remark (Construction commentary, from author). My motivation was to write an inequality where Titu could be applied creatively to yield a telescoping sum. This can be difficult because most of the time, such a reverse-engineered inequality will be so loose it's trivial anyways. My first attempt was the not-so-amazing inequality  $$ \frac{n^{2}+3 n}{2}=\sum_{1}^{n} i+1=\sum_{1}^{n} \frac{\frac{1}{i}}{\frac{1}{i(i+1)}}>\left(\sum_{1}^{n} \frac{1}{\sqrt{i}}\right)^{2} $$  which is really not surprising given that $\sum \frac{1}{\sqrt{i}} \ll \frac{n}{\sqrt{2}}$. The key here is that we need "near-equality" as dictated by the Cauchy-Schwarz equality case, i.e. the square root of the numerators should be approximately proportional to the denominators.  This motivates using $\frac{1}{i^{4}}$ as the numerator, which works like a charm. After working out the resulting statement, the LHS and RHS even share a sum, which adds to the simplicity of the problem.  The final touch was to unrestrict the starting value of the sum, since this allows the strength of the estimate $\frac{1}{i^{2}} \approx \frac{1}{i(i+1)}$ to be fully exploited.


solution: 【 Second approach by inducting down, Luke Robitaille and Carl Schildkraut. Fix $n$; we'll induct downwards on $m$. For the base case of $n=m$ the result is easy, since the left side is $\frac{m+1}{m^{3}}$ and the right side is $\frac{m}{m^{4}}=\frac{1}{m^{3}}$.  For the inductive step, suppose we have shown the result for $m+1$. Let  $$ A=\sum_{k=m+1}^{n} \frac{1}{k^{2}} \quad \text { and } \quad B=\sum_{k=m+1}^{n} \frac{1}{k^{3}} $$  We know $A+B \geq(m+1) A^{2}$, and we want to show  $$ \left(A+\frac{1}{m^{2}}\right)+\left(B+\frac{1}{m^{3}}\right) \geq m\left(A+\frac{1}{m^{2}}\right)^{2} $$  Indeed,  $$ \begin{aligned} \left(A+\frac{1}{m^{2}}\right)+\left(B+\frac{1}{m^{3}}\right)-m\left(A+\frac{1}{m^{2}}\right)^{2} & =A+B+\frac{m+1}{m^{3}}-m A^{2}-\frac{2 A}{m}-\frac{1}{m^{3}} \\ & =\left(A+B-(m+1) A^{2}\right)+\left(A-\frac{1}{m}\right)^{2} \geq 0 \end{aligned} $$  and we are done.


solution: I Third approach by reducing $n \rightarrow \infty$, Michael Ren and Carl Schildkraut. First, we give:  Claim (Reduction to $n \rightarrow \infty$ ) - If the problem is true when $n \rightarrow \infty$, it is true for all $n$.   However, the region is bounded by a convex curve, and the sequence of points $(0,0)$, $\left(\frac{1}{m^{2}}, \frac{1}{m^{3}}\right),\left(\frac{1}{m^{2}}+\frac{1}{(m+1)^{2}}, \frac{1}{m^{3}}+\frac{1}{(m+1)^{3}}\right), \ldots$ has successively decreasing slopes between consecutive points. Thus it suffices to check that the inequality is true when $n \rightarrow \infty$.  Set $n=\infty$ henceforth. Let  $$ A=\sum_{k=m}^{\infty} \frac{1}{k^{2}} \text { and } B=\sum_{k=m}^{\infty} \frac{1}{k^{3}} $$  we want to show $B \geq m A^{2}-A$, which rearranges to  $$ 1+4 m B \geq(2 m A-1)^{2} $$  Write  $$ C=\sum_{k=m}^{\infty} \frac{1}{k^{2}(2 k-1)(2 k+1)} \text { and } D=\sum_{k=m}^{\infty} \frac{8 k^{2}-1}{k^{3}(2 k-1)^{2}(2 k+1)^{2}} $$  Then  $$ \frac{2}{2 k-1}-\frac{2}{2 k+1}=\frac{1}{k^{2}}+\frac{1}{k^{2}(2 k-1)(2 k+1)} $$  and  $$ \frac{2}{(2 k-1)^{2}}-\frac{2}{(2 k+1)^{2}}=\frac{1}{k^{3}}+\frac{8 k^{2}-1}{k^{3}(2 k-1)^{2}(2 k+1)^{2}}, $$  so that  $$ A=\frac{2}{2 m-1}-C \text { and } B=\frac{2}{(2 m-1)^{2}}-D $$  Our inequality we wish to show becomes  $$ \frac{2 m+1}{2 m-1} C \geq D+m C^{2} $$  We in fact show two claims: Claim - We have  $$ \frac{2 m+1 / 2}{2 m-1} C \geq D $$   $$ \frac{2 m+1 / 2}{2 m-1} \cdot \frac{1}{k^{2}(2 k-1)(2 k+1)} \geq \frac{8 k^{2}-1}{k^{3}(2 k-1)^{2}(2 k+1)^{2}} $$  for $k \geq m$. It suffices to show  $$ \frac{2 k+1 / 2}{2 k-1} \cdot \frac{1}{k^{2}(2 k-1)(2 k+1)} \geq \frac{8 k^{2}-1}{k^{3}(2 k-1)^{2}(2 k+1)^{2}} $$  which is equivalent to $k(2 k+1 / 2)(2 k+1) \geq 8 k^{2}-1$. This holds for all $k \geq 1$.  ## Claim - We have  $$ \frac{1 / 2}{2 m-1} C \geq m C^{2} $$   $$ \frac{1}{2 m(2 m-1)}=\sum_{k=m}^{\infty}\left(\frac{1}{2 k(2 k-1)}-\frac{1}{2(k+1)(2 k+1)}\right)=\sum_{k=m}^{\infty} \frac{4 k+1}{2 k(2 k-1)(k+1)(2 k+1)} $$  comparing term-wise with the definition of $C$ and using the inequality $k(4 k+1) \geq 2(k+1)$ for $k \geq 1$ gives the desired result.


