The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is: $\textbf{(A)}\ 12\quad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 6\sqrt{2}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 3\sqrt{2}$

Five people are standing in a straight line, and the distance between any two people is a unique positive integer number of units. Find the least possible distance between the leftmost and rightmost people in the line in units.

Given a point $p$ inside a convex polyhedron $P$. Show that there is a face $F$ of $P$ such that the foot of the perpendicular from $p$ to $F$ lies in the interior of $F$.

Let $P$ be the intersection point of the diagonals of the quadrangle $ABCD$, $M$ the intersection point of the lines connecting the midpoints of its opposite sides, $O$ the intersection point of the perpendicular bisectors of the diagonals, $H$ the intersection point of the lines connecting the orthocenters of the triangles $APD$ and $BCP$, $APB$ and $CPD$. Prove that $M$ is the midpoint of $OH$.

Point $A$ lies inside an angle with vertex $M$. A ray issuing from point $A$ is reflected in one side of the angle at point $B$, then in the other side at point $C$ and then returns back to point $A$ (the ordinary rule of reflection holds). Prove that the center of the circle circumscribed about triangle $\triangle BCM$ lies on line $AM$.

In acute triangle $ABC$ ($\triangle ABC$ is not an isosceles triangle), $I$ is its incentre, and circle $ \omega$ is its inscribed circle. $\odot\omega$ touches $BC, CA, AB$ at $D, E, F$ respectively. $AD$ intersects with $\odot\omega$ at $J$ ($J\neq D$), and the circumcircle of $\triangle BCJ$ intersects $\odot\omega$ at $K$ ($K\neq J$). The circumcircle of $\triangle BFK$ and $\triangle CEK$ meet at $L$ ($L\neq K$). Let $M$ be the midpoint of the major arc $BAC$. Prove that $M, I, L$ are collinear.

An isosceles triangle $ABC$ with base $AC$ is given. On the rays $CA$, $AB$ and $BC$, the points $D, E$ and $F$ were marked, respectively, in such a way that $AD = AC$, $BE = BA$ and $CF = CB$. Find the sum of the angles $\angle ADB$, $\angle BEC$ and $\angle CFA$.

Given any $60$ points on a circle of radius $1$, prove that there is a point on the circle the sum of whose distances to these $60$ points is at most $80$.

Quadrilateral $ABCD$ is given such that $$\angle DAC = \angle CAB = 60^\circ,$$ and $$AB = BD - AC.$$ Lines $AB$ and $CD$ intersect each other at point $E$. Prove that \[ \angle ADB = 2\angle BEC. \] [i]Proposed by Iman Maghsoudi[/i]

In acute triangle $ ABC,$ points $ P,Q$ lie on its sidelines $ AB,AC,$ respectively. The circumcircle of triangle $ ABC$ intersects of triangle $ APQ$ at $ X$ (different from $ A$). Let $ Y$ be the reflection of $ X$ in line $ PQ.$ Given $ PX>PB.$ Prove that $ S_{\bigtriangleup XPQ}>S_{\bigtriangleup YBC}.$ Where $ S_{\bigtriangleup XYZ}$ denotes the area of triangle $ XYZ.$

$ABC$ is a triangle in the plane. Find the locus of point $P$ for which $PA,PB,PC$ form a triangle whose area is equal to one third of the area of triangle $ABC$.

The diameters of two circles are $ 8$ inches and $ 12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is: $ \textbf{(A)}\ \frac{2}{3} \qquad\textbf{(B)}\ \frac{4}{9} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{1}{2} \qquad\textbf{(E)}\ \text{none of these}$

2 Points $ P\left(a,\ \frac{1}{a}\right),\ Q\left(2a,\ \frac{1}{2a}\right)\ (a > 0)$ are on the curve $ C: y \equal{}\frac{1}{x}$. Let $ l,\ m$ be the tangent lines at $ P,\ Q$ respectively. Find the area of the figure surrounded by $ l,\ m$ and $ C$.

A $(3n + 1) \times (3n + 1)$ table $(n \in \mathbb{N})$ is given. Prove that deleting any one of its squares yields a shape cuttable into pieces of the following form and its rotations: ''L" shape formed by cutting one square from a $2 \times 2$ squares.

