Found problems: 3585
Let $\Gamma_1$ be a circle with centre $A$ and $\Gamma_2$ be a circle with centre $B$, with $A$ lying on $\Gamma_2$. On $\Gamma_2$ there is a (variable) point $P$ not lying on $AB$. A line through $P$ is a tangent of $\Gamma_1$ at $S$, and it intersects $\Gamma_2$ again in $Q$, with $P$ and $Q$ lying on the same side of $AB$. A different line through $Q$ is tangent to $\Gamma_1$ at $T$. Moreover, let $M$ be the foot of the perpendicular to $AB$ through $P$. Let $N$ be the intersection of $AQ$ and $MT$.
Show that $N$ lies on a line independent of the position of $P$ on $\Gamma_2$.
The incircle of triangle $ABC$ for which $AB\neq AC$, is tangent to sides $BC,CA$ and $AB$ in points $D,E$ and $F$ respectively. Perpendicular from $D$ to $EF$ intersects side $AB$ at $X$, and the second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX\perp TF$.
[i]Proposed By Pedram Safaei[/i]
In an acute traingle $ABC$ with $AB< BC$ let $BH_b$ be its altitude, and let $O$ be the circumcenter. A line through $H_b$ parallel to $CO$ meets $BO$ at $X$. Prove that $X$ and the midpoints of $AB$ and $AC$ are collinear.
Consider $\triangle XPQ$ and $\triangle YPQ$ such that $X$ and $Y$ are on the opposite sides of $PQ$ and the circumradius of $\triangle XPQ$ and the circumradius of $\triangle YPQ$ are the same. $I$ and $J$ are the incenters of $\triangle XPQ$ and $\triangle YPQ$ respectively. Let $M$ be the midpoint of $PQ$. Suppose $I, M, J$ are collinear. Prove that $XPYQ$ is a parallelogram.
Let $ABCD$ be a convex cyclic quadrilateral with the diagonal intersection $S$. Let further be $P$ the circumcenter of the triangle $ABS$ and $Q$ the circumcenter of the triangle $BCS$. The parallel to $AD$ through $P$ and the parallel to $CD$ through $Q$ intersect at point $R$. Prove that $R$ is on $BD$.
(Karl Czakler)
A semicircle has diameter $AB$ with $AB = 100$. Points $C$ and $D$ lie on the semicircle such that $AC = 28$ and $BD = 60$. Find $CD$.
Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.
An acute triangle $ABC$ in which $AB<AC$ is given. The bisector of $\angle BAC$ crosses $BC$ at $D$. Point $M$ is the midpoint of $BC$. Prove that the line though centers of circles escribed on triangles $ABC$ and $ADM$ is parallel to $AD$.
Let $P$ be an arbitrary point on the diagonal $AC$ of cyclic quadrilateral $ABCD$, and $PK, PL, PM, PN, PO$ be the perpendiculars from $P$ to $AB, BC, CD, DA, BD$ respectively. Prove that the distance from $P$ to $KN$ is equal to the distance from $O$ to $ML$.
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.
[i]Proposed by Charles Leytem, Luxembourg[/i]
Let $K$ be the intersection of diagonals of cyclic quadrilateral $ABCD$, where $|AB|=|BC|$, $|BK|=b$, and $|DK|=d$. What is $|AB|$?
$
\textbf{(A)}\ \sqrt{d^2 + bd}
\qquad\textbf{(B)}\ \sqrt{b^2+bd}
\qquad\textbf{(C)}\ \sqrt{2bd}
\qquad\textbf{(D)}\ \sqrt{2(b^2+d^2-bd)}
\qquad\textbf{(E)}\ \sqrt{bd}
$
Let $ABC$ be a triangle with incenter $I$. Let $U$, $V$ and $W$ be the intersections of the angle bisectors of angles $A$, $B$, and $C$ with the incircle, so that $V$ lies between $B$ and $I$, and similarly with $U$ and $W$. Let $X$, $Y$, and $Z$ be the points of tangency of the incircle of triangle $ABC$ with $BC$, $AC$, and $AB$, respectively. Let triangle $UVW$ be the [i]David Yang triangle[/i] of $ABC$ and let $XYZ$ be the [i]Scott Wu triangle[/i] of $ABC$. Prove that the David Yang and Scott Wu triangles of a triangle are congruent if and only if $ABC$ is equilateral.
[i]Proposed by Owen Goff[/i]
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.
Prove $P$ is the incentre of triangle $CDE$.
