A merida path of order $2n$ is a lattice path in the first quadrant of $xy$- plane joining $(0,0)$ to $(2n,0)$ using three kinds of steps $U=(1,1)$, $D= (1,-1)$ and $L= (2,0)$, i.e. $U$ joins $x,y)$ to $(x+1,y+1)$ etc... An ascent in a merida path is a maximal string of consecutive steps of the form $U$. If $S(n,k)$ denotes the number of merdia paths of order $2n$ with exactly $k$ ascents, compute $S(n,1)$ and $S(n,n-1)$.

In an election between $\text{A}$ and $\text{B}$, during the counting of the votes, neither candidate was more than $2$ votes ahead, and the vote ended in a tie, $6$ votes to $6$ votes. Two votes for the same candidate are indistinguishable. In how many orders could the votes have been counted? One possibility is $\text{AABBABBABABA}$.

The volleyball championship with $16$ teams was held in one round (each team played with each exactly one times, there are no draws in volleyball). It turned out that some two teams won the same number of matches. Prove there are the three teams that beat each other in a round robin (i.e. A beat B, B beat C, and C beat A).

The vertices of a triangle are lattice points (they have integer coordinates). There are no other lattice points on the boundary of the triangle, but there is exactly one lattice point inside the triangle. Show that it must be the centroid.

Show that $\vert 12^m -5^n\vert \ge 7$ for all $m, n \in \mathbb{N}$.

For which $n\in\{3900, 3901,\cdots, 3909\}$ can the set $\{1, 2, . . . , n\}$ be partitioned into (disjoint) triples in such a way that in each triple one of the numbers equals the sum of the other two?

In $\triangle ABC, \angle A = 100^\circ, \angle B = 50^\circ, \angle C = 30^\circ, \overline{AH}$ is an altitude, and $\overline{BM}$ is a median. Then $\angle MHC =$ [asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); label("A", (6,6), N); label("B", (0,0), W); label("C", (16,0), E); label("H", (6,0), S); label("M", (11,3), NE);[/asy] $\text{(A)} \ 15^\circ \qquad \text{(B)} \ 22.5^\circ \qquad \text{(C)} \ 30^\circ \qquad \text{(D)} \ 40^\circ \qquad \text{(E)} \ 45^\circ$

Let $F_n$ be the set of all bijective functions $f:\left\{1,2,\ldots,n\right\}\rightarrow\left\{1,2,\ldots,n\right\}$ that satisfy the conditions: 1. $f(k)\leq k+1$ for all $k\in\left\{1,2,\ldots,n\right\}$ 2. $f(k)\neq k$ for all $k\in\left\{2,3,\ldots,n\right\}$ Find the probability that $f(1)\neq1$ for $f$ randomly chosen from $F_n$.

The set of integers is coloured in $100$ colours in such a way that all the colours are used and the following is true. For any choice of intervals $[a, b]$ and $[c,d]$ of equal length and with integral endpoints, if a and c as well as $b$ and $d$, respectively, have the same colour, then the whole intervals $[a, b]$ and $[c,d]$ are identically coloured in that, for any integer $x$, $0 \le x \le b - a$, the numbers $a + x$ and $c + x$ are of the same colour. Prove that $-1982$ and $1982$ are of different colours

For reals $1\leq a\leq b \leq c \leq d \leq e \leq f$ prove inequality $(af + be + cd)(af + bd + ce) \leq (a + b^2 + c^3 )(d + e^2 + f^3 )$.

In a triangle $ ABC$, take point $ D$ on $ BC$ such that $ DB \equal{} 14, DA \equal{} 13, DC \equal{} 4$, and the circumcircle of $ ADB$ is congruent to the circumcircle of $ ADC$. What is the area of triangle $ ABC$?

We call a coloring of an $8\times 8$ board in three colors good if in any corner of five cells contains cells of all three colors. (A five-square corner is a shape made from a $3 \times 3$ square by cutting square $ 2\times 2$.) Prove that the number of good colorings is not less than than $68$.

In triangle $ABC$, point $I$ is the center of the inscribed circle. $AT$ is a segment tangent to the circle circumscribed around the triangle $BIC$ . On the ray $AB$ beyond the point$ B$ and on the ray $AC$ beyond the point $C$, we draw the segments $BD$ and $CE$, respectively, such that $BD = CE = AT$. Let the point $F$ be such that $ABFC$ is a parallelogram. Prove that points $D, E$ and $F$ lie on the same line. (Dmitry Prokopenko)

Let $ABC$ be a triangle with $I$ as incenter.The incircle touches $BC$ at $D$.Let $D'$ be the antipode of $D$ on the incircle.Make a tangent at $D'$ to incircle.Let it meet $(ABC)$ at $X,Y$ respectively.Let the other tangent from $X$ meet the other tangent from $Y$ at $Z$.Prove that $(ZBD)$ meets $IB$ at the midpoint of $IB$

Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$ Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $$B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$$ Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$.

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted? [asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); label("$4$", (2,0), N); label("$3$", (0,1.5), E); label("$2$", (.8,1), E); label("$S$", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); [/asy] $\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} $