While it’s still fresh in my mind I guess.

A1. Determine all possible values of $A^3+B^3+C^3-3ABC$ where $A$, $B$, and $C$ are nonnegative integers.

Kinda standard question, just factorize $$X^3 + Y^3 + Z^3 -3XYZ = (X+Y+Z)\cdot \frac{1}{2}((X-Y)^2 + (Y-Z)^2 + (Z-X)^2)$$ then check 3 divides the left term iff it divides the right term. Then construct the rest.

A2. In the triangle $\triangle ABC$, let $G$ be the centroid, and let $I$ be the center of the inscribed circle. Let $\alpha$ and $\beta$ be the angles at the vertices $A$ and $B$, respectively. Suppose that the segment $IG$ is parallel to $AB$ and that $\beta = 2\tan^{-1}(1/3)$. Find $\alpha$.

Interpret the parallel condition using area $[AIB] = [AGB]$, then we get $a+b = 2c$. Using the tangent condition we get $s-b = 3r$, then just use $$[ABC] = rs = \sqrt{s(s-a)(s-b)(s-c)}$$ to get $s-a = r$, so $A$ is a right angle.

A3. Given real numbers $b_0,b_1,\ldots, b_{2019}$ with $b_{2019}\neq 0$, let $z_1,z_2,\ldots, z_{2019}$ be the roots in the complex plane of the polynomial $$ P(z) = \sum_{k=0}^{2019}b_kz^k. $$ Let $\mu = (|z_1|+ \cdots + |z_{2019}|)/2019$ be the average of the distances from $z_1,z_2,\ldots, z_{2019}$ to the origin. Determine the largest constant $M$ such that $\mu\geq M$ for all choices of $b_0,b_1,\ldots, b_{2019}$ that satisfy $$ 1\leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019. $$

This looked hard until I came back to it later. You should suspect the equality case has to do with roots of unity (something like $1+x+x^2 + ... +x^{2019}$), and you can even shrink it inwards by some appropriate factor $\lambda$ (to get $1 + \lambda x + \lambda ^2 x^2 + ...$). Then this is when you suspect a very trivial bound works: indeed we have the product of the absolute values, so just AM-GM.

A4. Let $f$ be a continuous real-valued function on $\mathbb R^3$. Suppose that for every sphere $S$ of radius $1$, the integral of $f(x,y,z)$ over the surface of $S$ equals zero. Must $f(x,y,z)$ be identically zero?

Somewhat unfair question. If happen to know that the surface area of a horizontal slice of a sphere only depends on the height of the slice, then this is pretty easy.

A5. Let $p$ be an odd prime number, and let $\mathbb{F}_p$ denote the field of integers modulo $p$. Let $\mathbb{F}_p[x]$ be the ring of polynomials over $\mathbb{F}_p$, and let $q(x) \in \mathbb{F}_p[x]$ be given by $q(x) = \sum_{k=1}^{p-1} a_k x^k$ where $a_k = k^{(p-1)/2}$ mod $p$. Find the greatest nonnegative integer $n$ such that $(x-1)^n$ divides $q(x)$ in $\mathbb{F}_p[x]$.

Small cases suggests that the right exponent is $(p-1)/2$. We just have to check that $q(x)^2 \pmod{x^p - 1}$ is a multiple of $(x^{p-1} + ... + 1)$. The $[x^l]$ coefficient of $q(x)^2\pmod{x^p-1}$ corresponds to $(\chi\ast \chi)(l)$, where $\chi$ is the legendre symbol mod $l$ and it’s fairly fast to check that this equals to $-\left(\frac{-1}{p}\right)$.

A6. Let $g$ be a real-valued function that is continuous on the closed interval $[0,1]$ and twice differentiable on the open interval $(0,1)$. Suppose that for some real number $r>1$, $$\lim_{x\to 0^+}\frac{g(x)}{x^r} = 0.$$ Prove that either $$ \lim_{x\to 0^+}g'(x) = 0\qquad\text{or}\qquad \limsup_{x\to 0^+}x^r|g''(x)|= \infty.$$

There’s only one trick to use in situations like these, and that is to use Taylor expansion with remainder.

B1. Denote by $\mathbb Z^2$ the set of all points $(x,y)$ in the plane with integer coordinates. For each integer $n\geq 0$, let $P_n$ be the subset of $\mathbb Z^2$ consisting of the point $(0,0)$ together with all points $(x,y)$ such that $x^2+y^2=2^k$ for some integer $k\leq n$. Determine, as a function of $n$, the number of four-point subsets of $P_n$ whose elements are the vertices of a square.

