Does there exist a triangle in which two sides are integer multiples of the median to that side? Does there exist a triangle in which every side is an integer multiple of the median to that side?

Two circles $k_1$ and $k_2$ with centers $O_1$ and $O_2$ are tangent to each other externally at point $O$. Points $X$ and $Y$ on $k_1$ and $k_2$ respectively are such that rays $O_1X$ and $O_2Y$ are parallel and codirectional. Prove that two tangents from $X$ to $k_2$ and two tangents from $Y$ to $k_1$ touch the same circle passing through $O$. (V. Yasinsky)

The sides of an equilateral triangle $ABC$ are divided into $n$ equal parts $(n \geq 2) .$ For each point on a side, we draw the lines parallel to other sides of the triangle $ABC,$ e.g. for $n=3$ we have the following diagram: [asy] unitsize(150); defaultpen(linewidth(0.7)); int n = 3; /* # of vertical lines, including AB */ pair A = (0,0), B = dir(-30), C = dir(30); draw(A--B--C--cycle,linewidth(2)); dot(A,UnFill(0)); dot(B,UnFill(0)); dot(C,UnFill(0)); label("$A$",A,W); label("$C$",C,NE); label("$B$",B,SE); for(int i = 1; i < n; ++i) { draw((i*A+(n-i)*B)/n--(i*A+(n-i)*C)/n); draw((i*B+(n-i)*A)/n--(i*B+(n-i)*C)/n); draw((i*C+(n-i)*A)/n--(i*C+(n-i)*B)/n); } [/asy] For each $n \geq 2,$ find the number of existing parallelograms.

Let $\triangle ABC$ be a triangle and let $X$ be a point in the interior of the triangle. The second intersection points of the lines $XA,XB$ and $XC$ with the circumcircle of $\triangle ABC$ are $P,Q$ and $R$. Let $U$ be a point on the ray $XP$ (these are the points on the line $XP$ such that $P$ and $U$ lie on the same side of $X$). The line through $U$ parallel to $AB$ intersects $BQ$ in $V$ . The line through $U$ parallel to $AC$ intersects $CR$ in $W$. Prove that $Q, R, V$ , and $W$ lie on a circle.

On a chalkboard, Benji draws a square with side length $6.$ He then splits each side into $3$ equal segments using $2$ points for a total of $12$ segments and $8$ points. After trying some shapes, Benji finds that by using a circle, he can connect all $8$ points together. What is the area of this circle? [center] [img] https://cdn.artofproblemsolving.com/attachments/d/f/ead2d0c85f474612d49d3a023a43215e11066b.png [/img] [/center]

Let $ABC$ be a triangle with circumcircle $(O)$. Let $M,N$ be the midpoint of arc $AB,AC$ which does not contain $C,B$ and let $M',N'$ be the point of tangency of incircle of $\triangle ABC$ with $AB,AC$. Suppose that $X,Y$ are foot of perpendicular of $A$ to $MM',NN'$. If $I$ is the incenter of $\triangle ABC$ then prove that quadrilateral $AXIY$ is cyclic if and only if $b+c=2a$.

Let B,C be fixed points and A be roving point. Let H, G be orthecentre and centroid of triagle ABC. Known midpoint of HG lies on BC, find locus of A

The points $D$ and $E$ on the circumcircle of an acute triangle $ABC$ are such that $AD=AE = BC$. Let $H$ be the common point of the altitudes of triangle $ABC$. It is known that $AH^{2}=BH^{2}+CH^{2}$. Prove that $H$ lies on the segment $DE$. [i]Proposed by D. Shiryaev[/i]

The diagonals of a convex quadrilateral $ABCD$ meet at point $L$. The orthocenter $H$ of the triangle $LAB$ and the circumcenters $O_1, O_2$, and $O_3$ of the triangles $LBC, LCD$, and $LDA$ were marked. Then the whole configuration except for points $H, O_1, O_2$, and $O_3$ was erased. Restore it using a compass and a ruler.

In triangle $\vartriangle ABC$ with orthocenter $H$, the internal angle bisector of $\angle BAC$ intersects $\overline{BC}$ at $Y$ . Given that $AH = 4$, $AY = 6$, and the distance from $Y$ to $\overline{AC}$ is $\sqrt{15}$, compute $BC$.

It is known about the triangle $ABC$ that $3 BC = CA + AB$. Let the $A$-symmedian of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at point $D$. Prove that $\frac{1}{BD}+ \frac{1}{CD}= \frac{6}{AD}$. (Ercole Suppa, Italy)

Let $ABCD$ be a tetrahedron with $\angle BAD = 60^{\cdot}$, $\angle BAC = 40^{\cdot}$, $\angle ABD = 80^{\cdot}$, $\angle ABC = 70^{\cdot}$. Prove that the lines $AB$ and $CD$ are perpendicular.

