exam: USAJMO

problem: Find all positive integers $n$ such that $2^{n}+12^{n}+2011^{n}$ is a perfect square.

solution: The answer $n=1$ works, because $2^{1}+12^{1}+2011^{1}=45^{2}$. We prove it's the only one.  - If $n \geq 2$ is even, then modulo 3 we have $2^{n}+12^{n}+2011^{n} \equiv 1+0+1 \equiv 2(\bmod 3)$ so it is not a square. - If $n \geq 3$ is odd, then modulo 4 we have $2^{n}+12^{n}+2011^{n} \equiv 0+0+3 \equiv 3(\bmod 4)$ so it is not a square.


solution: The condition becomes $2 \geq a^{2}+b^{2}+c^{2}+a b+b c+c a$. Therefore,  $$ \begin{aligned} \sum_{\text {cyc }} \frac{2 a b+2}{(a+b)^{2}} & \geq \sum_{\text {cyc }} \frac{2 a b+\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\right)}{(a+b)^{2}} \\ & =\sum_{\text {cyc }} \frac{(a+b)^{2}+(c+a)(c+b)}{(a+b)^{2}} \\ & =3+\sum_{\text {cyc }} \frac{(c+a)(c+b)}{(a+b)^{2}} \\ & \geq 3+3 \sqrt[3]{\prod_{\text {cyc }} \frac{(c+a)(c+b)}{(a+b)^{2}}}=3+3=6 \end{aligned} $$  with the last line by AM-GM. This completes the proof.


problem: For a point $P=\left(a, a^{2}\right)$ in the coordinate plane, let $\ell(P)$ denote the line passing through $P$ with slope $2 a$. Consider the set of triangles with vertices of the form $P_{1}=\left(a_{1}, a_{1}^{2}\right), P_{2}=\left(a_{2}, a_{2}^{2}\right), P_{3}=\left(a_{3}, a_{3}^{2}\right)$, such that the intersection of the lines $\ell\left(P_{1}\right), \ell\left(P_{2}\right), \ell\left(P_{3}\right)$ form an equilateral triangle $\Delta$. Find the locus of the center of $\Delta$ as $P_{1} P_{2} P_{3}$ ranges over all such triangles.

solution: The answer is the line $y=-1 / 4$. I did not find this problem inspiring, so I will not write out most of the boring calculations since most solutions are just going to be "use Cartesian coordinates and grind all the way through".  The "nice" form of the main claim is as follows (which is certainly overkill for the present task, but is too good to resist including):  Claim (Naoki Sato) - In general, the orthocenter of $\Delta$ lies on the directrix $y=-1 / 4$ of the parabola (even if the triangle $\Delta$ is not equilateral).   $$ \left(\frac{a_{1}+a_{2}}{2}, a_{1} a_{2}\right) ; \quad\left(\frac{a_{2}+a_{3}}{2}, a_{2} a_{3}\right) ; \quad\left(\frac{a_{3}+a_{1}}{2}, a_{3} a_{1}\right) . $$  The coordinates of the orthocenter can be checked explicitly to be  $$ H=\left(\frac{a_{1}+a_{2}+a_{3}+4 a_{1} a_{2} a_{3}}{2},-\frac{1}{4}\right) . $$  This claim already shows that every point lies on $y=-1 / 4$. We now turn to showing that, even when restricted to equilateral triangles, we can achieve every point on $y=-1 / 4$. In what follows $a=a_{1}, b=a_{2}, c=a_{3}$ for legibility.  Claim - Lines $\ell(a), \ell(b), \ell(c)$ form an equilateral triangle if and only if  $$ \begin{aligned} a+b+c & =-12 a b c \\ a b+b c+c a & =-\frac{3}{4} . \end{aligned} $$  Moreover, the $x$-coordinate of the equilateral triangle is $\frac{1}{3}(a+b+c)$.  $$ \left(\frac{a+b+c}{3}, \frac{a b+b c+c a}{3}\right)=G=H=\left(\frac{a+b+c+4 a b c}{2},-\frac{1}{4}\right) . $$  Setting the $x$ and $y$ coordinates equal, we derive the claimed equations.  Let $\lambda$ be any real number. We are tasked to show that  $$ P(X)=X^{3}-3 \lambda \cdot X^{2}-\frac{3}{4} X+\frac{\lambda}{4} $$  has three real roots (with multiplicity); then taking those roots as $(a, b, c)$ yields a valid equilateral-triangle triple whose $x$-coordinate is exactly $\lambda$, be the previous claim.  To prove that, pick the values  $$ \begin{aligned} P(-\sqrt{3} / 2) & =-2 \lambda \\ P(0) & =\frac{1}{4} \lambda \\ P(\sqrt{3} / 2) & =-2 \lambda . \end{aligned} $$  The intermediate value theorem (at least for $\lambda \neq 0$ ) implies that $P$ should have at least two real roots now, and since $P$ has degree 3, it has all real roots. That's all.