solution: 『 Fourth approach by bashing, Carl Schildkraut. With a bit more work, the third approach can be adapted to avoid the $n \rightarrow \infty$ reduction. Similarly to before, define  $$ A=\sum_{k=m}^{n} \frac{1}{k^{2}} \text { and } B=\sum_{k=m}^{n} \frac{1}{k^{3}} $$  we want to show $1+4 m B \geq(2 m A-1)^{2}$. Writing  $$ C=\sum_{k=m}^{n} \frac{1}{k^{2}(2 k-1)(2 k+1)} \text { and } D=\sum_{k=m}^{n} \frac{8 k^{2}-1}{k^{3}(2 k-1)^{2}(2 k+1)^{2}} $$  We compute  $$ A=\frac{2}{2 m-1}-\frac{2}{2 n+1}-C \text { and } B=\frac{2}{(2 m-1)^{2}}-\frac{2}{(2 n+1)^{2}}-D . $$  Then, the inequality we wish to show reduces (as in the previous solution) to  $$ \frac{2 m+1}{2 m-1} C+\frac{2(2 m+1)}{(2 m-1)(2 n+1)} \geq D+m C^{2}+\frac{2(2 m+1)}{(2 n+1)^{2}}+\frac{4 m}{2 n+1} C $$  We deal first with the terms not containing the variable $n$, i.e. we show that  $$ \frac{2 m+1}{2 m-1} C \geq D+m C^{2} $$  For this part, the two claims from the previous solution go through exactly as written above, and we have $C \leq 1 /(2 m(2 m-1))$. We now need to show  $$ \frac{2(2 m+1)}{(2 m-1)(2 n+1)} \geq \frac{2(2 m+1)}{(2 n+1)^{2}}+\frac{4 m}{2 n+1} C $$  (this is just the inequality between the remaining terms); our bound on $C$ reduces this to proving  $$ \frac{4(2 m+1)(n-m+1)}{(2 m-1)(2 n+1)^{2}} \geq \frac{2}{(2 m-1)(2 n+1)} $$  Expanding and writing in terms of $n$, this is equivalent to  $$ n \geq \frac{1+2(m-1)(2 m+1)}{4 m}=m-\frac{2 m+1}{4 m} $$  which holds for all $n \geq m$.


problem: Find all positive integers $n$ for which it is possible to color some cells of an infinite grid of unit squares red, such that each rectangle consisting of exactly $n$ cells (and whose edges lie along the lines of the grid) contains an odd number of red cells.

solution: We claim that this is possible for all positive integers $n$. Call a positive integer for which such a coloring is possible good. To show that all positive integers $n$ are good we prove the following: (i) If $n$ is good and $p$ is an odd prime, then $p n$ is good; (ii) For every $k \geq 0$, the number $n=2^{k}$ is good.  Together, (i) and (ii) imply that all positive integers are good.  Thus every coloring that works for $n$ automatically also works for $p n$.  Claim - For each of these $k+1$ shapes, there exists a coloring with two properties:  - Every rectangle with $n$ cells and shape $2^{m} \times 2^{k-m}$ contains an odd number of red cells. - Every rectangle with $n$ cells and a different shape contains an even number of red cells.  ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-13.jpg?height=395&width=401&top_left_y=2084&top_left_x=833)  A $2^{m} \times 2^{k-m}$ rectangle contains every possible pair $\left(x \bmod 2^{m}, y \bmod 2^{k-m}\right)$ exactly once, so such a rectangle will contain one red cell (an odd number).  On the other hand, consider a $2^{\ell} \times 2^{k-\ell}$ rectangle with $\ell>m$. The set of cells this covers is $(x, y)$ where $x$ covers a range of size $2^{\ell}$ and $y$ covers a range of size $2^{k-\ell}$. The number of red cells is the count of $x$ with $x \equiv 0 \bmod 2^{m}$ multiplied by the count of $y$ with $y \equiv 0 \bmod 2^{k-m}$. The former number is exactly $2^{\ell-k}$ because $2^{k}$ divides $2^{\ell}$ (while the latter is 0 or 1) so the number of red cells is even. The $\ell<m$ case is similar.  Finally, given these $k+1$ colorings, we can add them up modulo 2, i.e. a cell will be colored red if it is red in an odd number of these $k+1$ coloring. We illustrate $n=4$ as an example; the coloring is 4 -periodic in both axes so we only show one $4 \times 4$ cell. ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-14.jpg?height=215&width=1103&top_left_y=675&top_left_x=479)  This solves the problem. Remark. The final coloring can be described as follows: color $(x, y)$ red if  $$ \max \left(0, \min \left(\nu_{2}(x), k\right)+\min \left(\nu_{2}(y), k\right)-k+1\right) $$  is odd.  Remark (Luke Robitaille). Alternatively for (i), if $n=2^{e} k$ for odd $k$ then one may dissect an $a \times b$ rectangle with area $n$ into $k$ rectangles of area $2^{e}$, each $2^{\nu_{2}(a)} \times 2^{\nu_{2}(b)}$. This gives a way to deduce the problem from (ii) without having to consider odd prime numbers.   $$ f(x, y)=\sum_{i=0}^{2^{k}-1} \sum_{j=0}^{2^{k}-1} \lambda_{i, j} x^{i} y^{j} $$  denote its generating function, where $f \in \mathbb{F}_{2}[x, y]$. For this to be valid, we need that for any $2^{p} \times 2^{q}$ rectangle with area $n$, the sum of the coefficients of $f$ over it should be one, modulo $x^{2^{k}}=y^{2^{k}}=1$. In other words, whenever $p+q=k$, we must have  $$ f(x, y)\left(1+\cdots+x^{2^{p}-1}\right)\left(1+\cdots+y^{2^{q}-1}\right)=\left(1+\cdots+x^{2^{k}-1}\right)\left(1+\cdots+y^{2^{k}-1}\right) $$  taken modulo $x^{2^{k}}=y^{2^{k}}=1$. The idea is to rewrite these expressions: because we're in characteristic 2, the given assertion is $(x+1)^{2^{k}}=(y+1)^{2^{k}}=0$, and the requested property is  $$ f(x, y)(x+1)^{2^{p}-1}(y+1)^{2^{q}-1}=(x+1)^{2^{k}-1}(y+1)^{2^{k}-1} . $$  This suggests the substitution $g(x, y)=f(x+1, y+1)$ : then we can replace $(x+1, y+1) \mapsto$ $(x, y)$ to simplify the requested property significantly:  Whenever $p+q=k$, we must have  $$ g(x, y) x^{2^{p}-1} y^{2^{q}-1}=x^{2^{k}-1} y^{2^{k}-1} $$  modulo $x^{2^{k}}$ and $y^{2^{k}}$.  However, now the construction of $g$ is very simple: for example, the choice  $$ g(x, y)=\sum_{p+q=k} x^{2^{k}-2^{p}} y^{2^{k}-2^{q}} $$  works. The end. Remark. Unraveling the substitutions seen here, it's possible to show that this is actually the same construction provided in the first solution.


problem: Let $n \geq 3$ be an integer and let $K_{n}$ be the complete graph on $n$ vertices. Each edge of $K_{n}$ is colored either red, green, or blue. Let $A$ denote the number of triangles in $K_{n}$ with all edges of the same color, and let $B$ denote the number of triangles in $K_{n}$ with all edges of different colors. Prove that  $$ B \leq 2 A+\frac{n(n-1)}{3} . $$

solution: $  Consider all unordered pairs of different edges which share exactly one vertex (call these vees for convenience). Assign each vee a charge of +2 if its edge colors are the same, and a charge of -1 otherwise.  We compute the total charge in two ways.  ## 【 Total charge by summing over triangles. Note that  - each monochromatic triangle has a charge of +6 , - each bichromatic triangle has a charge of 0 , and - each trichromatic triangle has a charge of -3 .  Since each vee contributes to exactly one triangle, we obtain that the total charge is $6 A-3 B$.  【 Total charge by summing over vertices. We can also calculate the total charge by examining the centers of the vees. If a vertex has $a$ red edges, $b$ green edges, and $c$ blue edges, the vees centered at that vertex contribute a total charge of  $$ \begin{aligned} & 2\left[\binom{a}{2}+\binom{b}{2}+\binom{c}{2}\right]-(a b+a c+b c) \\ = & \left(a^{2}-a+b^{2}-b+c^{2}-c\right)-(a b+a c+b c) \\ = & \left(a^{2}+b^{2}+c^{2}-a b-a c-b c\right)-(a+b+c) \\ = & \left(a^{2}+b^{2}+c^{2}-a b-a c-b c\right)-(n-1) \\ \geq & -(n-1) . \end{aligned} $$  In particular, the total charge is at least $-n(n-1)$. đ Conclusion. Thus, we obtain  $$ 6 A-3 B \geq-n(n-1) \Longleftrightarrow B \leq 2 A+\frac{n(n-1)}{3} $$  as desired.


