Divide a regular hexagon into $24$ identical small equilateral triangles as shown. To each of the $19$ points that are vertices of at least one of the equilateral triangles, a different number is assigned. Find the maximum possible number of small equilateral triangles with their vertices having numbers in ascending order going clockwise. [asy] size(5.5cm); int n=6; pair[] V= sequence(new pair(int i){return dir(360*i/n);}, n); V.cyclic=true; for(int i=0;i<n;++i){ draw(V[i]--V[i+1],black+0.7bp); draw(V[i]--V[i+3],black+0.7bp); draw(midpoint(V[i]--V[i+1])--midpoint(V[i+2]--V[i+3]),black+0.7bp); } [/asy]

Let $ABCD$ be a parallelogram. $E$ and $F$ are on $BC, CD$ respectively such that the triangles $ABE$ and $BCF$ have the same area. Let $BD$ intersect $AE, AF$ at $M, N$ respectively. Prove there exists a triangle whose side lengths are $BM, MN, ND$.

Find all functions $f: R \rightarrow R$ such that $f(x^{2015} + (f(y))^{2015}) = (f(x))^{2015} + y^{2015}$ holds for all reals $x, y$

Let $k$ be a positive integer. Determine the least positive integer $n$, with $n\geq k+1$, for which the game below can be played indefinitely: Consider $n$ boxes, labelled $b_1,b_2,...,b_n$. For each index $i$, box $b_i$ contains exactly $i$ coins. At each step, the following three substeps are performed in order: [b](1)[/b] Choose $k+1$ boxes; [b](2)[/b] Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add to the remaining box, if labelled $b_i$, a number of $i$ coins. [b](3)[/b] If one of the boxes is left empty, the game ends; otherwise, go to the next step. [i]Proposed by Demetres Christofides, Cyprus[/i]

In triangle $ABC, \angle A = 90^o$. $M$ is the midpoint of $BC$ and $H$ is the foot of the altitude from $A$ to $BC$. The line passing through $M$ and perpendicular to $AC$ meets the circumcircle of triangle $AMC$ again at $P$. If $BP$ intersects $AH$ at $K$, prove that $AK = KH$.

$\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$.

What is the maximum number of bishops that can be placed on an $ 8 \times 8 $ chessboard such that at most three bishops lie on any diagonal?

In Kacs Aladár street, houses are only found on one side of the road, so that only odd house numbers are found along the street. There are an odd number of allotments, as well. The middle three allotments belong to Scrooge McDuck, so he only put up the smallest of the three house numbers. The numbering of the other houses is standard, and the numbering begins with $1$. What is the largest number in the street if the sum of house numbers put up is $3133$?

Let $b_1,b_2,... $ be a sequence of positive real numbers such that for each $ n\ge 1$, $$b_{n+1}^2 \ge \frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+...+\frac{b_n^2}{n^3}$$ Show that there is a positive integer $M$ such that $$\sum_{n=1}^M \frac{b_{n+1}}{b_1+b_2+...+b_n} > \frac{2013}{1013}$$

[b]1A.[/b] The iTest, by virtue of being the first national internet-based high school math competition, saves a lot of paper every year. The quantity of trees saved (“$a$”) is determined by the following formula: $a = x^2 + 3x + 9$, where $x$ is the number of participating students in the competition. If $x$ is the correct answer from short answer [hide=problem 22]x=20[/hide], then find $a$. [i](1 point)[/i] [b]1B.[/b] Let $q$ be the sum of the digits of $a$. If $q = b! - (b-1)! + (b-2)! - (b-3)!$, find $b$. [i](2 points)[/i] [b]1C.[/b] Find the number of the following statements that are false: [i] (4 points)[/i] 1. $q$ is the first prime number resulting from the sum of cubes of distinct fractions, where both the numerator and denominator are primes. 2. $q$ is composite. 3. $q$ is composite and is the sum of the first four prime numbers and $1$. 4. $q$ is the smallest prime equal to the difference of cubes of two consecutive primes. 5. $q$ is not the smallest prime equal to the product of twin primes plus their arithmetic mean. 6. The sum of $q$ consecutive Fibonacci numbers, starting from the $q^{th}$ Fibonacci number, is prime. 7. $q$ is the largest prime factor of $1bbb$. 8. $q$ is the $8^{th}$ largest prime number. 9. $a$ is composite. 10. $a + q + b = q^2$. 11. The decimal expansion of $q^q$ begins with $q$. 12. $q$ is the smallest prime equal to the sum of three distinct primes. 13. $q^5 + q^2 + q^1 + q^3 + q^5 + q^6 + q^4 + q^0 = 52135640$. 14. $q$ is not the smallest prime such that $q$ and $q^2$ have the same sum of their digits. 15. $q$ is the smallest prime such that $q$ = (the product of its digits + the sum of its digits). [hide=ANSWER KEY]1A. 469 1B. 4 1C. 6[/hide]

Solve $|x-1|-2|x+5|>3+x$.

Evaluate \[ \log_{10}(\tan 1^{\circ})+ \log_{10}(\tan 2^{\circ})+ \log_{10}(\tan 3^{\circ})+ \cdots + \log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}). \] $ \textbf{(A)}\ 0 \qquad\textbf{(B)}\ \frac{1}{2}\log_{10}(\frac{\sqrt{3}}{2}) \qquad\textbf{(C)}\ \frac{1}{2}\log_{10}2 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ \text{none of these} $

Kailey starts with the number $0$, and she has a fair coin with sides labeled $1$ and $2$. She repeatedly flips the coin, and adds the result to her number. She stops when her number is a positive perfect square. What is the expected value of Kailey’s number when she stops? If E is your estimate and A is the correct answer, you will receive $\left\lfloor 25e^{-\frac{5|E-A|}{2} }\right\rfloor$ points.

Two circles are tangent to each other internally at a point $\ T $. Let the chord $\ AB $ of the larger circle be tangent to the smaller circle at a point $\ P $. Prove that the line $\ TP $ bisects $\ \angle ATB $.

For a convex polygon with $n$ edges $F$, if all its diagonals have the equal length, then $\text{(A)}F\in \{\text{quadrilaterals}\}$ $\text{(B)}F\in \{\text{pentagons}\}$ $\text{(C)}F\in \{\text{pentagons}\} \cup\{\text{quadrilaterals}\}$ $\text{(D)}F\in \{\text{convex polygons that have all edges' length equal}\} \cup\{\text{convex polygons that have all inner angles equal}\}$

Jane and Josh wish to buy a candy. However Jane needs seven more cents to buy the candy, while John needs one more cent. They decide to buy only one candy together, but discover that they do not have enough money. How much does the candy cost?