$ABC$ is an acute angle triangle such that $AB>AC$ and $\hat{BAC}=60^{\circ}$. Let's denote by $O$ the center of the circumscribed circle of the triangle and $H$ the intersection of altitudes of this triangle. Line $OH$ intersects $AB$ in point $P$ and $AC$ in point $Q$. Find the value of the ration $\frac{PO}{HQ}$.

In the equilateral $\bigtriangleup ABC$, $D$ is a point on side $BC$. $O_1$ and $I_1$ are the circumcenter and incenter of $\bigtriangleup ABD$ respectively, and $O_2$ and $I_2$ are the circumcenter and incenter of $\bigtriangleup ADC$ respectively. $O_1I_1$ intersects $O_2I_2$ at $P$. Find the locus of point $P$ as $D$ moves along $BC$.

See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url] Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.

[color=darkred]Quadrilateral $ ABCD$ is inscribed in the circle of diameter $ BD$. Point $ A_1$ is reflection of point $ A$ wrt $ BD$ and $ B_1$ is reflection of $ B$ wrt $ AC$. Denote $ \{P\}\equal{}CA_1 \cap BD$ and $ \{Q\}\equal{}DB_1\cap AC$. Prove that $ AC\perp PQ$.[/color]

Let $H$, $I$, $O$ be the orthocenter, incenter and circumcenter of a triangle. Show that $2 \cdot IO \geq IH$. When does the equality hold ?

Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_1$, $I_2$ and $I_3$ the incenters of $\triangle ABM$, $\triangle MNC$ and $\triangle NDA$, respectively. Prove that the orthocenter of $\triangle I_1I_2I_3$ lies on $g$. [i]Proposed by Nikolay Beluhov, Bulgaria[/i]

Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$. Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.

Let $ ABC$ be a non-equilateral triangle. Denote by $ I$ the incenter and by $ O$ the circumcenter of the triangle $ ABC$. Prove that $ \angle AIO\leq\frac{\pi}{2}$ holds if and only if $ 2\cdot BC\leq AB\plus{}AC$.

Let $A'\in(BC),$ $B'\in(CA),C'\in(AB)$ be the points of tangency of the excribed circles of triangle $\triangle ABC$ with the sides of $\triangle ABC.$ Let $R'$ be the circumradius of triangle $\triangle A'B'C'.$ Show that \[ R'=\frac{1}{2r}\sqrt{2R\left(2R-h_{a}\right)\left(2R-h_{b}\right)\left(2R-h_{c}\right)}\] where as usual, $R$ is the circumradius of $\triangle ABC,$ r is the inradius of $\triangle ABC,$ and $h_{a},h_{b},h_{c}$ are the lengths of altitudes of $\triangle ABC.$

In triangle $ABC$, if $L,M,N$ are midpoints of $AB,AC,BC$. And $H$ is orthogonal center of triangle $ABC$, then prove that \[LH^{2}+MH^{2}+NH^{2}\leq\frac14(AB^{2}+AC^{2}+BC^{2})\]

Let $ABC $be a triangle, and $I $ the incenter, $M $ midpoint of $ BC $, $ D $ the touch point of incircle and $ BC $. Prove that perpendiculars from $M, D, A $ to $AI, IM, BC $ respectively are concurrent

Let $I$ be the incenter of the triangle $ABC$, et let $A',B',C'$ be the symmetric of $I$ with respect to the lines $BC,CA,AB$ respectively. It is known that $B$ belongs to the circumcircle of $A'B'C'$. Find $\widehat {ABC}$. Pierre.

In acute angled triangle $ABC$, $BC=a$,$CA=b$,$AB=c$, and $a>b>c$. $I,O,H$ are the incentre, circumcentre and orthocentre of $\triangle{ABC}$ respectively. Point $D \in BC$, $E \in CA$ and $AE=BD$, $CD+CE=AB$. Let the intersectionf of $BE$ and $AD$ be $K$. Prove that $KH \parallel IO$ and $KH = 2IO$.

Let $I$ be the incentre of a triangle $ABC$ and $\Gamma_a$ be the excircle opposite $A$ touching $BC$ at $D$. If $ID$ meets $\Gamma_a$ again at $S$, prove that $DS$ bisects $\angle BSC$.

