Found problems: 847
A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$.
[i]Proposed by Tomasz Ciesla, Poland[/i]
Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$.
[i]Proposed by David Monk, United Kingdom[/i]
Let $ABC$ be a triangle and $X,Y,Z$ be the tangency points of its inscribed circle with the sides $BC, CA, AB$, respectively. Suppose that $C_1, C_2, C_3$ are circle with chords $YZ, ZX, XY$, respectively, such that $C_1$ and $C_2$ intersect on the line $CZ$ and that $C_1$ and $C_3$ intersect on the line $BY$. Suppose that $C_1$ intersects the chords $XY$ and $ZX$ at $J$ and $M$, respectively; that $C_2$ intersects the chords $YZ$ and $XY$ at $L$ and $I$, respectively; and that $C_3$ intersects the chords $YZ$ and $ZX$ at $K$ and $N$, respectively. Show that $I, J, K, L, M, N$ lie on the same circle.
Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. The lines tangent to $\Gamma$ through $B$ and $C$ meet at $P$. Let $M$ be a point on the arc $AC$ that does not contain $B$ such that $M \neq A$ and $M \neq C$, and $K$ be the point where the lines $BC$ and $AM$ meet. Let $R$ be the point symmetrical to $P$ with respect to the line $AM$ and $Q$ the point of intersection of lines $RA$ and $PM$. Let $J$ be the midpoint of $BC$ and $L$ be the intersection point of the line $PJ$ and the line through $A$ parallel to $PR$. Prove that $L, J, A, Q,$ and $K$ all lie on a circle.
Let $AA_1 , BB_1$, and $CC_1$ be the altitudes of the non-isosceles acute-angled triangle $ABC$. The circles circumscibred around the triangles $ABC$ and $A_1 B_1 C$ intersect again at the point $P , Z$ is the intersection point of the tangents to the circumscribed circle of the triangle $ABC$ conducted at points $A$ and $B$ . Prove that lines $AP , BC$ and $ZC_1$ are concurrent.
An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.
a) Prove that $N$ and $N'$ coincide;
b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$;
c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.
Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.
[i]Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea[/i]
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
[i]Proposed by Nazar Serdyuk, Ukraine[/i]
In a triangle $ABC$, $\angle A$ is $60^\circ$. On sides $AB$ and $AC$ we make two equilateral triangles (outside the triangle $ABC$) $ABK$ and $ACL$. $CK$ and $AB$ intersect at $S$ , $AC$ and $BL$ intersect at $R$ , $BL$ and $CK$ intersect at $T$. Prove the radical centre of circumcircle of triangles $BSK, CLR$ and $BTC$ is on the median of vertex $A$ in triangle $ABC$.
[i]Proposed by Ali Zamani[/i]
Symedians of an acute-angled non-isosceles triangle $ABC$ intersect at a point at point $L$, and $AA_1$, $BB_1$ and $CC_1$ are its altitudes. Prove that you can construct equilateral triangles $A_1B_1C'$, $B_1C_1A'$ and $C_1A_1B'$ not lying in the plane $ABC$, so that lines $AA' , BB'$ and $CC'$ and also perpendicular to the plane $ABC$ at point $L$ intersected at one point.
Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$.
[i]Alex Zhu.[/i]
Let $\Delta ABC$ be an acute angled scalene triangle with circumcircle $\omega$. Let $O$ and $H$ be the circumcenter and orthocenter of $\Delta ABC,$ respectively. Let $E,F$ and $Q$ be points on segments $AB,AC$ and $\omega$, respectively, such that
$$\angle BHE=\angle CHF=\angle AQH=90^\circ.$$
Prove that $OQ$ and $AH$ intersect on the circumcircle of $\Delta AEF$.
[i] Proposed by Antareep Nath [/i]
Given triangle $ABC$ variable points $X$ and $Y$ are chosen on segments $AB$ and $AC$, respectively. Point $Z$ on line $BC$ is chosen such that $ZX=ZY$. The circumcircle of $XYZ$ cuts the line $BC$ for the second time at $T$. Point $P$ is given on line $XY$ such that $\angle PTZ = 90^ \circ$. Point $Q$ is on the same side of line $XY$ with $A$ furthermore $\angle QXY = \angle ACP$ and $\angle QYX = \angle ABP$. Prove that the circumcircle of triangle $QXY$ passes through a fixed point (as $X$ and $Y$ vary).
Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$.
[i]Proposed by Ankan Bhattacharya[/i]
In an acute triangle $ABC$, its inscribed circle touches the sides $AB,BC$ at the points $C_1,A_1$ respectively. Let $M$ be the midpoint of the side $AC$, $N$ be the midpoint of the arc $ABC$ on the circumcircle of triangle $ABC$, and $P$ be the projection of $M$ on the segment $A_1C_1$.
Prove that the points $P,N$ and the incenter $I$ of the triangle $ABC$ lie on the same line.
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
[i]Author: Farzan Barekat, Canada[/i]
Consider a semicircle of center $O$ and diameter $AB$. A line intersects $AB$ at $M$ and the semicircle at $C$ and $D$ s.t. $MC>MD$ and $MB<MA$. The circumcircles od the $AOC$ and $BOD$ intersect again at $K$. Prove that $MK\perp KO$.
Let $ABC$ be an acute scalene triangle with circumcenter $O$, and let $T$ be on line $BC$ such that $\angle TAO = 90^{\circ}$. The circle with diameter $\overline{AT}$ intersects the circumcircle of $\triangle BOC$ at two points $A_1$ and $A_2$, where $OA_1 < OA_2$. Points $B_1$, $B_2$, $C_1$, $C_2$ are defined analogously.
[list=a][*] Prove that $\overline{AA_1}$, $\overline{BB_1}$, $\overline{CC_1}$ are concurrent.
[*] Prove that $\overline{AA_2}$, $\overline{BB_2}$, $\overline{CC_2}$ are concurrent on the Euler line of triangle $ABC$. [/list][i]Evan Chen[/i]
Let $ABC$ be a triangle and $M$ be the midpoint of $BC$, and let $AM$ meet the circumcircle of $ABC$ again at $R$. A line passing through $R$ and parallel to $BC$ meet the circumcircle of $ABC$ again at $S$. Let $U$ be the foot from $R$ to $BC$, and $T$ be the reflection of $U$ in $R$. $D$ lies in $BC$ such that $AD$ is an altitude. $N$ is the midpoint of $AD$. Finally let $AS$ and $MN$ meets at $K$. Prove that $AT$ bisector $MK$.
Let $ABC$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{EF}$. Suppose that the circumcircle of $\triangle AIB$ meets the incircle at two distinct points $C_1$ and $C_2$, while the circumcircle of $\triangle AIC$ meets the incircle at two distinct points $B_1$ and $B_2$. Prove that the radical axis of the circumcircles of $\triangle BB_1B_2$ and $\triangle CC_1C_2$ passes through the midpoint $M$ of $\overline{DK}$.
[i]Proposed by Danielle Wang[/i]
Lines tangent to circle $O$ in points $A$ and $B$, intersect in point $P$. Point $Z$ is the center of $O$. On the minor arc $AB$, point $C$ is chosen not on the midpoint of the arc. Lines $AC$ and $PB$ intersect at point $D$. Lines $BC$ and $AP$ intersect at point $E$. Prove that the circumcentres of triangles $ACE$, $BCD$, and $PCZ$ are collinear.
The tangents at $A$ and $B$ to the circumcircle of the acute triangle $ABC$ intersect the tangent at $C$ at the points $D$ and $E$, respectively. The line $AE$ intersects $BC$ at $P$ and the line $BD$ intersects $AC$ at $R$. Let $Q$ and $S$ be the midpoints of the segments $AP$ and $BR$ respectively. Prove that $\angle ABQ=\angle BAS$.
Let $ABC$ be a triangle with circumcircle $\Omega$. Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$). Let $I$ be the incenter of $ABC$. Let $\omega_b$ be the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$, and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$. Show that the line $IN_a$, and the lines through the intersections of $\omega_b$ and $\omega_c$, meet on $\Omega$.
Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.
[i]Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea[/i]
A right triangle $ABC$ with $\angle C=90^o$ is inscribed in a circle. Suppose that $K$ is the midpoint of the arc $BC$ that does not contain $A$. Let $N$ be the midpoint of the segment $AC$, and $M$ be the intersection point of the ray $KN$ and the circle.The tangents to the circle drawn at $A$ and $C$ meet at $E$. prove that $\angle EMK = 90^o$