problem: Suppose $a, b$, and $c$ are three complex numbers with product 1 . Assume that none of $a, b$, and $c$ are real or have absolute value 1 . Define  $$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text { and } \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} . $$  Given that both $p$ and $q$ are real numbers, find all possible values of the ordered pair $(p, q)$.

solution: We show $(p, q)=(-3,3)$ is the only possible ordered pair. ## 【T First solution.   $$ \begin{aligned} p+3 & =3+\sum_{\text {cyc }}\left(\frac{x}{y}+\frac{y}{x}\right)=3+\frac{x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)}{x y z} \\ & =\frac{(x+y+z)(x y+y z+z x)}{x y z} \\ q-3 & =-3+\sum_{\text {cyc }} \frac{y^{2}}{z x}=\frac{x^{3}+y^{3}+z^{3}-3 x y z}{x y z} \\ & =\frac{(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)}{x y z} . \end{aligned} $$  It follows that  $$ \begin{aligned} \mathbb{R} & \ni 3(p+3)+(q-3) \\ & =\frac{(x+y+z)\left(x^{2}+y^{2}+z^{2}+2(x y+y z+z x)\right)}{x y z} \\ & =\frac{(x+y+z)^{3}}{x y z} \end{aligned} $$  Now, note that if $x+y+z=0$, then $p=-3, q=3$ so we are done.  Scale $x, y, z$ in such a way that $x+y+z$ is nonzero and real; hence so is $x y z$. Thus, as $p+3 \in \mathbb{R}$, we conclude $x y+y z+z x \in \mathbb{R}$ as well. Hence, $x, y, z$ are the roots of a cubic with real coefficients. Thus,  - either all three of $\{x, y, z\}$ are real (which implies $a, b, c \in \mathbb{R}$ ), - or two of $\{x, y, z\}$ are a complex conjugate pair (which implies one of $a, b, c$ has absolute value 1). Both of these were forbidden by hypothesis.  Construction As we saw in the setup, $(p, q)=(-3,3)$ will occur as long as $x+y+z=0$, and no two of $x, y, z$ to share the same magnitude or are collinear with the origin. This is easy to do; for example, we could choose $(x, y, z)=(3,4 i,-(3+4 i))$. Hence $a=\frac{3}{4 i}$, $b=-\frac{4 i}{3+4 i}, c=-\frac{3+4 i}{3}$ satisfies the hypotheses of the problem statement.


solution: We show $(p, q)=(-3,3)$ is the only possible ordered pair. I Second solution, found by contestants. The main idea is to make the substitution  $$ x=a+\frac{1}{c}, \quad y=b+\frac{1}{a}, \quad z=c+\frac{1}{b} . $$  Then we can check that  $$ \begin{aligned} x+y+z & =p \\ x y+y z+z x & =p+q+3 \\ x y z & =p+2 . \end{aligned} $$  Therefore $x, y, z$ are the roots of a cubic with real coefficients. As in the previous solution, we note that this cubic must either have all real roots, or a complex conjugate pair of roots. We also have the relation $a(y+1)=a b+a+1=x+1$, and likewise $b(z+1)=y+1, c(x+1)=z+1$. This means that if any of $x, y, z$ are equal to -1, then all are equal to -1 .  Assume for the sake of contradiction that none are equal to -1 . In the case where the cubic has three real roots, $a=\frac{x+1}{y+1}$ would be real. On the other hand, if there is a complex conjugate pair (without loss of generality, $x$ and $y$ ) then $a$ has magnitude 1. Therefore this cannot occur.  We conclude that $x=y=z=-1$, so $p=-3$ and $q=3$. The solutions $(a, b, c)$ can then be parameterized as $\left(a,-1-\frac{1}{a},-\frac{1}{1+a}\right)$. To construct a solution, we need to choose a specific value of $a$ such that none of the wrong conditions hold; when $a=2 i$, say, we obtain the solution $\left(2 i,-1+\frac{i}{2}, \frac{-1+2 i}{5}\right)$.