The easy part is to figure out where all the points are. Call all points at distance $2^{k/2}$ level $k$. Then we can show that no square mixes between two non-adjacent levels, so all squares are either the levels themselves or something involving $(0,0)$.

B2. For all $n\ge 1$, let $a_n=\sum_{k=1}^{n-1}\frac{\sin(\frac{(2k-1)\pi}{2n})}{\cos^2(\frac{(k-1)\pi}{2n})\cos^2(\frac{k\pi}{2n})}$. Determine $\lim_{n\rightarrow \infty}\frac{a_n}{n^3}$.

This is missing a constant term from the telescope $$\sum \left(\frac{1}{\cos^2(\cdot)} - \frac{1}{\cos^2(\cdot)}\right)$$ and after that is fairly easy.

B3. Let $Q$ be an $n$-by-$n$ real orthogonal matrix, and let $u\in \mathbb{R}^n$ be a unit column vector (that is, $u^Tu=1$). Let $P=I-2uu^T$, where $I$ is the $n$-by-$n$ identity matrix. Show that if $1$ is not an eigenvalue of $Q$, then $1$ is an eigenvalue of $PQ$.

I used the Sylvester trick for this: $$\begin{aligned}\det(I-PQ) &= \det(I-Q)\det(I - 2(I-Q)^{-1}uu^TQ)\\ &= \det(I-Q)(1 + 2u^T Q(I-Q)^{-1}u) \end{aligned}$$ so we just have to check $1 + 2u^T Q(I-Q)^{-1}u = 0$ for any unit length $u$. This sounds false, and I was stuck here for a while.

Breakthrough was that I noticed $\langle (I-Q)v, (I-Q)v\rangle = 2\langle v, (I-Q)v\rangle$ because of weird $Q$ pushing behaviors (like $\langle v, Q^Tv \rangle = \langle v, Qv\rangle$. Eventually this works out (I still don’t have an intuitive explanation though).

B4. Let $\mathcal F$ be the set of functions $f(x,y)$ that are twice continuously differentiable for $x\geq 1$, $y\geq 1$ and that satisfy the following two equations (where subscripts denote partial derivatives): $$xf_x + yf_y = xy\ln(xy),$$ $$x^2f_{xx} + y^2f_{yy} = xy.$$ For each $f\in\mathcal F$, let $$m(f) = \min_{s\geq 1}\left(f(s+1,s+1) - f(s+1,s)-f(s,s+1) + f(s,s)\right).$$ Determine $m(f)$, and show that it is independent of the choice of $f$.

This a terrible question. Start with $$F(x,y) = \frac{f_x}{y} + \frac{f_y}{x} - \ln(xy) = 0$$ then you check that $xy(F_x + F_y)$ looks really like the second equation. This spits out the exact form of $f_{xy}$ which you can integrate to get what you want.

B5. Let $F_m$ be the $m$’th Fibonacci number, defined by $F_1=F_2=1$ and $F_m = F_{m-1}+F_{m-2}$ for all $m \geq 3$. Let $p(x)$ be the polynomial of degree 1008 such that $p(2n+1)=F_{2n+1}$ for $n=0,1,2,\ldots,1008$. Find integers $j$ and $k$ such that $p(2019) = F_j - F_k$.

Use finite differences. Taking $\Delta^{1008}p(2n+1)$ you get a connstant, then work your way back up to get the answer.

B6. Let $\mathbb{Z}^n$ be the integer lattice in $\mathbb{R}^n$. Two points in $\mathbb{Z}^n$ are called {\em neighbors} if they differ by exactly 1 in one coordinate and are equal in all other coordinates. For which integers $n \geq 1$ does there exist a set of points $S \subset \mathbb{Z}^n$ satisfying the following two conditions? (1) If $p$ is in $S$, then none of the neighbors of $p$ is in $S$. (2) If $p \in \mathbb{Z}^n$ is not in $S$, then exactly one of the neighbors of $p$ is in $S$.

The only problem I didn’t get during the test because I thought it would be a lot harder. Essentially, let $f(x_1,...,x_n) = x_1+2x_2+...+nx_n \mod{2n+1}$, then put your points where $f = 0$.