Let $\triangle ABC$ be an acute triangle with $AB>AC$, let $I$ be the center of the incircle. Let $M,N$ be the midpoint of $AC$ and $AB$ respectively. $D,E$ are on $AC$ and $AB$ respectively such that $BD\parallel IM$ and $CE\parallel IN$. A line through $I$ parallel to $DE$ intersects $BC$ in $P$. Let $Q$ be the projection of $P$ on line $AI$. Prove that $Q$ is on the circumcircle of $\triangle ABC$.

Let $ABC$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $BC$, $CA$, $AB$ are drawn on the outside of the triangle $ABC$. The circle tangent to all of these three semicircles has radius $t$. Prove that \[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \] [i]Alternative formulation.[/i] In a triangle $ABC$, construct circles with diameters $BC$, $CA$, and $AB$, respectively. Construct a circle $w$ externally tangent to these three circles. Let the radius of this circle $w$ be $t$. Prove: $\frac{s}{2}<t\le\frac{s}{2}+\frac12\left(2-\sqrt3\right)r$, where $r$ is the inradius and $s$ is the semiperimeter of triangle $ABC$. [i]Proposed by Dirk Laurie, South Africa[/i]

Let $O$ be the circumcenter and $G$ be the centroid of a triangle $ABC$. If $R$ and $r$ are the circumcenter and incenter of the triangle, respectively, prove that \[ OG \leq \sqrt{ R ( R - 2r ) } . \] [i]Greece[/i]

Let $ABC$ be a triangle with circumcircle $\omega$ . Point $D$ lies on the arc $BC$ of $\omega$ and is different than $B,C$ and the midpoint of arc $BC$. Tangent of $\Gamma$ at $D$ intersects lines $BC$, $CA$, $AB$ at $A',B',C'$, respectively. Lines $BB'$ and $CC'$ intersect at $E$. Line $AA'$ intersects the circle $\omega$ again at $F$. Prove that points $D,E,F$ are collinear. (Saudi Arabia)

Arbitrary triangle $ABC$ is given (with $AB<BC$). Let $M$ - midpoint of $AC$, $N$- midpoint of arc $AC$ of circumcircle $ABC$, which is contains point $B$. Let $I$ - in-center of $ABC$. Proved, that $ \angle IMA = \angle INB$

Let $P$ be a point in the interior of the acute-angled triangle $ABC$. Prove that if the reflections of $P$ with respect to the sides of the triangle lie on the circumcircle of the triangle, then $P$ is the orthocenter of $ABC$.

In an acute-angled triangle $ ABC,$ let $ AD,BE$ be altitudes and $ AP,BQ$ internal bisectors. Denote by $ I$ and $ O$ the incenter and the circumcentre of the triangle, respectively. Prove that the points $ D, E,$ and $ I$ are collinear if and only if the points $ P, Q,$ and $ O$ are collinear.

(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A_c$, $ B_c$; $ C_1$ is the common point of $ AA_c$ and $ BB_c$. Points $ A_1$, $ B_1$ are defined similarly. Prove that circle $ A_1B_1C_1$ passes through the circumcenter of triangle $ ABC$.

A point $D$ lies inside triangle $ABC$. Let $A_1, B_1, C_1$ be the second intersection points of the lines $AD$, $BD$, and $CD$ with the circumcircles of $BDC$, $CDA$, and $ADB$, respectively. Prove that $$\frac{AD}{AA_1} + \frac{BD}{BA_1} + \frac{CD}{CC_1} = 1.$$

Let $ABC$ be an acute triangle with circumcenter $O$, and select $E$ on $\overline{AC}$ and $F$ on $\overline{AB}$ so that $\overline{BE} \perp \overline{AC}$, $\overline{CF} \perp \overline{AB}$. Suppose $\angle EOF - \angle A = 90^{\circ}$ and $\angle AOB - \angle B = 30^{\circ}$. If the maximum possible measure of $\angle C$ is $\tfrac mn \cdot 180^{\circ}$ for some positive integers $m$ and $n$ with $m < n$ and $\gcd(m,n)=1$, compute $m+n$. [i]Proposed by Evan Chen[/i]

Let $AD$ be the base of trapezoid $ABCD$. It is known that the circumcenter of triangle $ABC$ lies on $BD$. Prove that the circumcenter of triangle $ABD$ lies on $AC$. [i]Proposed by Ye.Bakayev[/i]

Consider an acute triangle $ABC$ of area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$, respectively $MND$. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$. [i]Proposed by Marcin E. Kuczma, Poland[/i]

Let $ I$ be incenter of triangle $ ABC$, $ M$ be midpoint of side $ BC$, and $ T$ be the intersection point of $ IM$ with incircle, in such a way that $ I$ is between $ M$ and $ T$. Prove that $ \angle BIM\minus{}\angle CIM\equal{}\frac{3}2(\angle B\minus{}\angle C)$, if and only if $ AT\perp BC$.