solution: We show $(p, q)=(-3,3)$ is the only possible ordered pair. 【 Third solution by Luke Robitaille and Daniel Zhu. The answer is $p=-3$ and $q=3$. Let's first prove that no other $(p, q)$ work.  Let $e_{1}=a+b+c$ and $e_{2}=a^{-1}+b^{-1}+c^{-1}=a b+a c+b c$. Also, let $f=e_{1} e_{2}$. Note that $p=e_{1}+e_{2}$.  Our main insight is to consider the quantity $q^{\prime}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$. Note that $f=q+q^{\prime}+3$. Also,  $$ \begin{aligned} q q^{\prime} & =3+\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}+\frac{b c}{a^{2}}+\frac{a c}{b^{2}}+\frac{a b}{c^{2}} \\ & =3+a^{3}+b^{3}+c^{3}+a^{-3}+b^{-3}+c^{-3} \\ & =9+a^{3}+b^{3}+c^{3}-3 a b c+a^{-3}+b^{-3}+c^{-3}-3 a^{-1} b^{-1} c^{-1} \\ & =9+e_{1}\left(e_{1}^{2}-3 e_{2}\right)+e_{2}\left(e_{2}^{2}-3 e_{1}\right) \\ & =9+e_{1}^{3}+e_{2}^{3}-6 e_{1} e_{2} \\ & =9+p\left(p^{2}-3 f\right)-6 f \\ & =p^{3}-(3 p+6) f+9 . \end{aligned} $$  As a result, the quadratic with roots $q$ and $q^{\prime}$ is $x^{2}-(f-3) x+\left(p^{3}-(3 p+6) f+9\right)$, which implies that  $$ q^{2}-q f+3 q+p^{3}-(3 p+6) f+9=0 \Longleftrightarrow(3 p+q+6) f=p^{3}+q^{2}+3 q+9 $$  At this point, two miracles occur. The first is the following claim:  Claim - $f$ is not real.  - $e_{1}$ and $e_{2}$ are real. Then, $a, b$, and $c$ are the roots of $x^{3}-e_{1} x^{2}+e_{2} x-1$, and since every cubic with real coefficients has at least one real root, at least one of $a, b$, and $c$ is real, contradiction. - $e_{1}$ and $e_{2}$ are conjugates. Then, the polynomial $x^{3}-\bar{e}_{2} x^{2}+\bar{e}_{1} x-1$, which has roots $\bar{a}^{-1}, \bar{b}^{-1}$, and $\bar{c}^{-1}$, is the same as the polynomial with $a, b, c$ as roots. We conclude that the multiset $\{a, b, c\}$ is invariant under inversion about the unit circle, so one of $a, b$, and $c$ must lie on the unit circle. This is yet another contradiction.  As a result, we know that $3 p+q+6=p^{3}+q^{2}+3 q+9=0$. The second miracle is that substituting $q=-3 p-6$ into $q^{2}+3 q+p^{3}+9=0$, we get  $$ 0=p^{3}+9 p^{2}+27 p+27=(p+3)^{3} $$  so $p=-3$. Thus $q=3$. It remains to construct valid $a, b$, and $c$. To do this, let's pick some $e_{1}$, let $e_{2}=-3-e_{1}$, and let $a, b$, and $c$ be the roots of $x^{3}-e_{1} x^{2}+e_{2} x-1$. It is clear that this guarantees $p=-3$. By our above calculations, $q$ and $q^{\prime}$ are the roots of the quadratic $x^{2}-(f-3) x+(3 f-18)$, so one of $q$ and $q^{\prime}$ must be 3 ; by changing the order of $a, b$, and $c$ if needed, we can guarantee this to be $q$. It suffices to show that for some choice of $e_{1}$, none of $a, b$, or $c$ are real or lie on the unit circle.  To do this, note that we can rewrite $x^{3}-e_{1} x^{2}+\left(-3-e_{1}\right) x-1=0$ as  $$ e_{1}=\frac{x^{3}-3 x-1}{x^{2}+x} $$  so all we need is a value of $e_{1}$ that is not $\frac{x^{3}-3 x-1}{x^{2}+x}$ for any real $x$ or $x$ on the unit circle. One way to do this is to choose any nonreal $e_{1}$ with $\left|e_{1}\right|<1 / 2$. This clearly rules out any real $x$. Also, if $|x|=1$, by the triangle inequality $\left|x^{3}-3 x-1\right| \geq|3 x|-\left|x^{3}\right|-|1|=1$ and $\left|x^{2}+x\right| \leq 2$, so $\left|\frac{x^{3}-3 x-1}{x^{2}+x}\right| \geq \frac{1}{2}$.


problem: Let $A B C$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle P A Q, \angle P B Q$, and $\angle P C Q$ are the altitudes of triangle $A B C$. Prove that the midpoint of $\overline{P Q}$ lies on the Euler line of $A B C$.

solution: 『 Solution 1 (Ankit Bisain). Let $H$ be the orthocenter of $A B C$, and construct $P^{\prime}$ using the following claim.  Claim - There is a point $P^{\prime}$ for which  $$ \measuredangle A P H+\measuredangle A P^{\prime} H=\measuredangle B P H+\measuredangle B P^{\prime} H=\measuredangle C P H+\measuredangle C P^{\prime} H=0 . $$   Now, let $X, Y$, and $Z$ be the reflections of $P$ over $\overline{A H}, \overline{B H}$, and $\overline{C H}$ respectively. Additionally, let $Q^{\prime}$ be the image of $Q$ under inversion about ( $P X Y Z$ ). ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-20.jpg?height=675&width=989&top_left_y=1399&top_left_x=539)  $$ \text { Claim }-A B C P^{\prime} \approx X Y Z Q^{\prime} $$   $$ \measuredangle Y X Z=\measuredangle Y P Z=\measuredangle(\overline{B H}, \overline{C H})=-\measuredangle B A C $$  and cyclic variants, triangles $A B C$ and $X Y Z$ are similar. Additionally,  $$ \measuredangle H Q^{\prime} X=-\measuredangle H X Q=-\measuredangle H X A=\measuredangle H P A=-\measuredangle H P^{\prime} A $$  and cyclic variants, so summing in pairs gives $\measuredangle Y Q^{\prime} Z=-\measuredangle B P^{\prime} C$ and cyclic variants; this implies the similarity.  Claim - $Q^{\prime}$ lies on the Euler line of triangle $X Y Z$.   To finish the problem, let $G_{1}$ be the centroid of $A B C$ and $G_{2}$ be the centroid of $X Y Z$. Then with signed areas,  $$ \begin{aligned} {\left[G_{1} H P\right]+\left[G_{1} H Q\right] } & =\frac{[A H P]+[B H P]+[C H P]}{3}+\frac{[A H Q]+[B H Q]+[C H Q]}{3} \\ & =\frac{[A H Q]-[A H X]+[B H Q]-[B H Y]+[C H Q]-[C H Z]}{3} \\ & =\frac{[H Q X]+[H Q Y]+[H Q Z]}{3} \\ & =\left[Q G_{2} H\right] \\ & =0 \end{aligned} $$  where the last line follows from the last claim. Therefore $\overline{G_{1} H}$ bisects $\overline{P Q}$, as desired.