In the triangle $ABC$, the side $BC$ is equal to $a$. Point $F$ is midpoint of $AB$, $I$ is the point of intersection of the bisectors of triangle $ABC$. It turned out that $\angle AIF = \angle ACB$. Find the perimeter of the triangle $ABC$. (Grigory Filippovsky)

In a parallelogram $\Box ABCD$ $(AB < BC)$ The incircle of $\triangle ABC$ meets $\overline {BC}$ and $\overline {CA}$ at $P, Q$. The incircle of $\triangle ACD$ and $\overline {CD}$ meets at $R$. Let $S$ = $PQ$ $\cap$ $AD$ $U$ = $AR$ $\cap$ $CS$ $T$, a point on $\overline {BC}$ such that $\overline {AB} = \overline {BT}$ Prove that $AT, BU, PQ$ are concurrent

Let $ABC$ be an acute-angled triangle with $\angle BAC = 60^{\circ}$ and with incenter $I$ and circumcenter $O$. Let $H$ be the point diametrically opposite(antipode) to $O$ in the circumcircle of $\triangle BOC$. Prove that $IH=BI+IC$.

Let $ABC$ be a triangle with $AB<AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$ [img]https://cdn.artofproblemsolving.com/attachments/0/5/4639d82d183247b875128a842a013ed7415fba.jpg[/img] [hide=.][url=http://artofproblemsolving.com/community/c6h602657p10667541]source[/url], translated by Antreas Hatzipolakis in fb, corrected by me in order to be compatible with it's figure[/hide]

Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $I$ be the incentre of triangle $ABC$, and let $D,E,F$ be the points where the incircle touches the sides $BC,CA,AB,$ respectively. Let $BI,CI$ meet the line $EF$ at $Y,X$ respectively. Further assume that both $X$ and $Y$ are outside the triangle $ABC$. Prove that $\text{(i)}$ $B,C,Y,X$ are concyclic. $\text{(ii)}$ $I$ is also the incentre of triangle $DYX$.

Let $AA _1$ and $CC_1$ be altitudes of an acute triangle $ABC$. Let $I$ and $J$ be the incenters of the triangles $AA_1C$ and $AC_1C$ respectively. The $C_1J$ and $A_1 I$ lines cut into $T$. Prove that lines $AT$ and $TC$ are perpendicular.

The perpendicular bisectors of the sides AB and BC of a triangle ABC meet the lines BC and AB at the points X and Z, respectively. The angle bisectors of the angles XAC and ZCA intersect at a point B'. Similarly, define two points C' and A'. Prove that the points A', B', C' lie on one line through the incenter I of triangle ABC. [i]Extension:[/i] Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC. Darij

Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then, we must have: $\text{(A)}\ CD=BD=O'D \qquad \text{(B)}\ AO=CO=OD \qquad \text{(C)}\ CD=CO=BD \qquad\\ \text{(D)}\ CD=OD=BD \qquad \text{(E)}\ O'B=O'C=OD $ [asy] size(200); defaultpen(linewidth(0.8)+fontsize(12pt)); pair A=origin,B=(15,0),C=(5,9),O=incenter(A,B,C),Op=circumcenter(A,B,C); path incirc = incircle(A,B,C),circumcirc = circumcircle(A,B,C),line=A--3*O; pair D[]=intersectionpoints(circumcirc,line); draw(A--B--C--A--D[0]^^incirc^^circumcirc); dot(O^^Op,linewidth(4)); label("$A$",A,dir(185)); label("$B$",B,dir(355)); label("$C$",C,dir(95)); label("$D$",D[0],dir(O--D[0])); label("$O$",O,NW); label("$O'$",Op,E);[/asy]

Given triangle $ ABC$ with $ C$ a right angle, show that the diameter of the incenter is $ a\plus{}b\minus{}c$, where $ a\equal{}BC$, $ b\equal{}CA$, and $ c\equal{}AB$.

Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ C(O,R)$ . Denote $ D\in BC$ for which $ AD\perp BC$ and $ AD \equal{} h_a$ . Prove that $ DI^2 \equal{} (2R \minus{} h_a)(h_a \minus{} 2r)$ .

In a triangle $ABC, A_1,B_1,C_1$ are the midpoints of segments $BC,CA,AB, A_2,B_2,C_2$ are the midpoints of segments $B_1C_1,C_1A_1,A_1B_1$, and $A_3,B_3,C_3$ are the incenters of triangles $B_1AC_1,C_1BA_1,A_1CB_1$, respectively. Show that the lines $A_2A_3,B_2B_3$ and $C_2C_3$ are concurrent.