Triangle $ABC$ is given. Let $M$ be the midpoint of the segment $AB$ and $T$ be the midpoint of the arc $BC$ not containing $A$ of the circumcircle of $ABC.$ The point $K$ inside the triangle $ABC$ is such that $MATK$ is an isosceles trapezoid with $AT\parallel MK.$ Show that $AK = KC.$

Let $AEF$ be a triangle with $EF = 20$ and $AE = AF = 21$. Let $B$ and $D$ be points chosen on segments $AE$ and $AF,$ respectively, such that $BD$ is parallel to $EF.$ Point $C$ is chosen in the interior of triangle $AEF$ such that $ABCD$ is cyclic. If $BC = 3$ and $CD = 4,$ then the ratio of areas $\tfrac{[ABCD]}{[AEF]}$ can be written as $\tfrac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$.

In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD$; $M$ is the midpoint of $AB$; $N$ is the midpoint of $BC$; and $AN$ and $CM$ intersect at $O$. The ratio of the area of $AOCD$ to the area of $ABCD$ is [asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); label("A", (0,0), S); label("B", (2,0), S); label("C", (2,2), N); label("D", (0,2), N); label("M", (1,0), S); label("N", (2,1), E); label("O", (1.2, .8)); [/asy] $ \textbf{(A)}\ \frac{5}{6} \qquad\textbf{(B)}\ \frac{3}{4} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(E)}\ \frac{(\sqrt{3}-1)}{2} $

Let $ABCD$ be an isosceles trapezoid, $AD=BC$, $AB \parallel CD$. The diagonals of the trapezoid intersect at the point $O$, and the point $M$ is the midpoint of the side $AD$. The circle circumscribed around the triangle $BCM$ intersects the side $AD$ at the point $K$. Prove that $OK \parallel AB$.

The sum of the base-$ 10$ logarithms of the divisors of $ 10^n$ is $ 792$. What is $ n$? $ \textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

Given $\triangle ABC$ inscribed in $(O)$ an let $I$ and $I_a$ be it's incenter and $A$-excenter ,respectively. Tangent lines to $(O)$ at $C,B$ intersect the angle bisector of $A$ at $M,N$ ,respectively. Second tangent lines through $M,N$ intersect $(O)$ at $X,Y$. Prove that $XYII_a$ is cyclic.

An artist has $14$ cubes, each with an edge of $1$ meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surface of the sculpture. How many square meters does she paint? $\text{(A)}\ 21 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 33 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 42$ [asy] draw((0,0)--(2.35,-.15)--(2.44,.81)--(.09,.96)--cycle); draw((.783333333,-.05)--(.873333333,.91)--(1.135,1.135)); draw((1.566666667,-.1)--(1.656666667,.86)--(1.89,1.1)); draw((2.35,-.15)--(4.3,1.5)--(4.39,2.46)--(2.44,.81)); draw((3,.4)--(3.09,1.36)--(2.61,1.4)); draw((3.65,.95)--(3.74,1.91)--(3.29,1.94)); draw((.09,.96)--(.76,1.49)--(.71,1.17)--(2.2,1.1)--(3.6,2.2)--(3.62,2.52)--(4.39,2.46)); draw((.76,1.49)--(.82,1.96)--(2.28,1.89)--(2.2,1.1)); draw((2.28,1.89)--(3.68,2.99)--(3.62,2.52)); draw((1.455,1.135)--(1.55,1.925)--(1.89,2.26)); draw((2.5,2.48)--(2.98,2.44)--(2.9,1.65)); draw((.82,1.96)--(1.55,2.6)--(1.51,2.3)--(2.2,2.26)--(2.9,2.8)--(2.93,3.05)--(3.68,2.99)); draw((1.55,2.6)--(1.59,3.09)--(2.28,3.05)--(2.2,2.26)); draw((2.28,3.05)--(2.98,3.59)--(2.93,3.05)); draw((1.59,3.09)--(2.29,3.63)--(2.98,3.59)); [/asy]

A square $ABCD$ with sude length 1 is given and a circle with diameter $AD$. Find the radius of the circumcircle of this figure.

Area of convex quadrilateral $ABCD$ is $1$. Prove that we can find four points on its side (vertex included) or inside, satisfying: area of triangles comprised of any three points of the four points is larger than $\frac{1}{4}$.

Points $D$ and $E$ are chosen on the sides $AB$ and $AC$ of the triangle $ABC$ in such a way that if $F$ is the intersection point of $BE$ and $CD$, then $AE + EF = AD + DF$. Prove that $AC + CF = AB + BF.$

Let $C$ be an interior point of line segment $AB$. Equilateral triangles $ADC$ and $CEB$ are constructed to the same side from $AB$. Find all points which can be the midpoint of the segment $DE$.

A scalene acute triangle has angles whose measures (in degrees) are whole numbers. What is the smallest possible measure of one of the angles, in degrees?

