Found problems: 87
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 $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that
\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that
\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]
Let $ABCD$ be a convex quadrilateral with side lengths satisfying the equality:
$$ AB \cdot CD = AD \cdot BC = AC \cdot BD.$$
Determine the sum of the acute angles of quadrilateral $ABCD$.
[i]Proposed by Zaza Meliqidze, Georgia[/i]
A transformation of the first quadrant of the coordinate plane maps each point $(x,y)$ to the point $(\sqrt{x},\sqrt{y}).$ The vertices of quadrilateral $ABCD$ are $A=(900,300), B=(1800,600), C=(600,1800),$ and $D=(300,900).$ Let $k$ be the area of the region enclosed by the image of quadrilateral $ABCD.$ Find the greatest integer that does not exceed $k.$
A tetrahedron $ABCD$ is give. A line $\ell$ meets the planes $ABC,BCD,CDA,DAB$ at points $D_0,A_0,B_0,C_0$ respectively. Let $P$ be an arbitrary point not lying on $\ell$ and the planes of the faces, and $A_1,B_1,C_1,D_1$ be the second common points of lines $PA_0,PB_0,PC_0,PD_0$ with the spheres $PBCD,PCDA,PDAB,PABC$ respectively. Prove $P,A_1,B_1,C_1,D_1$ lie on a circle.
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.
[i](Karl Czakler)[/i]
A line $l$ parallel to the side $BC$ of triangle $ABC$ touches its incircle and meets its circumcircle at points $D$ and $E$. Let $I$ be the incenter of $ABC$. Prove that $AI^2 = AD \cdot AE$.
Let $\Delta ABC$ be an acute angled triangle. $D$ is a point on side $BC$ such that $AD$ bisects angle $\angle BAC$. A line $l$ is tangent to the circumcircles of triangles $ADB$ and $ADC$ at point $K$ and $L$, respectively. Let $M$, $N$ and $P$ be its midpoints of $BD$, $DC$ and $KL$, respectively. Prove that $l$ is tangent to the circumcircle of $\Delta MNP$.
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 $ABC$ be a scalene triangle. The incircle of $ABC$ touches $\overline{BC}$ at $D$. Let $P$ be a point on $\overline{BC}$ satisfying $\angle BAP = \angle CAP$, and $M$ be the midpoint of $\overline{BC}$. Define $Q$ to be on $\overline{AM}$ such that $\overline{PQ} \perp \overline{AM}$. Prove that the circumcircle of $\triangle AQD$ is tangent to $\overline{BC}$.
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]
A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that:
$\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$
Find the locus of the centers of the inversions that transform two points $A, B$ of a given circle $\gamma$ , at diametrically opposite points of the inverse circles of $\gamma$ .
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)
[i]Proposed by Ivan Borsenco[/i]
Let $ABC$ be a triangle and $M$ be the midpoint of side $BC$. The perpendicular bisector of $BC$ intersects the circumcircle of $\triangle ABC$ at points $K$ and $L$ ($K$ and $A$ lie on the opposite sides of $BC$). A circle passing through $L$ and $M$ intersects $AK$ at points $P$ and $Q$ ($P$ lies on the line segment $AQ$). $LQ$ intersects the circumcircle of $\triangle KMQ$ again at $R$. Prove that $BPCR$ is a cyclic quadrilateral.
Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$.
Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.
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[i]Edited by Orl.[/i]
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There is a circle $k$ in a plane with center $M$ and radius $r$. The following illustration, through which every point $P \ne M$., is called a “reflection on the circle $k$” from $\varepsilon$ a point $P'$ from $\varepsilon$ is assigned:
(1) $P'$ lies on the ray emanating from$ M$ and passing through $P$.
(2) It is $MP \cdot MP' = r^2$.
a) Construct the mirror point $ P'$ for any given point $P \ne M$ inside $k$.
b) Let another circle $k_1$ be given arbitrarily, but such that $M$ lies outside $k_1$.Construct $k'_1$ , i.e. the set of all mirror points $P'$ of the points $P$ of $k_1$.
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Inside the circle $\omega$ through points $A, B$ point $C$ is chosen. An arbitrary point $X$ is selected on the segment $BC$. The ray $AX$ cuts the circle in $Y$. Prove that all circles $CXY$ pass through a two fixed points that is they intersect and are coaxial, independent of the position of $X$.
The interior bisector of $\angle A$ from $\triangle ABC$ intersects the side $BC$ and the circumcircle of $\Delta ABC$ at $D,M$, respectively. Let $\omega$ be a circle with center $M$ and radius $MB$. A line passing through $D$, intersects $\omega$ at $X,Y$. Prove that $AD$ bisects $\angle XAY$.