Found problems: 87
2019 IFYM, Sozopol, 3
$\Delta ABC$ is isosceles with a circumscribed circle $\omega (O)$. Let $H$ be the foot of the altitude from $C$ to $AB$ and let $M$ be the middle point of $AB$. We define a point $X$ as the second intersection point of the circle with diameter $CM$ and $\omega$ and let $XH$ intersect $\omega$ for a second time in $Y$. If $CO\cap AB=D$, then prove that the circumscribed circle of $\Delta YHD$ is tangent to $\omega$.
2010 Indonesia TST, 2
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
\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]
1999 Yugoslav Team Selection Test, Problem 2
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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2019 IFYM, Sozopol, 8
We are given a $\Delta ABC$. Point $D$ on the circumscribed circle k is such that $CD$ is a symmedian in $\Delta ABC$. Let $X$ and $Y$ be on the rays $\overrightarrow{CB}$ and $\overrightarrow{CA}$, so that $CX=2CA$ and $CY=2CB$. Prove that the circle, tangent externally to $k$ and to the lines $CA$ and $CB$, is tangent to the circumscribed circle of $\Delta XDY$.
1967 Spain Mathematical Olympiad, 6
An equilateral triangle $ABC$ with center $O$ and radius $OA = R$ is given, and consider the seven regions that the lines of the sides determine on the plane. It is asked to draw and describe the region of the plane transformed from the two shaded regions in the attached figure, by the inversion of center $O$ and power $R^2$.
[img]https://cdn.artofproblemsolving.com/attachments/e/c/bf1cb12c961467d216d54885f3387b328ce744.png[/img]
2002 Turkey Team Selection Test, 2
Two circles are internally tangent at a point $A$. Let $C$ be a point on the smaller circle other than $A$. The tangent line to the smaller circle at $C$ meets the bigger circle at $D$ and $E$; and the line $AC$ meets the bigger circle at $A$ and $P$. Show that the line $PE$ is tangent to the circle through $A$, $C$, and $E$.
2022 Bolivia Cono Sur TST, P2
On $\triangle ABC$ if there existed a point $D$ in $AC$ such that $\angle CBD=\angle ABD+60$ and $\angle BDC=30$ and $AB \cdot BC=BD^2$, then find the angles inside the triangle $\triangle ABC$
2015 USA TSTST, 2
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]
2025 Austrian MO National Competition, 2
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]
2017 Estonia Team Selection Test, 3
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 $BDE$.
2010 ELMO Shortlist, 4
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.
[i]Amol Aggarwal.[/i]
2008 IMO Shortlist, 3
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]
2024 Thailand TSTST, 9
Let triangle \( ABC \) be an acute-angled triangle. Square \( AEFB \) and \( ADGC \) lie outside triangle \( ABC \). \( BD \) intersects \( CE \) at point \( H \), and \( BG \) intersects \( CF \) at point \( I \). The circumcircle of triangle \( BFI \) intersects the circumcircle of triangle \( CGI \) again at point \( K \). Prove that line segment \( HK \) bisects \( BC \).
2014 ELMO Shortlist, 8
In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear circles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$.
[i]Proposed by Sammy Luo[/i]
2014 Taiwan TST Round 3, 4
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 $QB = QR$.
2023 HMIC, P3
Triangle $ABC$ has incircle $\omega$ and $A$-excircle $\omega_A.$ Circle $\gamma_B$ passes through $B$ and is externally tangent to $\omega$ and $\omega_A.$ Circle $\gamma_C$ passes through $C$ and is externally tangent to $\omega$ and $\omega_A.$ If $\gamma_B$ intersects line $BC$ again at $D,$ and $\gamma_C$ intersects line $BC$ again at $E,$ prove that $BD=EC.$
2023 Sharygin Geometry Olympiad, 24
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.
2017 Germany Team Selection Test, 3
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 $BDE$.
2018 Taiwan TST Round 1, 6
Given six points $ A, B, C, D, E, F $ such that $ \triangle BCD \stackrel{+}{\sim} \triangle ECA \stackrel{+}{\sim} \triangle BFA $ and let $ I $ be the incenter of $ \triangle ABC. $ Prove that the circumcenter of $ \triangle AID, \triangle BIE, \triangle CIF $ are collinear.
[i]Proposed by Telv Cohl[/i]
1967 Spain Mathematical Olympiad, 2
Determine the poles of the inversions that transform four collienar points $A,B, C, D$, aligned in this order, at four points $A' $, $B' $, $C'$ , $D'$ that are vertices of a rectangle, and such that $A'$ and $C'$ are opposite vertices.
1999 AIME Problems, 6
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 $k.$
2009 Belarus Team Selection Test, 2
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]
2021 Romanian Master of Mathematics Shortlist, G2
Let $ABC$ be a triangle with incenter $I$. The line through $I$, perpendicular to $AI$, intersects the circumcircle of $ABC$ at points $P$ and $Q$. It turns out there exists a point $T$ on the side $BC$ such that $AB + BT = AC + CT$ and $AT^2 = AB \cdot AC$. Determine all possible values of the ratio $IP/IQ$.
2010 Indonesia TST, 2
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
\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]
2018 USAMO, 5
In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.
[i]Proposed by Kada Williams[/i]