Let $a,b,c,$ and $d$ be the lengths of sides $MN,NP,PQ,$ and $QM$, respectively, of quadrilateral $MNPQ$. If $A$ is the area of $MNPQ$, then $\textbf{(A) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is convex}$ $\textbf{(B) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(C) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(D) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$ $\textbf{(E) }A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$

[b]p1[/b]. Prove that there is no interger $n$ such that $n^2 +1$ is divisible by $7$. [b]p2.[/b] Find all solutions of the system $$x^2-yz=1$$ $$y^2-xz=2$$ $$z^2-xy=3$$ [b]p3.[/b] A triangle with long legs is an isoceles triangle in which the length of the two equal sides is greater than or equal to the length of the remaining side. What is the maximum number, $n$ , of points in the plane with the property that every three of them form the vertices of a triangle with long legs? Prove all assertions. [b]p4.[/b] Prove that the area of a quadrilateral of sides $a, b, c, d$ which can be inscribed in a circle and circumscribed about another circle is given by $A=\sqrt{abcd}$ [b]p5.[/b] Prove that all of the squares of side length $$\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},...,\frac{1}{n},...$$ can fit inside a square of side length $1$ without overlap. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

A perpendicular bisector to the side $AC$ of triangle $ABC$ meets $BC,AB$ at points $A_1$ and $C_1$ respectively. Points $O,O_1$ are the circumcenters of triangles $ABC$ and $A_1BC_1$ respectively. Prove that $C_1O_1\perp AO$.

In the right triangle $ABC,$ $CF$ is the altitude based on the hypotenuse $AB.$ The circle centered at $B$ and passing through $F$ and the circle with centre $A$ and the same radius intersect at a point of $CB.$ Determine the ratio $FB : BC.$

The points $A,B,C,D$ lie, in this order, on a circle $\omega$, where $AD$ is a diameter of $\omega$. Furthermore, $AB=BC=a$ and $CD=c$ for some relatively prime integers $a$ and $c$. Show that if the diameter $d$ of $\omega$ is also an integer, then either $d$ or $2d$ is a perfect square.

Let $ABC$ be an acute triangle with the circumcircle $k$ and incenter $I$. The perpendicular through $I$ in $CI$ intersects segment $[BC]$ in $U$ and $k$ in $V$. In particular $V$ and $A$ are on different sides of $BC$. The parallel line through $U$ to $AI$ intersects $AV$ in $X$. Prove: If $XI$ and $AI$ are perpendicular to each other, then $XI$ intersects segment $[AC]$ in its midpoint $M$. [i](Notation: $[\cdot]$ denotes the line segment.)[/i]

Let $ABCDEF$ be a convex hexagon in which diagonals $AD, BE, CF$ are concurrent at $O$. Suppose $[OAF]$ is geometric mean of $[OAB]$ and $[OEF]$ and $[OBC]$ is geometric mean of $[OAB]$ and $[OCD]$. Prove that $[OED]$ is the geometric mean of $[OCD]$ and $[OEF]$. (Here $[XYZ]$ denotes are of $\triangle XYZ$)

$I_b$ is the $B$-excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$. $MI_b$ meets $\omega$ at $T\not =M$. Prove that $$ TB\cdot TC=TI_b^2.$$

In space, we are given the points $A,B,C$ and a sphere with center $O$ and radius $1$. Find the point $X$ from the sphere for which the sum $f(X)=|XA|^2+|XB|^2+|XC|^2$ attains its maximal and minimal value. Prove that if the segments $OA,OB,OC$ are pairwise perpendicular and $d$ is the distance from the center $O$ to the centroid of the triangle $ABC$ then: (a) the maximum of $f(X)$ is equal to $9d^2+3+6d$; (b) the minimum of $f(X)$ is equal to $9d^2+3-6d$. [i]K. Dochev and I. Dimovski[/i]

Given two internally tangent circles; in the bigger one we inscribe an equilateral triangle. From each of the vertices of this triangle, we draw a tangent to the smaller circle. Prove that the length of one of these tangents equals the sum of the lengths of the two other tangents.

In the Cartesian plane, each point with integer coordinates is colored either red, green, or blue. It is possible to form right isosceles triangles ($45^\circ - 90^\circ - 45^\circ$) using colored points as vertices. Prove that regardless of how the coloring is done, there always exists a right isosceles triangle such that all its vertices are either the same color or all different colors.

In the isosceles triangle $ABC$, with $AB=AC$, $D$ is a point on the extension of $CA$ such that $DB$ is perpendicular to $BC$, $E$ is a point on the extension of $BC$ such that $CE=2BC$, and $F$ is a point on $ED$ such that $FC$ is parallel to $AB$. Prove that $FA$ is parallel to $BC$.

A)Let $x,y$ be two complex numbers on the unit circle so that: $\frac{\pi }{3} \le \arg (x)-\arg (y) \le \frac{5 \pi }{3}$ Prove that for any $z \in \mathbb{C}$ we have: $|z|+|z-x|+|z-y| \ge |zx-y|$ B)Let $x,y$ be two complex numbers so that: $\frac{\pi }{3} \le \arg (x)-\arg (y) \le \frac{2 \pi }{3}$ Prove that for any $z \in \mathbb{C}$ we have: $|z|+|z-y|+|z-x| \ge | \frac{\sqrt{3}}{2} x +(y-\frac{x}{2})i|$