Given an acute triangle $ABC$. $\Gamma _{B}$ is a circle that passes through $AB$, tangent to $AC$ at $A$ and centered at $O_{B}$. Define $\Gamma_C$ and $O_C$ the same way. Let the altitudes of $\triangle ABC$ from $B$ and $C$ meets the circumcircle of $\triangle ABC$ at $X$ and $Y$, respectively. Prove that $A$, the midpoint of $XY$ and the midpoint of $O_{B}O_{C}$ is collinear.

Given is the convex quadrilateral $ ABCD$. Assume that there exists a point $ P$ inside the quadrilateral for which the triangles $ ABP$ and $ CDP$ are both isosceles right triangles with the right angle at the common vertex $ P$. Prove that there exists a point $ Q$ for which the triangles $ BCQ$ and $ ADQ$ are also isosceles right triangles with the right angle at the common vertex $ Q$.

In $\triangle ABC$, $\angle A = 60$, $AB>AC$, point $O$ is the circumcenter and $H$ is the intersection point of two altitudes $BE$ and $CF$. Points $M$ and $N$ are on the line segments $BH$ and $HF$ respectively, and satisfy $BM=CN$. Determine the value of $\frac{MH+NH}{OH}$.

Let $A,\ B$ and $C$ be given points on a circumference $K$ such that the triangle $\triangle{ABC}$ is acute. Let $P$ be a point in the interior of $K$. $X,\ Y$ and $Z$ be the other intersection of $AP, BP$ and $CP$ with the circumference. Determine the position of $P$ such that $\triangle{XYZ}$ is equilateral.

The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $ABC$ meet at $X$. Let $M$ be the midpoint of $BC$. Prove that [i](a)[/i] $\angle BAM = \angle CAX$, and [i](b)[/i] $\frac{AM}{AX} = \cos\angle BAC.$

A rectangular piece of paper $ABCD$ has sides of lengths $AB = 1$, $BC = 2$. The rectangle is folded in half such that $AD$ coincides with $BC$ and $EF$ is the folding line. Then fold the paper along a line $BM$ such that the corner $A$ falls on line $EF$. How large, in degrees, is $\angle ABM$? [asy] size(180); pathpen = rgb(0,0,0.6)+linewidth(1); pointpen = black+linewidth(3); pointfontpen = fontsize(10); pen dd = rgb(0,0,0.6) + linewidth(0.7) + linetype("4 4"), dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1); pair A=(0,1), B=(0,0), C=(2,0), D=(2,1), E=A/2, F=(2,.5), M=(1/3^.5,1), N=reflect(B,M)*A; D(B--M--D("N",N,NE)--B--D("C",C,SE)--D("D",D,NE)--M); D(D("M",M,plain.N)--D("A",A,NW)--D("B",B,SW),dd); D(D("E",E,W)--D("F",F,plain.E),dd); [/asy]

Let $ABC$ be a triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $D$ be the reflection of $A$ in $B$ and let $E$ the reflection of $A$ in $C$. Let $M$ be the midpoint of segment $DE$. Show that the tangent to $\Gamma$ in $A$ is perpendicular to $HM$.

In an acute triangle $ABC$ with $\overline{AB} < \overline{AC}$, angle bisector of $A$ and perpendicular bisector of $\overline{BC}$ intersect at $D$. Let $P$ be an interior point of triangle $ABC$. Line $CP$ meets the circumcircle of triangle $ABP$ again at $K$. Prove that $B, D, K$ are collinear if and only if $AD$ and $BC$ meet on the circumcircle of triangle $APC$.

If $I$ is the incenter of a triangle $ABC$ and $R$ is the radius of its circumcircle then \[AI+BI+CI\leq 3R\]

Let $ABC$ be a scalene triangle and $l_0$ be a line that is tangent to the circumcircle of $ABC$ at point $A$. Let $l$ be a variable line which is parallel to line $l_0$. Let $l$ intersect segment $AB$ and $AC$ at the point $X$, $Y$ respectively. $BY$ and $CX$ intersects at point $T$ and the line $AT$ intersects the circumcirle of $ABC$ at $Z$. Prove that as $l$ varies, circumcircle of $XYZ$ passes through a fixed point.

There is a set of $N\geq 3$ points in the plane, such that no three of them are collinear. Three points $A$, $B$, $C$ in the set are said to form a [i]Baltic triangle[/i] if no other point in the set lies on the circumcircle of triangle $ABC$. Assume that there exists at least one Baltic triangle. Show that there exist at least $\displaystyle\frac{N}{3}$ Baltic triangles.

