A triangle $ABC$ is given with $AC = BC = 5$. The altitude of this triangle drawn from vertex $A$ has length $4$. Calculate the length of the altitude of $ABC$ drawn from vertex $C$.

Find all pairs of positive integers $ (m,n)$ such that $ mn \minus{} 1$ divides $ (n^2 \minus{} n \plus{} 1)^2$. [i]Aaron Pixton.[/i]

In decimal representation $$\text {34!=295232799039a041408476186096435b0000000}.$$ Find the numbers $a$ and $b$.

In a convex quadrilateral $ABCD$ suppose $AC \cap BD = O$ and $M$ is the midpoint of $BC$. Let $MO \cap AD = E$. Prove that $\frac{AE}{ED} = \frac{S_{\triangle ABO}}{S_{\triangle CDO}}$.

A positive integer $N$ is [i]interoceanic[/i] if its prime factorization $$N=p_1^{x_1}p_2^{x_2}\cdots p_k^{x_k}$$ satisfies $$x_1+x_2+\dots +x_k=p_1+p_2+\cdots +p_k.$$ Find all interoceanic numbers less than 2020.

Let $n\not\equiv 2\pmod{3}$. Is $\sqrt{\lfloor n+\tfrac {2n}{3}\rfloor+7},\forall n \in \mathbb {N}$, a natural number?

Let $q$ be an odd positive integer, and let $N_q$ denote the number of integers $a$ such that $0<a<q/4$ and $\gcd(a,q)=1.$ Show that $N_q$ is odd if and only if $q$ is of the form $p^k$ with $k$ a positive integer and $p$ a prime congruent to $5$ or $7$ modulo $8.$

Prove: If $x, y, z$ are the lengths of the angle bisectors of a triangle with perimeter 6, than we have: \[\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \geq 1.\]

In the accompanying figure , $y=f(x)$ is the graph of a one-to-one continuous function $f$ . At each point $P$ on the graph of $y=2x^2$ , assume that the areas $OAP$ and $OBP$ are equal . Here $PA,PB$ are the horizontal and vertical segments . Determine the function $f$. [asy] Label f; xaxis(0,60,blue); yaxis(0,60,blue); real f(real x) { return (x^2)/60; } draw(graph(f,0,53),red); label("$y=x^2$",(30,15),E); real f(real x) { return (x^2)/25; } draw(graph(f,0,38),red); label("$y=2x^2$",(37,37^2/25),E); real f(real x) { return (x^2)/10; } draw(graph(f,0,25),red); label("$y=f(x)$",(24,576/10),W); label("$O(0,0)$",(0,0),S); dot((20,400/25)); dot((20,400/60)); label("$P$",(20,400/25),E); label("$B$",(20,400/60),SE); dot(((4000/25)^(0.5),400/25)); label("$A$",((4000/25)^(0.5),400/25),W); draw((20,400/25)..((4000/25)^(0.5),400/25)); draw((20,400/25)..(20,400/60)); [/asy]

Forty-two cubes with 1 cm edges are glued together to form a solid rectangular block. If the perimeter of the base of the block is 18 cm, then the height, in cm, is $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \dfrac{7}{3} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Let $B$ be an integer greater than 10 such that everyone of its digits belongs to the set $\{1,3,7,9\}$. Show that $B$ has a [b]prime divisor[/b] greater than or equal to 11.

Let $f : \mathbb R \to \mathbb R$ be a continuous function. Suppose that the restriction of $f$ to the set of irrational numbers is injective. What can we say about $f$? Answer the analogous question if $f$ is restricted to rationals.

Let $n \ge 2$ and let $x_1, x_2, ..., x_n$ be real numbers satisfying $x_1 +x_2 +...+x_n \ge 0$ and $x_1^2+x_2^2+...+x_n^2=1$. Let $M = max \{x_1, x_2,... , x_n\}$. Show that $M \ge \frac{1}{\sqrt{n(n-1)}}$ (1) .When does equality hold in (1)?

Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.

Which of the following represents the result when the figure shown below is rotated clockwise $120^\circ$ about its center? [asy] unitsize(6); draw(circle((0,0),5)); draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle); draw(circle((-2.5,-1.5),1)); draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle); [/asy] [asy] unitsize(6); for (int i = 0; i < 5; ++i) { draw(circle((12*i,0),5)); } draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle); draw(circle((-2.5,-1.5),1)); draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle); draw((14,-2)--(16,-2)--(15,-2+sqrt(3))--cycle); draw(circle((12,3),1)); draw((10.5,-1)--(9,0)--(8,-1.5)--(9.5,-2.5)--cycle); draw((22,-2)--(20,-2)--(21,-2+sqrt(3))--cycle); draw(circle((27,-1),1)); draw((24,1.5)--(22.75,2.75)--(24,4)--(25.25,2.75)--cycle); draw((35,2.5)--(37,2.5)--(36,2.5+sqrt(3))--cycle); draw(circle((39,-1),1)); draw((34.5,-1)--(33,0)--(32,-1.5)--(33.5,-2.5)--cycle); draw((50,-2)--(52,-2)--(51,-2+sqrt(3))--cycle); draw(circle((45.5,-1.5),1)); draw((48,1.5)--(46.75,2.75)--(48,4)--(49.25,2.75)--cycle); label("(A)",(0,5),N); label("(B)",(12,5),N); label("(C)",(24,5),N); label("(D)",(36,5),N); label("(E)",(48,5),N); [/asy]

On Misha's new phone, a passlock consists of six circles arranged in a $2\times 3$ rectangle. The lock is opened by a continuous path connecting the six circles; the path cannot pass through a circle on the way between two others (e.g. the top left and right circles cannot be adjacent). For example, the left path shown below is allowed but the right path is not. (Paths are considered to be oriented, so that a path starting at $A$ and ending at $B$ is different from a path starting at $B$ and ending at $A$. However, in the diagrams below, the paths are valid/invalid regardless of orientation.) How many passlocks are there consisting of all six circles? [asy] size(270); defaultpen(linewidth(0.8)); real r = 0.3, rad = 0.1, shift = 3.7; pen th = linewidth(5)+gray(0.2); for(int i=0; i<= 2;i=i+1) { for(int j=0; j<= 1;j=j+1) { fill(circle((i,j),r),gray(0.8)); fill(circle((i+shift,j),r),gray(0.8)); } draw((0,1)--(2-rad,1)^^(2,1-rad)--(2,rad)^^(2-rad,0)--(0,0),th); draw(arc((2-rad,1-rad),rad,0,90)^^arc((2-rad,rad),rad,270,360),th); draw((shift+1,0)--(shift+1,1-2*rad)^^(shift+1-rad,1-rad)--(shift+rad,1-rad)^^(shift+rad,1+rad)--(shift+2,1+rad),th); draw(arc((shift+1-rad,1-2*rad),rad,0,90)^^arc((shift+rad,1),rad,90,270),th); } [/asy]