problem: A word is defined as any finite string of letters. A word is a palindrome if it reads the same backwards and forwards. Let a sequence of words $W_{0}, W_{1}, W_{2}, \ldots$ be defined as follows: $W_{0}=a, W_{1}=b$, and for $n \geq 2, W_{n}$ is the word formed by writing $W_{n-2}$ followed by $W_{n-1}$. Prove that for any $n \geq 1$, the word formed by writing $W_{1}, W_{2}, W_{3}, \ldots, W_{n}$ in succession is a palindrome.

solution: $$ \begin{aligned} & W_{0}=a \\ & W_{1}=b \\ & W_{2}=a b \\ & W_{3}=b a b \\ & W_{4}=a b b a b \\ & W_{5}=b a b a b b a b \\ & W_{6}=a b b a b b a b a b b a b \\ & W_{7}=b a b a b b a b a b b a b b a b a b b a b \end{aligned} $$  We prove that $W_{1} W_{2} \ldots W_{n}$ is a palindrome by induction on $n$. The base cases $n=$ $1,2,3,4$ can be verified by hand.  For the inductive step, we let $\bar{X}$ denote the word $X$ written backwards. Then  $$ \begin{aligned} W_{1} W_{2} \ldots W_{n-3} W_{n-2} W_{n-1} W_{n} & \stackrel{\mathrm{IH}}{=}\left(\overline{W_{n-1} W_{n-2} W_{n-3}} \ldots \overline{W_{2} W_{1}}\right) W_{n} \\ & =\left(\overline{W_{n-1} W_{n-2} W_{n-3}} \ldots \overline{W_{2} W_{1}}\right) W_{n-2} W_{n-1} \\ & =\overline{W_{n-1} W_{n-2}}\left(\overline{W_{n-3}} \ldots \overline{W_{2} W_{1}}\right) W_{n-2} W_{n-1} \end{aligned} $$  with the first equality being by the induction hypothesis. By induction hypothesis again the inner parenthesized term is also a palindrome, and so this completes the proof.


problem: Points $A, B, C, D, E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $P B$ and $P D$ are tangent to $\omega$, (ii) $P, A, C$ are collinear, and (iii) $\overline{D E} \| \overline{A C}$. Prove that $\overline{B E}$ bisects $\overline{A C}$.

solution: 【 First solution using harmonic bundles. Let $M=\overline{B E} \cap \overline{A C}$ and let $\infty$ be the point at infinity along $\overline{D E} \| \overline{A C}$. ![](https://cdn.mathpix.com/cropped/2024_11_19_3c39ee86a3666408bff8g-8.jpg?height=646&width=817&top_left_y=962&top_left_x=628)  Note that $A B C D$ is harmonic, so  $$ -1=(A C ; B D) \stackrel{E}{=}(A C ; M \infty) $$  implying $M$ is the midpoint of $\overline{A C}$.


solution: I Second solution using complex numbers (Cynthia Du). Suppose we let $b, d, e$ be free on unit circle, so $p=\frac{2 b d}{b+d}$. Then $d / c=a / e$, and $a+c=p+a c \bar{p}$. Consequently,  $$ \begin{aligned} a c & =d e \\ \frac{1}{2}(a+c) & =\frac{b d}{b+d}+d e \cdot \frac{1}{b+d}=\frac{d(b+e)}{b+d} \\ \frac{a+c}{2 a c} & =\frac{(b+e)}{e(b+d)} \end{aligned} $$  From here it's easy to see  $$ \frac{a+c}{2}+\frac{a+c}{2 a c} \cdot b e=b+e $$  which is what we wanted to prove.


problem: Consider the assertion that for each positive integer $n \geq 2$, the remainder upon dividing $2^{2^{n}}$ by $2^{n}-1$ is a power of 4 . Either prove the assertion or find (with proof) a counterexample.

solution: We claim $n=25$ is a counterexample. Since $2^{25} \equiv 2^{0}\left(\bmod 2^{25}-1\right)$, we have  $$ 2^{2^{25}} \equiv 2^{2^{25} \bmod 25} \equiv 2^{7} \bmod 2^{25}-1 $$  and the right-hand side is actually the remainder, since $0<2^{7}<2^{25}$. But $2^{7}$ is not a power of 4.  Remark. Really, the problem is just equivalent for asking $2^{n}$ to have odd remainder when divided by $n$.