Circles $\omega_1$ and $\omega_2$ meet at points $A$ and $B$. Let points $K_1$ and $K_2 $ of $\omega_1$ and $\omega_2$ respectively be such that $K_1A$ touches $\omega_2$, and $K_2A$ touches $\omega_1$. The circumcircle of triangle $K_1BK_2$ meets lines $AK_1$ and $AK_2$ for the second time at points $L_1$ and $L_2$ respectively. Prove that $L_1$ and $L_2$ are equidistant from line $AB$.

For three points $ X,Y,Z$ let $ R_{XYZ}$ be the circumcircle radius of the triangle $ XYZ.$ If $ ABC$ is a triangle with incircle centre $ I$ then we have: \[ \frac{1}{R_{ABI}} \plus{} \frac{1}{R_{BCI}} \plus{} \frac{1}{R_{CAI}} \leq \frac{1}{\bar{AI}} \plus{} \frac{1}{\bar{BI}} \plus{} \frac{1}{\bar{CI}}.\]

If $ a$, $ b$ and $ c$ are positive reals prove inequality: \[ \left(1\plus{}\frac{4a}{b\plus{}c}\right)\left(1\plus{}\frac{4b}{a\plus{}c}\right)\left(1\plus{}\frac{4c}{a\plus{}b}\right) > 25.\]

Given a triangle $ ABC$. A circle passes through vertices $ B$ and $ C$ and intersects sides $ AB$ and $ AC$ at points $ D$ and $ E$, respectively. Segments $ CD$ and $ BE$ intersect at point $ O$. Denote the incenters of triangles $ ADE$ and $ ODE$ by $ M$ and $ N$, respectiely. Prove that the midpoint of the smaller arc $ DE$ lies on line $ MN$.

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $A_1,B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC,CBA$ and $ACB$ of $\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$.

Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/i]

Let $ABC$ be a triangle and let $P$ be a point in its interior. Denote by $D$, $E$, $F$ the feet of the perpendiculars from $P$ to the lines $BC$, $CA$, $AB$, respectively. Suppose that \[AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.\] Denote by $I_A$, $I_B$, $I_C$ the excenters of the triangle $ABC$. Prove that $P$ is the circumcenter of the triangle $I_AI_BI_C$. [i]Proposed by C.R. Pranesachar, India [/i]

If the inradius of a triangle is half of its circumradius, prove that the triangle is equilateral.

In triangle $ABC$, $\angle{BAC}=15^{\circ}$. The circumcenter $O$ of triangle $ABC$ lies in its interior. Find $\angle{OBC}$. [asy] size(3cm,0); dot((0,0)); draw(Circle((0,0),1)); draw(dir(70)--dir(220)); draw(dir(220)--dir(310)); draw(dir(310)--dir(70)); draw((0,0)--dir(220)); label("$A$",dir(70),NE); label("$B$",dir(220),SW); label("$C$",dir(310),SE); label("$O$",(0,0),NE); [/asy] $\text{(A) }30^{\circ}\qquad\text{(B) }75^{\circ}\qquad\text{(C) }45^{\circ}\qquad\text{(D) }60^{\circ}\qquad\text{(E) }15^{\circ}$

Given the equilateral triangle $ABC$. It is known that the radius of the inscribed circle is in this triangle is equal to $1$. The rectangle $ABDE$ is such that point $C$ belongs to its side $DE$. Find the radius of the circle circumscribed around the rectangle $ABDE$.

An isosceles triangle $ABC$ with $AC = BC$ is given. Let $M$ be the midpoint of the side $AB$ and let $P$ be a point inside the triangle such that $\angle PAB = \angle PBC$. Prove that $\angle APM + \angle BPC = 180 \textdegree $

Two circles are internally tangent at $N$. The chords $BA$ and $BC$ of the larger circle are tangent to the smaller circle at $K$ and $M$ respectively. $Q$ and $P$ are midpoint of arcs $AB$ and $BC$ respectively. Circumcircles of triangles $BQK$ and $BPM$ are intersect at $L$. Show that $BPLQ$ is a parallelogram.

In an isosceles trapezoid $ABCD$, the base $AB$ is twice as large as the base $CD$. Point $M$ is the midpoint of $AB$. It is known that the center of the circle inscribed in the triangle $MCB$ lies on the circle circumscribed around the triangle $MDC$. Find the angle $\angle MBC$. [img]https://cdn.artofproblemsolving.com/attachments/8/a/7af6a1d32c4e2affa49cb3eed9c10ba1e7ab71.png[/img]