A cube is made by soldering twelve $ 3$-inch lengths of wire properly at the vertices of the cube. If a fly alights at one of the vertices and then walks along the edges, the greatest distance it could travel before coming to any vertex a second time, without retracing any distance, is: $ \textbf{(A)}\ 24\text{ in.}\qquad \textbf{(B)}\ 12\text{ in.}\qquad \textbf{(C)}\ 30\text{ in.}\qquad \textbf{(D)}\ 18\text{ in.}\qquad \textbf{(E)}\ 36\text{ in.}$

On a line $\ell$ there exists $3$ points $A, B$, and $C$ where $B$ is located between $A$ and $C$. Let $\Gamma_1, \Gamma_2, \Gamma_3$ be circles with $AC, AB$, and $BC$ as diameter respectively; $BD$ is a segment, perpendicular to $\ell$ with $D$ on $\Gamma_1$. Circles $\Gamma_4, \Gamma_5, \Gamma_6$ and $\Gamma_7$ satisfies the following conditions: $\bullet$ $\Gamma_4$ touches $\Gamma_1, \Gamma_2$, and$ BD$. $\bullet$ $\Gamma_5$ touches $\Gamma_1, \Gamma_3$, and $BD$. $\bullet$ $\Gamma_6$ touches $\Gamma_1$ internally, and touches $\Gamma_2$ and $\Gamma_3$ externally. $\bullet$ $\Gamma_7$ passes through $B$ and the tangent points of $\Gamma_2$ with $\Gamma_6$, and $\Gamma_3$ with $\Gamma_6$. Show that the circles $\Gamma_4, \Gamma_5$, and $\Gamma_7$ are congruent.

Let $ ABCD $ be a convex quadrilateral with no pair of parallel sides, such that $ \angle ABC = \angle CDA $. Assume that the intersections of the pairs of neighbouring angle bisectors of $ ABCD $ form a convex quadrilateral $ EFGH $. Let $ K $ be the intersection of the diagonals of $ EFGH$. Prove that the lines $ AB $ and $ CD $ intersect on the circumcircle of the triangle $ BKD $.

Let $ABC$ be a triangle inscribed in circle $\omega$ with center $O$, let $\omega_A$ be its $A$-mixtilinear incircle, $\omega_B$ be its $B$-mixtilinear incircle, $\omega_C$ be its $C$-mixtilinear incircle, and $X$ be the radical center of $\omega_A$, $\omega_B$, $\omega_C$. Let $A'$, $B'$, $C'$ be the points at which $\omega_A$, $\omega_B$, $\omega_C$ are tangent to $\omega$. Prove that $AA'$, $BB'$, $CC'$ and $OX$ are concurrent. [i]Proposed by Robin Park[/i]

Let $n \geq 2$ be an integer. Find the smallest positive integer $m$ for which the following holds: given $n$ points in the plane, no three on a line, there are $m$ lines such that no line passes through any of the given points, and for all points $X \neq Y$ there is a line with respect to which $X$ and $Y$ lie on opposite sides

Into how many regions do $n$ circles divide the plane, if each pair of circles intersects in two points and no point lies on three circles?

Denote by $M$ and $N$ feets of perpendiculars from $A$ to angle bisectors of exterior angles at $B$ and $C,$ in triangle $\triangle ABC.$ Prove that the length of segment $MN$ is equal to semiperimeter of triangle $\triangle ABC.$

Three circles $\Gamma_A$, $\Gamma_B$ and $\Gamma_C$ share a common point of intersection $O$. The other common point of $\Gamma_A$ and $\Gamma_B$ is $C$, that of $\Gamma_A$ and $\Gamma_C$ is $B$, and that of $\Gamma_C$ and $\Gamma_B$ is $A$. The line $AO$ intersects the circle $\Gamma_A$ in the point $X \ne O$. Similarly, the line $BO$ intersects the circle $\Gamma_B$ in the point $Y \ne O$, and the line $CO$ intersects the circle $\Gamma_C$ in the point $Z \ne O$. Show that \[\frac{|AY |\cdot|BZ|\cdot|CX|}{|AZ|\cdot|BX|\cdot|CY |}= 1.\]

Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube? $ \textbf{(A)}\ \frac {1}{8}\qquad \textbf{(B)}\ \frac {3}{16}\qquad \textbf{(C)}\ \frac {1}{4} \qquad \textbf{(D)}\ \frac {3}{8}\qquad \textbf{(E)}\ \frac {1}{2}$

In an acute-angled triangle $ABC$, $M$ is the midpoint of side $BC$, and $D, E$ and $F$ the feet of the altitudes from $A, B$ and $C$, respectively. Let $H$ be the orthocenter of $\Delta ABC$, $S$ the midpoint of $AH$, and $G$ the intersection of $FE$ and $AH$. If $N$ is the intersection of the median $AM$ and the circumcircle of $\Delta BCH$, prove that $\angle HMA = \angle GNS$. [i]Proposed by Marko Djikic[/i]

Points $A$ and $B$ are given on a circle. With the help of a compass and a ruler, construct on this circle the points $C,$ $D$, $E$ that lie on one side of the straight line $AB$ and for which the pentagon with vertices $A$, $B$, $C$, $D$, $E$ has the largest possible area