Consider a triangle $ABC$ with $AB = 13, BC = 14, CA = 15$. A line perpendicular to $BC$ divides the interior of $\vartriangle BC$ into two regions of equal area. Suppose that the aforesaid perpendicular cuts $BC$ at $D$, and cuts $\vartriangle ABC$ again at $E$. If $L$ is the length of the line segment $DE$, find $L^2$.

Let $ABCD$ be a convex quadrilateral in which $\angle BAC = 50^{\circ}, \angle CAD = 60^{\circ}$and $\angle BDC = 25^{\circ}$. If $E$ is the point of intersection of $AC$ and $BD$, find $\angle AEB$.

Let $ABC$ be a triangle with $CA>BC>AB$. Let $O$ and $H$ be the circumcentre and orthocentre of triangle $ABC$ respectively. Denote by $D$ and $E$ the midpoints of the arcs $AB$ and $AC$ of the circumcircle of triangle $ABC$ not containing the opposite vertices. Let $D'$ be the reflection of $D$ about $AB$ and $E'$ the reflection of $E$ about $AC$. Prove that $O,H,D',E'$ are concylic if and only if $A,D',E'$ are collinear.

$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$. [i]Proposed by Giorgi Arabidze, Georgia.[/i]

A circle $\omega$ with center $O$ and radius $r$ is constructed. A point $P$ is chosen on the circumference $\omega$ and a point A is taken inside it, such that is outside the line that passes through $P$ and $O$. Point $B$ is constructed, the reflection of $A$ wrt $O$. and $P'$ is another point on the circumference such that the chord $PP'$ is perpendicular to $PA$. Let $Q$ be the point on the line $PP'$ that minimizes the sum of distances from $A$ to $Q$ and from $Q$ to $B$. Show that the value of the sum of the lengths $AQ+QB$ does not depend on the choice of points $P$ or $A$

Cut the non-equilateral triangle into four similar triangles, among which not all are the same.

Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points in the interiors of the sides $ BC$, $ CA$, $ AB$ of this triangle. Prove that the area of at least one of the three triangles $ AQR$, $ BRP$, $ CPQ$ is less than or equal to one quarter of the area of triangle $ ABC$. [i]Alternative formulation:[/i] Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Prove that $ \min\left\{\left|AQR\right|,\left|BRP\right|,\left|CPQ\right|\right\}\leq\frac14\cdot\left|ABC\right|$, where the abbreviation $ \left|P_1P_2P_3\right|$ denotes the (non-directed) area of an arbitrary triangle $ P_1P_2P_3$.

Find all positive integers $ n$ having the following properties:in two-dimensional Cartesian coordinates, there exists a convex $ n$ lattice polygon whose lengths of all sides are odd numbers, and unequal to each other. (where lattice polygon is defined as polygon whose coordinates of all vertices are integers in Cartesian coordinates.)

Let $ABC$ be an equilateral triangle and $n\ge 2$ be an integer. Denote by $\mathcal{A}$ the set of $n-1$ straight lines which are parallel to $BC$ and divide the surface $[ABC]$ into $n$ polygons having the same area and denote by $\mathcal{P}$ the set of $n-1$ straight lines parallel to $BC$ which divide the surface $[ABC]$ into $n$ polygons having the same perimeter. Prove that the intersection $\mathcal{A} \cap \mathcal{P}$ is empty. [i]Laurentiu Panaitopol[/i]

Let $A,B$ be adjacent vertices of a regular $n$-gon ($n\ge5$) with center $O$. A triangle $XYZ$, which is congruent to and initially coincides with $OAB$, moves in the plane in such a way that $Y$ and $Z$ each trace out the whole boundary of the polygon, with $X$ remaining inside the polygon. Find the locus of $X$.

Let $F$ be a closed convex region inside a circle $C$ with center $O$ and radius $1$. Furthermore, assume that from each point of $C$ one can draw two rays tangent to $F$ which form an angle of $60^o$. Prove that $F$ is the disc centered at $O$ with radius $1/2$.

Let $ I,O $ denote the incenter, respectively, the circumcenter of a triangle $ ABC. $ The $ A\text{-excircle} $ touches the lines $ AB,AC,BC $ at $ K,L, $ respectively, $ M. $ The midpoint of $ KL $ lies on the circumcircle of $ ABC. $ Show that the points $ I,M,O $ are collinear. [i]Павел Кожевников[/i]