Every natural number is coloured in one of the $ k$ colors. Prove that there exist four distinct natural numbers $ a, b, c, d$, all coloured in the same colour, such that $ ad \equal{} bc$, $ \displaystyle \frac b a$ is power of 2 and $ \displaystyle \frac c a$ is power of 3.

Let $ABC$ be a triangle with area $9$, and let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Let $P$ be the point in side $BC$ such that $PC = \frac{1}{3}BC$. Let $O$ be the intersection point between $PN$ and $CM$. Find the area of the quadrilateral $BPOM$.

Let $ABC$ be an acute-angled, not equilateral triangle, where vertex $A$ lies on the perpendicular bisector of the segment $HO$, joining the orthocentre $H$ to the circumcentre $O$. Determine all possible values for the measure of angle $A$. (U.S.A. - 1989 IMO Shortlist)

Point $O$ is the circumcenter of acute-angled triangle $ABC$, points $A_1,B_1, C_1$ are the bases of its altitudes. Points $A', B', C'$ lying on lines $OA_1, OB_1, OC_1$ respectively are such that quadrilaterals $AOBC', BOCA', COAB'$ are cyclic. Prove that the circumcircles of triangles $AA_1A', BB_1B', CC_1C'$ have a common point.

Let $ ABC$ be an equilateral triangle, take line $ t $ such that $ t\parallel BC $ and $ t $ passes through $ A $. Let point $ D $ be on side $ AC $ , the bisector of angle $ ABD $ intersects line $ t $ in point $ E $. Prove that $ BD = CD + AE $.

In the quadrilateral $ABCD$ the side $AB$ has length $7, BC$ length $14, CD$ length $26$, and $DA$ length $23$. Show that the diagonals are perpendicular. You may assume that the quadrilateral is convex (all internal angles are less than $180^o$).

The sphere inscribed in a tetrahedron $ABCD$ touches face $ABC$ at point $H$. Another sphere touches face $ABC$ at $O$ and the planes containing the other three faces at points exterior to the faces. Prove that if $O$ is the circumcenter of triangle $ABC$, then $H$ is the orthocenter of that triangle.

A triangle $ABC$ and spheres are given in space $S_1$ and $S_2$, each of which passes through points $A, B$ and $C$. For points $M$ spheres $S_1$ not lying in the plane of triangle $ABC$ are drawn lines $MA, MB$ and $MC$, intersecting the sphere $S_2$ for the second time at points $A_1,B_1$ and $C_1$, respectively. Prove that the planes passing through points $A_1, B_1$ and $C_1$, touch a fixed sphere or pass through a fixed point.

A line segment $B_0B_n$ is divided into $n$ equal parts at points $B_1,B_2,...,B_{n-1} $ and $A$ is a point such that $\angle B_0AB_n$ is a right angle. Prove that : $$\sum_{k=0}^{n} |AB_k|^{2} = \sum_{k=0}^{n} |B_0B_k|^2$$

Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.

The bisector $BL$ was drawn in the triangle $ABC$. Let the points $I_1$ and $I_2$ be centers of the circles inscribed in the triangles $ABL$ and $CBL$, and the points $J_1$ and $J_2$ be centers of the excircles of these triangles touching the side $BL$. Prove that the points $I_1$, $I_2$, $J_1$ and $J_2$ lie on the same circle.

We define a [i]chessboard polygon[/i] to be a polygon whose sides are situated along lines of the form $ x \equal{} a$ or $ y \equal{} b$, where $ a$ and $ b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $ 1 \times 2$ rectangles. Finally, a [i]tasteful tiling[/i] is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $ 3 \times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner. [asy]size(300); pathpen = linewidth(2.5); void chessboard(int a, int b, pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle); } chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12)); /* draw lines */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy] a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique.

The area of triangle $ XYZ$ is 8 square inches. Points $ A$ and $ B$ are midpoints of congruent segments $ \overline{XY}$ and $ \overline{XZ}$. Altitude $ \overline{XC}$ bisects $ \overline{YZ}$. What is the area (in square inches) of the shaded region? [asy]/* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy] $ \textbf{(A)}\ 1\frac12\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac12\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac12$

On the sides $AB,BC,CD$ and $DA$ of a unit square $ABCD$ points $P,Q,R$ and $S$ are chosen respectively. It turned out that the perimeter of $PQRS$ is $2\sqrt{2}$. Find the sum of perpendiculars from $A,B,C,D$ to $SP,PQ,QR,RS$ respectively.

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

Squares $ABKL$, $BCMN$, $CAOP$ are drawn externally on the sides of a triangle $ABC$. The line segments $KL$, $MN$, $OP$, when extended, form a triangle $A'B'C'$. Find the area of $A'B'C'$ if $ABC$ is an equilateral triangle of side length $2$.

The bisectors $AD{}$ and $BE{}$ were drawn in the triangle $ABC{}$ and they intersected at point $I{}.$ Then everything was erased, leaving only the points $D{}$ and $E{}.$ Find the set of possible positions of the point $I{}.$ [i]Proposed by M. Didin[/i]

Two circles $\omega_1,\omega_2$ intersect at $A,B$. An arbitrary line through $B$ meets $\omega_1,\omega_2$ at $C,D$ respectively. The points $E,F$ are chosen on $\omega_1,\omega_2$ respectively so that $CE=CB,\ BD=DF$. Suppose that $BF$ meets $\omega_1$ at $P$, and $BE$ meets $\omega_2$ at $Q$. Prove that $A,P,Q$ are collinear. [i]Proposed by Iman Maghsoudi[/i]

Let $ABCD$ be a quadrilateral with side lengths $AB = 2$, $BC = 3$, $CD = 5$, and $DA = 4$. What is the maximum possible radius of a circle inscribed in quadrilateral $ABCD$?

Let points $ A = (0,0) , \ B = (1,2), \ C = (3,3), $ and $ D = (4,0) $. Quadrilateral $ ABCD $ is cut into equal area pieces by a line passing through $ A $. This line intersects $ \overline{CD} $ at point $ \left (\frac{p}{q}, \frac{r}{s} \right ) $, where these fractions are in lowest terms. What is $ p + q + r + s $? $ \textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75 $

Three circles with radii $23$, $46$, and $69$ are tangent to each other as shown in the figure below (figure is not drawn to scale). Find the radius of the largest circle that can fit inside the shaded region. [img]https://cdn.artofproblemsolving.com/attachments/6/d/158abc178e4ddd72541580958a4ee2348b2026.png[/img]

The area of the smallest square that will contain a circle of radius 4 is $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 128$

Two straight lines $a$ and $b$ are given and also points $A$ and $B$. Point $X$ slides along the line $a$, and point $Y$ slides along the line $b$, so that $AX \parallel BY$. Find the locus of the intersection point of $AY$ with $XB$.

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is $ \textbf{(A)}\ \frac{\sqrt{3}}{12} \qquad\textbf{(B)}\ \frac{\sqrt{6}}{16} \qquad\textbf{(C)}\ \frac{1}{6} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{8} \qquad\textbf{(E)}\ \frac{1}{4} $

Let $D$ be a point inside the equilateral triangle $\triangle ABC$ such that $|AD|=\sqrt{2}, |BD|=3, |CD|=\sqrt{5}$. $m(\widehat{ADB}) = ?$ $\textbf{(A)}\ 120^{\circ} \qquad\textbf{(B)}\ 105^{\circ} \qquad\textbf{(C)}\ 100^{\circ} \qquad\textbf{(D)}\ 95^{\circ} \qquad\textbf{(E)}\ 90^{\circ}$