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.)

In a triangle $ABC$ with $B = 90^\circ$, $D$ is a point on the segment $BC$ such that the inradii of triangles $ABD$ and $ADC$ are equal. If $\widehat{ADB} = \varphi$ then prove that $\tan^2 (\varphi/2) = \tan (C/2)$.

Let $ \triangle ABC $ be an acute triangle with circumcenter $ O $ and centroid $ G .$ Let $ D $ be the midpoint of $ BC $ and $ E\in \odot (BC) $ be a point inside $ \triangle ABC $ such that $ AE \perp BC . $ Let $ F=EG \cap OD $ and $ K, L $ be the point lie on $ BC $ such that $ FK \parallel OB, FL \parallel OC . $ Let $ M \in AB $ be a point such that $ MK \perp BC $ and $ N \in AC $ be a point such that $ NL \perp BC . $ Let $ \omega $ be a circle tangent to $ OB, OC $ at $ B, C, $ respectively $ . $ Prove that $ \odot (AMN) $ is tangent to $ \omega $

Let $ABC$ be an acute angled triangle with $AC>AB$ and incircle $\omega$. Let $\omega$ touch the sides $BC, CA,$ and $AB$ at $D, E,$ and $F$ respectively. Let $X$ and $Y$ be points outside $\triangle ABC$ satisfying \[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\] Prove that the circumcircles of $\triangle AXY, \triangle AEF$ and $\triangle ABC$ meet at a point $Z\ne A$. [i]Proposed by Atul Shatavart Nadig and Shantanu Nene[/i]

Let $f_n=\left(1+\frac{1}{n}\right)^n\left((2n-1)!F_n\right)^{\frac{1}{n}}$. Find $\lim \limits_{n \to \infty}(f_{n+1} - f_n)$ where $F_n$ denotes the $n$th Fibonacci number (given by $F_0 = 0$, $F_1 = 1$, and by $F_{n+1} = F_n + F_{n-1}$ for all $n \geq 1$

Find all the triples of prime numbers $(p, q, r)$ such that $$p | 2qr + r \,\,\,, \,\,\,q |2pr + p \,\,\, and \,\,\, r | 2pq + q.$$

Compute the number of positive integers that divide at least two of the integers in the set $\{1^1,2^2,3^3,4^4,5^5,6^6,7^7,8^8,9^9,10^{10}\}$.

Let $ABC$ be a triangle, and let $D, E$ and $F$ be the feet of the perpendiculars from $A, B$ and $C$ to $BC, CA$ and $AB$ respectively. Let $P, Q, R$ and $S$ be the feet of the perpendiculars from $D$ to $BA, BE, CF$ and $CA$ respectively. Prove that $P, Q, R$ and $S$ are collinear.

Let $n, k$ be natural numbers, $1 \leq k < n$. In each vertex of a regular polygon with $n$ sides is written $1$ or $-1$. At each step we choose $k$ consecutive vertices and change their signs. Is it possible that, starting from a certain configuration and by doing the operation a few times to obtain any other configuration?