Given that points $ D,E$ lie on the sidelines $ AB,BC$ of triangle $ ABC$, respectively, point $ P$ is in interior of triangle $ ABC$ such that $ PE \equal{} PC$ and $ \bigtriangleup DEP\sim \bigtriangleup PCA.$ Prove that $ BP$ is tangent of the circumcircle of triangle $ PAD.$

Let points $A_1$, $A_2$ and $A_3$ lie on the circle $\Gamma$ in a counter-clockwise order, and let $P$ be a point in the same plane. For $i \in \{1,2,3\}$, let $\tau_i$ denote the counter-clockwise rotation of the plane centred at $A_i$, where the angle of rotation is equial to the angle at vertex $A_i$ in $\triangle A_1A_2A_3$. Further, define $P_i$ to be the point $\tau_{i+2}(\tau_{i}(\tau_{i+1}(P)))$, where the indices are taken modulo $3$ (i.e., $\tau_4 = \tau_1$ and $\tau_5 = \tau_2$). Prove that the radius of the circumcircle of $\triangle P_1P_2P_3$ is at most the radius of $\Gamma$. [i]Proposed by Anant Mudgal[/i]

The sphere $ \omega $ passes through the vertex $S$ of the pyramid $SABC$ and intersects with the edges $SA,SB,SC$ at $A_1,B_1,C_1$ other than $S$. The sphere $ \Omega $ is the circumsphere of the pyramid $SABC$ and intersects with $ \omega $ circumferential, lies on a plane which parallel to the plane $(ABC)$. Points $A_2,B_2,C_2$ are symmetry points of the points $A_1,B_1,C_1$ respect to midpoints of the edges $SA,SB,SC$ respectively. Prove that the points $A$, $B$, $C$, $A_2$, $B_2$, and $C_2$ lie on a sphere.

A trapezoid $ABCD$ with bases $AB$ and $CD$ is such that the circumcircle of the triangle $BCD$ intersects the line $AD$ in a point $E$, distinct from $A$ and $D$. Prove that the circumcircle oF the triangle $ABE$ is tangent to the line $BC$.

Let $A',\,B',\,C'$ be points, in which excircles touch corresponding sides of triangle $ABC$. Circumcircles of triangles $A'B'C,\,AB'C',\,A'BC'$ intersect a circumcircle of $ABC$ in points $C_1\ne C,\,A_1\ne A,\,B_1\ne B$ respectively. Prove that a triangle $A_1B_1C_1$ is similar to a triangle, formed by points, in which incircle of $ABC$ touches its sides.

Given three points $P$, $A$, $B$ and a circle such that the lines $PA$ and $PB$ are tangent to the circle at the points $A$ and $B$, respectively. A line through the point $P$ intersects that circle at two points $C$ and $D$. Through the point $B$, draw a line parallel to $PA$; let this line intersect the lines $AC$ and $AD$ at the points $E$ and $F$, respectively. Prove that $BE = BF$.

In a triangle with sides $a,b,c,t_a,t_b,t_c$ are the corresponding medians and $D$ the diameter of the circumcircle. Prove that \[\frac{a^2+b^2}{t_c}+\frac{b^2+c^2}{t_a}+\frac{c^2+a^2}{t_b}\le 6D\]

Three circles $\mathcal{K}_1$, $\mathcal{K}_2$, $\mathcal{K}_3$ of radii $R_1,R_2,R_3$ respectively, pass through the point $O$ and intersect two by two in $A,B,C$. The point $O$ lies inside the triangle $ABC$. Let $A_1,B_1,C_1$ be the intersection points of the lines $AO,BO,CO$ with the sides $BC,CA,AB$ of the triangle $ABC$. Let $ \alpha = \frac {OA_1}{AA_1} $, $ \beta= \frac {OB_1}{BB_1} $ and $ \gamma = \frac {OC_1}{CC_1} $ and let $R$ be the circumradius of the triangle $ABC$. Prove that \[ \alpha R_1 + \beta R_2 + \gamma R_3 \geq R. \]

Pentagon $ABCDE$ with $CD\parallel BE$ is inscribed in circle $\omega$. Tangent to $\omega$ through $B$ intersects line $AC$ at $F$ in a way that $A$ lies between $C$ and $F$. Lines $BD$ and $AE$ intersect at $G$. Prove that $FG$ is tangent to the circumcircle of $\triangle ADG$.

In an acute triangle $ABC$ the angle bisector $AL$, $L \in BC$, intersects its circumcircle at $N$. Let $K$ and $M$ be the projections of $L$ onto sides $AB$ and $AC$. Prove that triangle $ABC$ and quadrilateral $A K N M$ have equal areas.