Point $P$ lies in the interior of a triangle $ABC$. Lines $AP$, $BP$, and $CP$ meet the opposite sides of triangle $ABC$ at $A$', $B'$, and $C'$ respectively. Let $P_A$ the midpoint of the segment joining the incenters of triangles $BPC'$ and $CPB'$, and define points $P_B$ and $P_C$ analogously. Show that if
\[ AB'+BC'+CA'=AC'+BA'+CB' \]
then points $P,P_A,P_B,$ and $P_C$ are concyclic.
[i]Nikolai Beluhov[/i]
Let O be the circumcenter of triangle ABC.
Let H be the orthocenter of triangle ABC.
the perpendicular bisector of AB meet AC,BC at D,E.
the circumcircle of triangle DEH meet AC,BC,OH again at F,G,L.
CH meet FG at T,and ABCT is concyclic.
Prove that LHBC is concyclic.
graph: https://cdn.luogu.com.cn/upload/image_hosting/w6z6mvm4.png
A line parallel to the side $BC$ of a triangle $ABC$ meets the sides $AB$ and $AC$ at points $P$ and $Q$, respectively. A point $M$ is chosen inside the triangle $APQ$. The segments $MB$ and $MC$ meet the segment $PQ$ at points $E$ and $F$, respectively. Let $N$ be the second intersection point of the circumcircles of the triangles $PMF$ and $QME$. Prove that the points $A,M,N$ are collinear.
Let $ AA_1 $ and $ CC_1 $ be the altitudes of the acute-angled triangle $ ABC $. A line passing through the centers of the inscribed circles the triangles $ AA_1C $ and $ CC_1A $ intersect the sides of $ AB $ and $ BC $ triangle $ ABC $ at points $ X $ and $ Y $. Prove that $ BX = BY $.
Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$. The line through $D$ perpendicular to $BC$ intersects the lines $AO, AC,$ and $AB$ at $W, X,$ and $Y,$ respectively. The circumcircles of triangles $AXY$ and $ABC$ intersect again at $Z \ne A$.
Prove that if $W \ne D$ and $OW = OD,$ then $DZ$ is tangent to the circle $AXY.$
Let $\omega$ be the circumcircle of triangle $ABC$. A point $T$ on the line $BC$ is such that $AT$ touches $\omega$. The bisector of angle $BAC$ meets $BC$ and $\omega$ at points $L$ and $A_0$ respectively. The line $TA_0$ meets $\omega$ at point $P$. The point $K$ lies on the segment $BC$ in such a way that $BL = CK$. Prove that $\angle BAP = \angle CAK$.
Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$. The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$. Let $M$ be the midpoint of $B_{1}B_{2}$. Let $M_{1}$, $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$, and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$. Show that $\angle MM_{1}B = \angle MM_{2}B$.
[i]Ciprus[/i]
Let $ABC$ be an equilateral triangle with side length $8.$ Let $X$ be on side $AB$ so that $AX=5$ and $Y$ be on side $AC$ so that $AY=3.$ Let $Z$ be on side $BC$ so that $AZ,BY,CX$ are concurrent. Let $ZX,ZY$ intersect the circumcircle of $AXY$ again at $P,Q$ respectively. Let $XQ$ and $YP$ intersect at $K.$ Compute $KX\cdot KQ.$
The bisector of $\angle BAD$ of a parallelogram $ABCD$ meets $BC$ at $K$. The point $L$ lies on $AB$ such that $AL=CK$. The lines $AK$ and $CL$ meet at $M$. Let $(ALM)$ meet $AD$ after $D$ at $N$. Prove that $\angle CNL=90^{o}$
The quadrilateral $ABCD$ is inscribed in circle $\omega$. $F$ is the intersection point of $AC$ and $BD$. $BA$ and $CD$ meet at $E$. Let the projection of $F$ on $AB$ and $CD$ be $G$ and $H$, respectively. Let $M$ and $N$ be the midpoints of $BC$ and $EF$, respectively. If the circumcircle of $\triangle MNG$ only meets segment $BF$ at $P$, and the circumcircle of $\triangle MNH$ only meets segment $CF$ at $Q$, prove that $PQ$ is parallel to $BC$.
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 $\angle AEB$.
In a non-equilateral triangle $ABC$ point $I$ is the incenter and point $O$ is the circumcenter. A line $s$ through $I$ is perpendicular to $IO$. Line $\ell$ symmetric to like $BC$ with respect to $s$ meets the segments $AB$ and $AC$ at points $K$ and $L$, respectively ($K$ and $L$ are different from $A$). Prove that the circumcenter of triangle $AKL$ lies on the line $IO$.
[i]Dušan Djukić[/i]