solution: Solution 2 using complex numbers (Carl Schildkraut and Milan Haiman). Let $(A B C)$ be the unit circle in the complex plane, and let $A=a, B=b, C=c$ such that $|a|=|b|=|c|=1$. Let $P=p$ and $Q=q$, and $O=0$ and $H=h=a+b+c$ be the circumcenter and orthocenter of $A B C$ respectively.  The first step is to translate the given geometric conditions into a single usable equation:  Claim - We have the equation  $$ (p+q) \sum_{\mathrm{cyc}} a^{3}\left(b^{2}-c^{2}\right)=(\bar{p}+\bar{q}) a b c \sum_{\mathrm{cyc}}\left(b c\left(b^{2}-c^{2}\right)\right) $$   $$ \begin{aligned} & \frac{(p-a)(q-a)}{(h-a)^{2}}=\frac{(p-a)(q-a)}{(b+c)^{2}} \in \mathbb{R} \\ \Longrightarrow & \frac{(p-a)(q-a)}{(b+c)^{2}}=\frac{\left(\frac{(p-a)(q-a)}{(b+c)^{2}}\right)}{}=\frac{(a \bar{p}-1)(a \bar{q}-1) b^{2} c^{2}}{(b+c)^{2} a^{2}} \\ \Longrightarrow & a^{2}(p-a)(q-a)=b^{2} c^{2}(a \bar{p}-1)(a \bar{q}-1) \\ \Longrightarrow & a^{2} p q-a^{2} b^{2} c^{2} \overline{p q}+\left(a^{4}-b^{2} c^{2}\right)=a^{3}(p+q)-a b^{2} c^{2}(\bar{p}+\bar{q}) . \end{aligned} $$  Writing the symmetric conditions that $\overline{B H}$ and $\overline{C H}$ bisect $\angle P B Q$ and $\angle P C Q$ gives three equations:  $$ \begin{aligned} & a^{2} p q-a^{2} b^{2} c^{2} \overline{p q}+\left(a^{4}-b^{2} c^{2}\right)=a^{3}(p+q)-a b^{2} c^{2}(\bar{p}+\bar{q}) \\ & b^{2} p q-a^{2} b^{2} c^{2} \overline{p q}+\left(b^{4}-c^{2} a^{2}\right)=b^{3}(p+q)-b c^{2} a^{2}(\bar{p}+\bar{q}) \end{aligned} $$  $$ c^{2} p q-a^{2} b^{2} c^{2} \overline{p q}+\left(c^{4}-a^{2} b^{2}\right)=c^{3}(p+q)-c a^{2} b^{2}(\bar{p}+\bar{q}) $$  Now, sum $\left(b^{2}-c^{2}\right)$ times the first equation, $\left(c^{2}-a^{2}\right)$ times the second equation, and $\left(a^{2}-b^{2}\right)$ times the third equation. On the left side, the coefficients of $p q$ and $\overline{p q}$ are 0 . Additionally, the coefficient of 1 (the parenthesized terms on the left sides of each equation) sum to 0 , since  $$ \sum_{\mathrm{cyc}}\left(a^{4}-b^{2} c^{2}\right)\left(b^{2}-c^{2}\right)=\sum_{\mathrm{cyc}}\left(a^{4} b^{2}-b^{4} c^{2}-a^{4} c^{2}+c^{4} b^{2}\right) $$  This gives (1) as desired. We can then factor (1): Claim - The left-hand side of (1) factors as  $$ -(p+q)(a-b)(b-c)(c-a)(a b+b c+c a) $$  while the right-hand side factors as  $$ -(\bar{p}+\bar{q})(a-b)(b-c)(c-a)(a+b+c) $$   Consider the cyclic sum on the left as a polynomial in $a, b$, and $c$. If $a=b$, then it simplifies as $a^{3}\left(a^{2}-c^{2}\right)+a^{3}\left(c^{2}-a^{2}\right)+c^{3}\left(a^{2}-a^{2}\right)=0$, so $a-b$ divides this polynomial. Similarly, $a-c$ and $b-c$ divide it, so it can be written as $f(a, b, c)(a-b)(b-c)(c-a)$ for some symmetric quadratic polynomial $f$, and thus it is some linear combination of $a^{2}+b^{2}+c^{2}$ and $a b+b c+c a$. When $a=0$, the whole expression is $b^{2} c^{2}(b-c)$, so $f(0, b, c)=-b c$, which implies that $f(a, b, c)=-(a b+b c+c a)$. Similarly, consider the cyclic sum on the right as a polynomial in $a, b$, and $c$. If $a=b$, then it simplifies as $a c\left(a^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)+a^{2}\left(a^{2}-a^{2}\right)=0$, so $a-b$ divides this polynomial. Similarly, $a-c$ and $b-c$ divide it, so it can be written as $g(a, b, c)(a-b)(b-c)(c-a)$ where $g$ is a symmetric linear polynomial; hence, it is a scalar multiple of $a+b+c$. When $a=0$, the whole expression is $b c\left(b^{2}-c^{2}\right)$, so $g(0, b, c)=-b-c$, which implies that $g(a, b, c)=-(a+b+c)$.  Since $A, B$, and $C$ are distinct, we may divide by $(a-b)(b-c)(c-a)$ to obtain  $$ (p+q)(a b+b c+c a)=(\bar{p}+\bar{q}) a b c(a+b+c) \Longrightarrow(p+q) \bar{h}=(\bar{p}+\bar{q}) h $$  This implies that $\frac{\frac{p+q}{h}-0}{h-0}$ is real, so the midpoint of $\overline{P Q}$ lies on line $\overline{O H}$.


solution: 【 Solution 3 also using complex numbers (Michael Ren). We use complex numbers as in the previous solution. The angle conditions imply that $\frac{(a-p)(a-q)}{(b-c)^{2}}, \frac{(b-p)(b-q)}{(c-a)^{2}}$, and $\frac{(c-p)(c-q)}{(a-b)^{2}}$ are real numbers. Take a linear combination of these with real coefficients $X$, $Y$, and $Z$ to be determined; after expansion, we obtain  $$ \begin{aligned} & {\left[\frac{X}{(b-c)^{2}}+\frac{Y}{(c-a)^{2}}+\frac{Z}{(a-b)^{2}}\right] p q } \\ - & {\left[\frac{a X}{(b-c)^{2}}+\frac{b Y}{(c-a)^{2}}+\frac{c Z}{(a-b)^{2}}\right](p+q) } \end{aligned} $$  $$ +\left[\frac{a^{2} X}{(b-c)^{2}}+\frac{b^{2} Y}{(c-a)^{2}}+\frac{c^{2} Z}{(a-b)^{2}}\right] $$  which is a real number. To get something about the midpoint of $P Q$, the $p q$ coefficient should be zero, which motivates the following lemma.  ## Lemma  There exist real $X, Y, Z$ for which  $$ \begin{aligned} & \frac{X}{(b-c)^{2}}+\frac{Y}{(c-a)^{2}}+\frac{Z}{(a-b)^{2}}=0 \text { and } \\ & \frac{a X}{(b-c)^{2}}+\frac{b Y}{(c-a)^{2}}+\frac{c Z}{(a-b)^{2}} \neq 0 . \end{aligned} $$   $$ \begin{aligned} & \frac{(b-a) Y}{(c-a)^{2}}+\frac{(c-a) Z}{(a-b)^{2}} \\ = & \frac{a X}{(b-c)^{2}}+\frac{b Y}{(c-a)^{2}}+\frac{c Z}{(a-b)^{2}}-a\left(\frac{X}{(b-c)^{2}}+\frac{Y}{(c-a)^{2}}+\frac{Z}{(a-b)^{2}}\right) \\ = & 0 \end{aligned} $$  We can easily check that $(Y, Z)=(0,0)$ is impossible, therefore $\frac{(b-a)^{3}}{(c-a)^{3}}=-\frac{Z}{Y}$ is real. This means $\angle B A C=60^{\circ}$ or $120^{\circ}$. By symmetry, the same is true of $\angle C B A$ and $\angle A C B$. This is impossible because $A B C$ is scalene.  With the choice of $X, Y, Z$ as in the lemma, there exist complex numbers $\alpha$ and $\beta$, depending only on $a, b$, and $c$, such that $\alpha \neq 0$ and $\alpha(p+q)+\beta$ is real. Therefore the midpoint of $P Q$, which corresponds to $\frac{p+q}{2}$, lies on a fixed line. It remains to show that this line is the Euler line. First, choose $P=Q$ to be the orthocenter to show that the orthocenter lies on the line. Secondly, choose $P$ and $Q$ to be the foci of the Steiner circumellipse to show that the centroid lies on the line. (By some ellipse properties, the external angle bisector of $\angle P A Q$ is the tangent to the circumellipse at $A$, which is the line through $A$ parallel to $B C$. Therefore these points are valid.) Therefore the fixed line of the midpoint is the Euler line.


