An equilateral triangle of side $n$ has been divided into little equilateral triangles of side $1$ in the usual way. We draw a path over the segments of this triangulation, in such a way that it visits exactly once each one of the $\frac{(n+1)(n+2)}{2}$ vertices. What is the minimum number of times the path can change its direction? The figure below shows a valid path on a triangular board of side $4$, with exactly $9$ changes of direction. [asy] unitsize(30); pair h = (1, 0); pair v = dir(60); pair d = dir(120); for(int i = 0; i < 4; ++i) { draw(i*v -- i*v + (4 - i)*h); draw(i*h -- i*h + (4 - i)*v); draw((i + 1)*h -- (i + 1)*h + (i + 1)*d); } draw(h + v -- v -- (0, 0) -- 2*h -- 2*h + v -- h + 2*v -- 2*v -- 4*v -- 3*h + v -- 3*h -- 4*h, linewidth(2)); draw(3*h -- 4*h, EndArrow); fill(circle(h + v, 0.1)); [/asy] [i]Proposed by Oriol Solé[/i]

Given $100$ integers, is it always possible to choose $15$ of them such that the difference of any two of the chosen numbers is divisible by $7$? What is the answer if $15$ is replaced by $16$?

In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$. [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]

Let $m$ and $n$ be positive integers and $p$ be a prime number. Find the greatest positive integer $s$ (as a function of $m,n$ and $p$) such that from a random set of $mnp$ positive integers we can choose $snp$ numbers, such that they can be partitioned into $s$ sets of $np$ numbers, such that the sum of the numbers in every group gives the same remainder when divided by $p$.

Let the real numbers $a, b, c$ be such that $a \ge b \ge c > 0$. Show that $$\frac{a^2-b^2}{c}+ \frac{c^2-b^2}{a}+ \frac{a^2-c^2}{b}\ge 3a - 4b + c.$$ When does equality hold?

Find the root that the following three polynomials have in common: \begin{align*} & x^3+41x^2-49x-2009 \\ & x^3 + 5x^2-49x-245 \\ & x^3 + 39x^2 - 117x - 1435\end{align*}

Given five points in the plane, no three of which are collinear, prove that there can be found at least two obtuse-angled triangles with vertices at the given points. Construct an example in which there are exactly two such triangles.

For positive integer n, define $a_n$ as the number of the triangles with integer length of every side and the length of the longest side being $2n.$ (1) Find $a_n$ in terms of $n;$ (2)If the sequence $\{ b_n\}$ satisfying for any positive integer $n,$ $\sum_{k=1}^n(-1)^{n-k}\binom {n}{k} b_k=a_n.$ Find the number of positive integer $n$ satisfying that $b_n\leq 2019a_n.$

Two semicircles touch the side of the rectangle, each other and the segment drawn in it as in the figure. What part of the whole rectangle is filled? [img]https://cdn.artofproblemsolving.com/attachments/3/e/70ca8b80240a282553294a58cb3ed807d016be.png[/img]

$ABCD $ is a cyclic quadrilateral such that the diagonals $AC $ and $BD $ are perpendicular and their intersection is $P $. Point $Q $ on the segment $CP$ is such that $CQ=AP $. Prove that the perimeter of triangle $BDQ $ is at least $2AC $. Tuymaada 2017 Q2 Juniors

Find the sum of the series $ x\plus{}\frac{x^3}{1\cdot 3}\plus{}\frac{x^5}{1\cdot 3\cdot 5}\plus{}\cdots\plus{}\frac{x^{2n\plus{}1}}{1\cdot 3\cdot 5\cdot \cdots \cdot (2n\plus{}1)}\plus{}\cdots$

In the parallelogram $ABCD$, $AC \cap BD = { O }$, and $M$ is the midpoint of $AB$. Let $P \in (OC)$ and $MP \cap BC = { Q }$. We draw a line parallel to $MP$ from $O$, which intersects line $CD$ at point $N$. Show that $A,N,Q$ are collinear if and only if $P$ is the midpoint of $OC$.

Let $S$ be the set of all subsets of the positive integers. Construct a function $f \colon \mathbb{R} \rightarrow S$ such that $f(a)$ is a proper subset of $f(b)$ whenever $a <b.$

Given three nonzero real numbers $a,b,c,$ such that $a>b>c$, prove the equation has at least one real root. $$\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}-\frac{3}{x}=0$$ @below sorry, I believe I fixed it with the added constraint.

Find all pair $(x, y)$ in degrees such that \begin{align*} &\sin (x + 150^\circ) = \cos (y - 75^\circ), \\ &\cos x + \sin (y - 225^\circ) + \frac{\sqrt3}{2} = 0. \end{align*}

If $ A \star B$ means $ \frac{A\plus{}B}{2}$, then $ (3 \star 5) \star 8$ is \[ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 30 \]

Given an acute triangle $ABC$. The longest line of altitude is the one from vertex $A$ perpendicular to $BC$, and it's length is equal to the length of the median of vertex $B$. Prove that $\angle ABC \le 60^o$

In every cell of an $n\times n$ board is written $1$ or $-1$. At each step we may choose any of the $4n-2$ diagonals of the board and change the signs of all the numbers on that diagonal. Determine the number of initial configurations from which, after a finite number of steps, we may arrive at a configuration where all products of numbers on rows and columns equal to $1$. [i]Proposed by Pavel Ciurea[/i]

In \( \triangle ABC \) with \( AB > BC \), a tangent to the circumcircle of \( \triangle ABC \) at point \( B \) intersects the extension of \( AC \) at point \( D \). \( E \) is the midpoint of \( BD \), and \( AE \) intersects the circumcircle of \( \triangle ABC \) at \( F \). Prove that \( \angle CBF = \angle BDF \).

Triangle $ABC$ has angle $A$ measuring $30^\circ$, angle $B$ measuring $60^\circ$, and angle $C$ measuring $90^\circ$. Show four different ways to divide triangle $ABC$ into four triangles, each similar to triangle $ABC$, but with one quarter of the area. Prove that the angles and sizes of the smaller triangles are correct.

Find the smallest prime which is not the difference (in some order) of a power of $2$ and a power of $3$.

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]

The vertices of a triangle have coordinates $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. For any integers $h$ and $k$, not both 0, both triangles whose vertices have coordinates $(x_1+h,y_1+k),(x_2+h,y_2+k)$ and $(x_3+h,y_3+k)$ has no common interior points with the original triangle. (a) Is it possible for the area of this triangle to be greater than $\tfrac{1}{2}$? (b) What is the maximum area of this triangle?

In the triangle $ABC$, $A'$ is the foot of the altitude to $A$, and $H$ is the orthocenter. $a)$ Given a positive real number $k = \frac{AA'}{HA'}$ , find the relationship between the angles $B$ and $C$, as a function of $k$. $b)$ If $B$ and $C$ are fixed, find the locus of the vertice $A$ for any value of $k$.

Let $ABC$ be a triangle, let $O$ be its circumcenter, and let $H$ be its orthocenter. Let $P$ be a point on the segment $OH$. Prove that $6r\leq PA+PB+PC\leq 3R$, where $r$ is the inradius and $R$ the circumradius of triangle $ABC$. [b]Moderator edit:[/b] This is true only if the point $P$ lies inside the triangle $ABC$. (Of course, this is always fulfilled if triangle $ABC$ is acute-angled, since in this case the segment $OH$ completely lies inside the triangle $ABC$; but if triangle $ABC$ is obtuse-angled, then the condition about $P$ lying inside the triangle $ABC$ is really necessary.)