Let $ABC$ be a non-isosceles triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Denote by $M$ the midpoint of $\overline{BC}$. Let $Q$ be a point on the incircle such that $\angle AQD = 90^{\circ}$. Let $P$ be the point inside the triangle on line $AI$ for which $MD = MP$. Prove that either $\angle PQE = 90^{\circ}$ or $\angle PQF = 90^{\circ}$. [i]Proposed by Evan Chen[/i]

Construct a triangle $ABC$ given its side $a = BC$, its circumradius $R \ (2R \geq a)$, and the difference $\frac{1}{k} = \frac{1}{c}-\frac{1}{b}$, where $c = AB$ and $ b = AC.$

The incircle of a triangle $ABC$ touches the sides $BC,CA,AB$ at $D,E,F$, respectively. Let $G$ be a point on the incircle such that $FG$ is a diameter. The lines $EG$ and $FD$ intersect at $H$. Prove that $CH\parallel AB$.

Let $ABCD$ be a convex circumscribed quadrilateral such that $\angle ABC+\angle ADC<180^{\circ}$ and $\angle ABD+\angle ACB=\angle ACD+\angle ADB$. Prove that one of the diagonals of quadrilateral $ABCD$ passes through the other diagonals midpoint.

Let ABC be a scalene triangle. Consider points D, E, F on segments AB, BC, CA, respectively, such that $\overline{AF}$=$\overline{DF}$ and $\overline{BE}$=$\overline{DE}$. Show that the circumcenter of ABC lies on the circumcircle of CEF.

Given an acute angled triangle $ ABC$ , $ O$ is the circumcenter and $ H$ is the orthocenter.Let $ A_1$,$ B_1$,$ C_1$ be the midpoints of the sides $ BC$,$ AC$ and $ AB$ respectively. Rays $ [HA_1$,$ [HB_1$,$ [HC_1$ cut the circumcircle of $ ABC$ at $ A_0$,$ B_0$ and $ C_0$ respectively.Prove that $ O$,$ H$ and $ H_0$ are collinear if $ H_0$ is the orthocenter of $ A_0B_0C_0$

Let $ ABC $ be an inscribed triangle in circle $ (O) $. Let $ D $ be the intersection of the two tangent lines of $ (O) $ at $ B $ and $ C $. The circle passing through $ A $ and tangent to $ BC $ at $ B $ intersects the median passing $ A $ of the triangle $ ABC $ at $ G $. Lines $ BG, CG $ intersect $ CD, BD $ at $ E, F $ respectively. a) The line passing through the midpoint of $ BE $ and $ CF $ cuts $ BF, CE $ at $ M, N $ respectively. Prove that the points $ A, D, M, N $ belong to the same circle. b) Let $ AD, AG $ intersect the circumcircle of the triangles $ DBC, GBC $ at $ H, K $ respectively. The perpendicular bisectors of $ HK, HE$, and $HF $ cut $ BC, CA$, and $AB $ at $ R, P$, and $Q $ respectively. Prove that the points $ R, P$, and $Q $ are collinear.

Let $AA _1$ and $BB_1$ be the altitudes of an isosceles acute-angled triangle $ABC, M$ the midpoint of $AB$. The circles circumscribed around the triangles $AMA_1$ and $BMB_1$ intersect the lines $AC$ and $BC$ at points $K$ and $L$, respectively. Prove that $K, M$, and $L$ lie on the same line.

Triangle I is equilateral with side $A$, perimeter $P$, area $K$, and circumradius $R$ (radius of the circumscribed circle). Triangle II is equilateral with side $a$, perimeter $p$, area $k$, and circumradius $r$. If $A$ is different from $a$, then: $ \textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad\textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad$ $\textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad\textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad$ $\textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes} $

Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point.

Inscribe in the triangle $ABC$ a triangle with minimum perimeter.

Acute triangle $\triangle ABC$ with $AB<AC$, circumcircle $\Gamma$ and circumcenter $O$ is given. Midpoint of side $AB$ is $D$. Point $E$ is chosen on side $AC$ so that $BE=CE$. Circumcircle of triangle $BDE$ intersects $\Gamma$ at point $F$ (different from point $B$). Point $K$ is chosen on line $AO$ satisfying $BK \perp AO$ (points $A$ and $K$ lie in different half-planes with respect to line $BE$). Prove that the intersection of lines $DF$ and $CK$ lies on $\Gamma$.

On side $AB$ of $\Delta ABC$ is chosen point $M$. A circle is tangent internally to the circumcircle of $\Delta ABC$ and segments $MB$ and $MC$ in points $P$ and $Q$ respectively. Prove that the center of the inscribed circle of $\Delta ABC$ lies on line $PQ$.

Let $ABC$ be a fixed triangle. Three similar (by point order) isosceles trapezoids are built on its sides: $ABXY, BCZW, CAUV$, such that the sides of the triangle are bases of the respective trapezoids. The circumcircles of triangles $XZU, YWV$ meet at two points $P, Q$. Prove that the line $PQ$ passes through a fixed point independent of the choice of trapezoids.

Let $ABC$ be a triangle with $AB \neq AC$ and with incenter $I$. Let $M$ be the midpoint of $BC$, and let $L$ be the midpoint of the circular arc $BAC$. Lines through $M$ parallel to $BI,CI$ meet $AB,AC$ at $E$ and $F$, respectively, and meet $LB$ and $LC$ at $P$ and $Q$, respectively. Show that $I$ lies on the radical axis of the circumcircles of triangles $EMF$ and $PMQ$. Proposed by [i]Andrew Wu[/i]