problem: The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of 2023 such coins alternately tails-up and heads-up, with the leftmost coin tails-up. In a move, Vera may flip over one of the coins in the row, subject to the following rules:  - On the first move, Vera may flip over any of the 2023 coins. - On all subsequent moves, Vera may only flip over a coin adjacent to the coin she flipped on the previous move. (We do not consider a coin to be adjacent to itself.)  Determine the smallest possible number of moves Vera can make to reach a state in which every coin is heads-up.

solution: The answer is 4044 . In general, replacing 2023 with $4 n+3$, the answer is $8 n+4$.  Bound. Observe that the first and last coins must be flipped, and so every coin is flipped at least once. Then, the $2 n+1$ even-indexed coins must be flipped at least twice, so they are flipped at least $4 n+2$ times.  The $2 n+2$ odd-indexed coins must then be flipped at least $4 n+1$ times. Since there are an even number of these coins, the total flip count must be even, so they are actually flipped a total of at least $4 n+2$ times, for a total of at least $8 n+4$ flips in all.  【 Construction. For $k=0,1, \ldots, n-1$, flip $(4 k+1,4 k+2,4 k+3,4 k+2,4 k+3,4 k+$ $4,4 k+3,4 k+4)$ in that order; then at the end, flip $4 n+1,4 n+2,4 n+3,4 n+2$. This is illustrated below for $4 n+3=15$. ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-24.jpg?height=641&width=1240&top_left_y=1898&top_left_x=411)  It is easy to check this works, and there are 4044 flips, as desired.


problem: Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line segments $B_{1} C_{2}, C_{1} A_{2}$, and $A_{1} B_{2}$ are concurrent, and the perimeters of triangles $A B_{2} C_{1}, B C_{2} A_{1}$, and $C A_{2} B_{1}$ are all equal. Find all possible values of this common perimeter.

solution: The only possible value of the common perimeter, denoted $p$, is 1 . 【 Synthetic approach (from author). We prove the converse of the problem first: Claim ( $p=1$ implies concurrence) - Suppose the six points are chosen so that triangles $A B_{2} C_{1}, B C_{2} A_{1}, C A_{2} B_{1}$ all have perimeter 1. Then lines $\overline{B_{1} C_{2}}, \overline{C_{1} A_{2}}$, and $\overline{A_{1} B_{2}}$ are concurrent.  ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-25.jpg?height=603&width=680&top_left_y=1389&top_left_x=688)  Hence the result follows by Brianchon's theorem. Now suppose $p \neq 1$. Let $\overline{B_{2}^{\prime} C_{1}^{\prime}}$ be the dilation of $\overline{B_{2} C_{1}}$ with ratio $\frac{1}{p}$ at center $A$, and define $C_{2}^{\prime}, A_{1}^{\prime}, A_{2}^{\prime}, B_{1}^{\prime}$ similarly. The following diagram showcases the situation $p<1$. ![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-26.jpg?height=774&width=861&top_left_y=244&top_left_x=603)  By the reasoning in the $p=1$ case, note that $\overline{B_{1}^{\prime} C_{2}^{\prime}}, \overline{C_{1}^{\prime} A_{2}^{\prime}}$, and $\overline{A_{1}^{\prime} B_{2}^{\prime}}$ are concurrent. However, $\overline{B_{1} C_{2}}, \overline{C_{1} A_{2}}, \overline{A_{1} B_{2}}$ lie in the interior of quadrilaterals $B C B_{1}^{\prime} C_{2}^{\prime}, C A C_{1}^{\prime} A_{2}^{\prime}$, and $A B A_{1}^{\prime} B_{2}^{\prime}$, and these quadrilaterals do not share an interior point, a contradiction.  Thus $p \geq 1$. Similarly, we can show $p \leq 1$, and so $p=1$ is forced (and achieved if, for example, the three triangles are equilateral with side length $1 / 3$ ).