Let $ABC$ be a triangle and let the points $D\in BC, E\in AC, F\in AB$, such that \[ \frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB} \] The half-lines $AD, BE,$ and $CF$ intersect the circumcircle of $ABC$ at points $M,N$ and $P$. Prove that the triangles $ABC$ and $MNP$ share the same centroid if and only if the areas of the triangles $BMC, CNA$ and $APB$ are equal.

Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic. [i]Proposed by Eric Chen[/i]

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles? $\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$

Each of three lines cuts chords of equal lengths in two given circles. The points of intersection of these lines form a triangle. Prove that its circumcircle passes through the midpoint of the segment joining the centres of the circles.

Points $M$, $N$, $K$ lie on the sides $BC$, $CA$, $AB$ of a triangle $ABC$, respectively, and are different from its vertices. The triangle $MNK$ is called[i] beautiful[/i] if $\angle BAC=\angle KMN$ and $\angle ABC=\angle KNM$. If in the triangle $ABC$ there are two beautiful triangles with a common vertex, prove that the triangle $ABC$ is right-angled. [i]Proposed by Nairi M. Sedrakyan, Armenia[/i]

Let $\triangle ABC$ be an acute triangle, and let $D$ be the projection of $A$ on $BC$. Let $M,N$ be the midpoints of $AB$ and $AC$ respectively. Let $\Gamma_1$ and $\Gamma_2$ be the circumcircles of $\triangle BDM$ and $\triangle CDN$ respectively, and let $K$ be the other intersection point of $\Gamma_1$ and $\Gamma_2$. Let $P$ be an arbitrary point on $BC$ and $E,F$ are on $AC$ and $AB$ respectively such that $PEAF$ is a parallelogram. Prove that if $MN$ is a common tangent line of $\Gamma_1$ and $\Gamma_2$, then $K,E,A,F$ are concyclic.

In a convex quadrilateral $ABCD$ we have $AB\cdot CD=BC\cdot DA$ and $2\angle A+\angle C=180^\circ$. Point $P$ lies on the circumcircle of triangle $ABD$ and is the midpoint of the arc $BD$ not containing $A$. It is known that the point $P$ lies inside the quadrilateral $ABCD$. Prove that $\angle BCA=\angle DCP$ [i]Proposed by S. Berlov[/i]

There are $110$ points in a unit square. Show that some four of these points reside in a circle whose radius is $1/8.$

Let $O$ be the circumcenter of triangle $ABC$. Choose a point $X$ on the circumcircle $\odot (ABC)$ such that $OX\parallel BC$. Assume that $\odot(AXO)$ intersects $AB, AC$ at $E, F$, respectively, and $OE, OF$ intersect $BC$ at $P, Q$, respectively. Furthermore, assume that $\odot(XP Q)$ and $\odot (ABC)$ intersect at $R$. Prove that $OR$ and $\odot (XP Q)$ are tangent to each other. (ltf0501)

We are given the square $ABCD$. On sides $AB$ and $CD$ we are given points $ K$ and $L$ respectively, and on segment $KL$ we are given point $M$ . Prove that the second intersection point (i.e. the one other than $M$) of the intersection points of circles circumscribed around triangles $AKM$ and $MLC$ lies on the diagonal $AC$. (V . N . Dubrovskiy)

$P$ is a point in the interior of acute triangle $ABC$. $D,E,F$ are the reflections of $P$ across $BC,CA,AB$ respectively. Rays $AP,BP,CP$ meet the circumcircle of $\triangle ABC$ at $L,M,N$ respectively. Prove that the circumcircles of $\triangle PDL,\triangle PEM,\triangle PFN$ meet at a point $T$ different from $P$.