solution: The only possible value of the common perimeter, denoted $p$, is 1 . 『 Barycentric solution (by Carl, Krit, Milan). We show that, if the common perimeter is 1 , then the lines concur. To do this, we use barycentric coordinates. Let $A=(1: 0: 0)$, $B=(0: 1: 0)$, and $C=(0: 0: 1)$. Let $A_{1}=\left(0: 1-a_{1}: a_{1}\right), A_{2}=\left(0: a_{2}: 1-a_{2}\right)$, $B_{1}=\left(b_{1}: 0: 1-b_{1}\right), B_{2}=\left(1-b_{2}: 0: b_{2}\right), C_{1}=\left(1-c_{1}: c_{1}: 0\right)$, and $C_{2}=\left(c_{2}: 1-c_{2}: 0\right)$. The line $B_{1} C_{2}$ is defined by the equation  $$ \operatorname{det}\left[\begin{array}{ccc} x & y & z \\ b_{1} & 0 & 1-b_{1} \\ c_{2} & 1-c_{2} & 0 \end{array}\right]=0 \text {; } $$  i.e.  $$ x\left(-\left(1-b_{1}\right)\left(1-c_{2}\right)\right)+y\left(\left(1-b_{1}\right) c_{2}\right)+z\left(b_{1}\left(1-c_{2}\right)\right)=0 . $$  Computing the equations for the other lines cyclically, we get that the lines $B_{1} C_{2}, C_{1} A_{2}$, and $A_{1} B_{2}$ concur if and only if  $$ \operatorname{det}\left[\begin{array}{ccc} -\left(1-b_{1}\right)\left(1-c_{2}\right) & \left(1-b_{1}\right) c_{2} & b_{1}\left(1-c_{2}\right) \\ c_{1}\left(1-a_{2}\right) & -\left(1-c_{1}\right)\left(1-a_{2}\right) & \left(1-c_{1}\right) a_{2} \\ \left(1-a_{1}\right) b_{2} & a_{1}\left(1-b_{2}\right) & -\left(1-a_{1}\right)\left(1-b_{2}\right) \end{array}\right]=0 . $$  Let this matrix be $M$. We also define the similar matrix  $$ N=\left[\begin{array}{ccc} -\left(1-b_{2}\right)\left(1-c_{1}\right) & \left(1-b_{2}\right) c_{1} & b_{2}\left(1-c_{1}\right) \\ c_{2}\left(1-a_{1}\right) & -\left(1-c_{2}\right)\left(1-a_{1}\right) & \left(1-c_{2}\right) a_{1} \\ \left(1-a_{2}\right) b_{1} & a_{2}\left(1-b_{1}\right) & -\left(1-a_{2}\right)\left(1-b_{1}\right) \end{array}\right] . $$  Geometrically, det $N=0$ if and only if $B_{2}^{\prime} C_{1}^{\prime}, C_{2}^{\prime} A_{1}^{\prime}$, and $A_{2}^{\prime} B_{1}^{\prime}$ concur, where for a point $P$ on a side of triangle $A B C, P^{\prime}$ denotes its reflection over that side's midpoint.  Claim - We have $\operatorname{det} M=\operatorname{det} N$.   We use the definition of the determinant as a sum over permutations. The even permutations give us the following three terms:  $$ \begin{aligned} -\left(1-b_{1}\right)\left(1-c_{2}\right)\left(1-c_{1}\right)\left(1-a_{2}\right)\left(1-a_{1}\right)\left(1-b_{2}\right) & =-\prod_{i=1}^{2}\left(\left(1-a_{i}\right)\left(1-b_{i}\right)\left(1-c_{i}\right)\right) \\ \left(1-a_{1}\right) b_{2}\left(1-b_{1}\right) c_{2}\left(1-c_{1}\right) a_{2} & =\left(\left(1-a_{1}\right)\left(1-b_{1}\right)\left(1-c_{1}\right)\right)\left(a_{2} b_{2} c_{2}\right) \\ c_{1}\left(1-a_{2}\right) a_{1}\left(1-b_{2}\right) b_{1}\left(1-c_{2}\right) & =\left(\left(1-a_{2}\right)\left(1-b_{2}\right)\left(1-c_{2}\right)\right)\left(a_{1} b_{1} c_{1}\right) . \end{aligned} $$  The first term is invariant under $\Psi$, while the second and third terms are swapped under $\Psi$. For the odd permutations, we have a contribution to the determinant of  $$ \sum_{\mathrm{cyc}}\left(1-b_{1}\right)\left(1-c_{2}\right)\left(1-c_{1}\right) a_{2} a_{1}\left(1-b_{2}\right) $$  each summand is invariant under $\Psi$. This finishes the proof of our claim. Now, it suffices to show that, if $A B_{2} C_{1}, B C_{2} A_{1}$, and $C A_{2} B_{1}$ each have perimeter 1, then  $$ \operatorname{det}\left[\begin{array}{ccc} -\left(1-b_{2}\right)\left(1-c_{1}\right) & \left(1-b_{2}\right) c_{1} & b_{2}\left(1-c_{1}\right) \\ c_{2}\left(1-a_{1}\right) & -\left(1-c_{2}\right)\left(1-a_{1}\right) & \left(1-c_{2}\right) a_{1} \\ \left(1-a_{2}\right) b_{1} & a_{2}\left(1-b_{1}\right) & -\left(1-a_{2}\right)\left(1-b_{1}\right) . \end{array}\right]=0 $$  Indeed, we have $A B_{2}=b_{2}$ and $A C_{1}=c_{1}$, so by the law of cosines,  $$ 1-b_{2}-c_{1}=1-A B_{2}-A C_{1}=B_{2} C_{1}=\sqrt{b_{2}^{2}+c_{1}^{2}-b_{2} c_{1}} $$  This gives  $$ \left(1-b_{2}-c_{1}\right)^{2}=b_{2}^{2}+c_{1}^{2}-b_{2} c_{1} \Longrightarrow 1-2 b_{2}-2 c_{1}+3 b_{2} c_{1}=0 $$  Similarly, $1-2 c_{2}-2 a_{1}+3 c_{2} a_{1}=0$ and $1-2 a_{2}-2 b_{1}+3 a_{2} b_{1}=0$. Now,  $$ \begin{aligned} N\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] & =\left[\begin{array}{l} -\left(1-b_{2}\right)\left(1-c_{1}\right)+\left(1-b_{2}\right) c_{1}+b_{2}\left(1-c_{1}\right) \\ -\left(1-c_{2}\right)\left(1-a_{1}\right)+\left(1-c_{2}\right) a_{1}+c_{2}\left(1-a_{1}\right) \\ -\left(1-a_{2}\right)\left(1-b_{1}\right)+\left(1-a_{2}\right) b_{1}+a_{2}\left(1-b_{1}\right) \end{array}\right] \\ & =\left[\begin{array}{l} -1+2 b_{2}+2 c_{1}-3 b_{2} c_{1} \\ -1+2 c_{2}+2 a_{1}-3 c_{2} a_{1} \\ -1+2 a_{2}+2 b_{1}-2 a_{2} b_{1} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] . \end{aligned} $$  So it follows $\operatorname{det} N=0$, as desired.


problem: Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e} \mathbb{Z}$ defined recursively by  $$ \begin{aligned} & \Delta^{0} f(n)=f(n) \\ & \Delta^{k} f(n)=\Delta^{k-1} f(n+1)-\Delta^{k-1} f(n) \quad \text { for } k=1,2, \ldots \end{aligned} $$  Determine the number of functions $f$ such that there exists some $k \geq 1$ for which $\Delta^{k} f=f$ 。

solution: Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e} \mathbb{Z}$ defined recursively by  $$ \begin{aligned} & \Delta^{0} f(n)=f(n) \\ & \Delta^{k} f(n)=\Delta^{k-1} f(n+1)-\Delta^{k-1} f(n) \quad \text { for } k=1,2, \ldots \end{aligned} $$  Determine the number of functions $f$ such that there exists some $k \geq 1$ for which $\Delta^{k} f=f$.  The answer is  $$ \left(p^{e}\right)^{a} \cdot p^{-e p^{\nu} p(a)}=p^{e\left(a-p^{\nu_{p}(a)}\right)} $$ 【 First solution by author. For convenience in what follows, set $d=\nu_{p}(a)$, let $a=p^{d} \cdot b$, and let a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ be essential if it equals one of its iterated finite differences.  The key claim is the following. Claim (Characterization of essential functions) - A function $f$ is essential if and only if  $$ f(x)+f\left(x+p^{d}\right)+\cdots+f\left(x+(b-1) p^{d}\right)=0 $$  for all $x$.  ## Lemma  Let $g: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ be any function, and let $h=\Delta^{p^{d}} g$. Then  $$ h(x)+h\left(x+p^{d}\right)+\cdots+h\left(x+(b-1) p^{d}\right) \equiv 0 \quad(\bmod p) $$  for all $x$.   $$ h(x)=\Delta^{p^{d}} g(x)=\sum_{k=0}^{p^{d}}(-1)^{k}\binom{p^{d}}{k} g\left(x+p^{d}-k\right) $$  However, it is known that $\binom{p^{d}}{k}$ is a multiple of $p$ if $1 \leq k \leq p^{d}-1$, so  $$ h(x) \equiv g\left(x+p^{d}\right)+(-1)^{p^{d}} g(x) \quad(\bmod p) . $$  Using this, we easily obtain  $$ \begin{aligned} & h(x)+h\left(x+p^{d}\right)+\cdots+h\left(x+(b-1) p^{d}\right) \\ \equiv & \begin{cases}0 & p>2 \\ 2\left(g(x)+g\left(x+p^{d}\right)+\cdots+g\left(x+(b-1) p^{d}\right)\right) & p=2\end{cases} \\ \equiv & 0 \quad(\bmod p) \end{aligned} $$  as desired.  ## Corollary  Let $g: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ be any function, and let $h=\Delta^{e p^{d}} g$. Then  $$ h(x)+h\left(x+p^{d}\right)+\cdots+h\left(x+(b-1) p^{d}\right)=0 $$  for all $x$.   $$ h_{1}(x)=\frac{h(x)+h\left(x+p^{d}\right)+\cdots+h\left(x+(b-1) p^{d}\right)}{p} . $$  Applying the lemma to $h_{1}$ shows the corollary for $e=2$, since $h_{1}(x)$ is divisible by $p$, hence the numerator is divisible by $p^{2}$. Continue in this manner to get the result for general $e>2$.  This immediately settles this direction, since $f$ is in the image of $\Delta^{e p^{d}}$.   We will show that $\Delta$ is injective on $\mathcal{S}$. Suppose otherwise, and consider two functions $f$, $g$ in $\mathcal{S}$ with $\Delta f=\Delta g$. Then, we obtain that $f$ and $g$ differ by a constant; say $g=f+\lambda$. However, then  $$ \begin{aligned} & g(0)+g\left(p^{e}\right)+\cdots+g\left((b-1) p^{e}\right) \\ = & (f(0)+\lambda)+\left(f\left(p^{e}\right)+\lambda\right)+\cdots+\left(f\left((b-1) p^{e}\right)+\lambda\right) \\ = & b \lambda \end{aligned} $$  This should also be zero. Since $p \nmid b$, we obtain $\lambda=0$, as desired.  Counting Finally, we can count the essential functions: all but the last $p^{d}$ entries can be chosen arbitrarily, and then each remaining entry has exactly one possible choice. This leads to a count of  $$ \left(p^{e}\right)^{a-p^{d}}=p^{e\left(a-p^{\nu_{p}(a)}\right)}, $$  as promised.


solution: Let $p$ be a fixed prime and let $a \geq 2$ and $e \geq 1$ be fixed integers. Given a function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ and an integer $k \geq 0$, the $k$ th finite difference, denoted $\Delta^{k} f$, is the function from $\mathbb{Z} / a \mathbb{Z}$ to $\mathbb{Z} / p^{e} \mathbb{Z}$ defined recursively by  $$ \begin{aligned} & \Delta^{0} f(n)=f(n) \\ & \Delta^{k} f(n)=\Delta^{k-1} f(n+1)-\Delta^{k-1} f(n) \quad \text { for } k=1,2, \ldots \end{aligned} $$  Determine the number of functions $f$ such that there exists some $k \geq 1$ for which $\Delta^{k} f=f$.  The answer is  $$ \left(p^{e}\right)^{a} \cdot p^{-e p^{\nu} p(a)}=p^{e\left(a-p^{\nu_{p}(a)}\right)} $$ II Second solution by Daniel Zhu. There are two parts to the proof: solving the $e=1$ case, and using the $e=1$ result to solve the general problem by induction on $e$. These parts are independent of each other.  The case $e=1 \quad$ Represent functions $f$ as elements  $$ \alpha_{f}:=\sum_{k \in \mathbb{Z} / a \mathbb{Z}} f(-k) x^{k} \in \mathbb{F}_{p}[x] /\left(x^{a}-1\right) $$  Then, since $\alpha_{\Delta f}=(x-1) \alpha_{f}$, we wish to find the number of $\alpha \in \mathbb{F}_{p}[x] /\left(x^{a}-1\right)$ such that $(x-1)^{m} \alpha=\alpha$ for some $m$.  Now, make the substitution $y=x-1$ and let $P(y)=(y+1)^{a}-1$; we want to find $\alpha \in \mathbb{F}_{p}[y] /(P(y))$ such that $y^{m} \alpha=\alpha$ for some $m$.  If we write $P(y)=y^{d} Q(y)$ with $Q(0) \neq 0$, then by the Chinese Remainder Theorem we have the ring isomorphism  $$ \mathbb{F}_{p}[y] /(P(y)) \cong \mathbb{F}_{p}[y] /\left(y^{d}\right) \times \mathbb{F}_{p}[y] /(Q(y)) $$  Note that $y$ is nilpotent in the first factor, while it is a unit in the second factor. So the $\alpha$ that work are exactly those that are zero in the first factor; thus there are $p^{a-d}$ such $\alpha$. We can calculate $d=p^{v_{p}(a)}$ (via, say, Lucas's Theorem), so we are done.  The general problem The general idea is as follows: call a $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z} e$-good if $\Delta^{m} f=f$ for some $m$. Our result above allows us to count the 1-good functions. Then, if $e \geq 1$, every $(e+1)$-good function, when reduced $\bmod p^{e}$, yields an $e$-good function, so we count $(e+1)$-good functions by counting how many reduce to any given $e$-good function.  Formally, we use induction on $e$, with the $e=1$ case being treated above. Suppose now we have solved the problem for a given $e \geq 1$, and we now wish to solve it for $e+1$. For any function $g: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e+1} \mathbb{Z}$, let $\bar{g}: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e} \mathbb{Z}$ be its reduction $\bmod p^{e}$. For a given $e$-good $f$, let $n(f)$ be the number of $(e+1)$-good $g$ with $\bar{g}=f$. The following two claims now finish the problem:  Claim - If $f$ is $e$-good, then $n(f)>0$.   $$ g, \Delta^{m} g, \Delta^{2 m} g, \ldots $$  Since there are finitely many functions $\mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p^{e+1} \mathbb{Z}$, there must exist $a<b$ such that $\Delta^{a m} g=\Delta^{b m} g$. We claim $\Delta^{a m} g$ is the desired $(e+1)$ good function. To see this, first note that since $\overline{\Delta^{k} g}=\Delta^{k} \bar{g}$, we must have $\overline{\Delta^{a m} g}=\Delta^{a m} f=f$. Moreover,  $$ \Delta^{(b-a) m}\left(\Delta^{a m} g\right)=\Delta^{b m} g=\Delta^{a m} g $$  so $\Delta^{a m} g$ is $(e+1)$-good.  Claim - If $f$ is $e$-good, and $n(f)>0$, then $n(f)$ is exactly the number of 1-good functions, i.e. $p^{a-p^{v_{p}(a)}}$.   To show that this condition is sufficient, note that $\overline{g+p^{e} h}=\bar{g}=f$. Moreover, if $\Delta^{m} g=g$ and $\Delta^{m^{\prime}} h=h$, then  $$ \Delta^{m m^{\prime}}\left(g+p^{e} h\right)=\Delta^{m m^{\prime}} g+p^{e} \Delta^{m m^{\prime}} h=g+p^{e} h . $$  To show that this condition is necessary, let $g_{1}$ be any $(e+1)$-good function such that $\bar{g}_{1}=f$. Then $g_{1}-g$ is also $(e+1)$-good, since if $\Delta^{m} g=g, \Delta^{m^{\prime}} g_{1}=g_{1}$, we have  $$ \Delta^{m m^{\prime}}\left(g_{1}-g\right)=\Delta^{m m^{\prime}} g_{1}-\Delta^{m m^{\prime}} g=g_{1}-g . $$  On the other hand, we also know that $g_{1}-g$ is divisible by $p^{e}$. This means that it must be $p^{e} h$ for some function $f: \mathbb{Z} / a \mathbb{Z} \rightarrow \mathbb{Z} / p \mathbb{Z}$, and it is not hard to show that $g_{1}-g$ being $(e+1)$-good means that $h$ is 1 -good.