geometry problems from Final Round of Philippine Mathematical Olympiads (PMO)
with aops links in the names
2008 - 2022
missing 2014, 2015
with aops links in the names
2008 - 2022
missing 2014, 2015
2008 Philippine MO P3
Let $P$ be a point outside a circle $\Gamma$, and let the two tangent lines through $P$ touch $\Gamma$ at $A$ and $B$. Let $C$ be on the minor arc $AB$, and let ray $PC$ intersect $\Gamma$ again at $D$. Let $\ell$ be the line through $B$ and parallel to $PA$. $\ell$ intersects $AC$ and $AD$ at $E$ and $F$, respectively. Prove that $B$ is the midpoint of $EF$.
Let $P$ be a point outside a circle $\Gamma$, and let the two tangent lines through $P$ touch $\Gamma$ at $A$ and $B$. Let $C$ be on the minor arc $AB$, and let ray $PC$ intersect $\Gamma$ again at $D$. Let $\ell$ be the line through $B$ and parallel to $PA$. $\ell$ intersects $AC$ and $AD$ at $E$ and $F$, respectively. Prove that $B$ is the midpoint of $EF$.
2009 Philippine MO P5
Segments $AC$ and $BD$ intersect at point $P$ such that $PA = PD$ and $PB = PC$. Let $E$ be the foot of the perpendicular from $P$ to the line $CD$. Prove that the line $PE$ and the perpendicular bisectors of the segments $PA$ and $PB$ are concurrent.
2010 Philippine MO P2
On a cyclic quadrilateral $ABCD$, there is a point $P$ on side $AD$ such that the triangle $CDP$ and the quadrilateral $ABCP$ have equal perimeters and equal areas. Prove that two sides of $ABCD$ have equal lengths.
2011 Philippine MO P2
In triangle $ABC$, let $X$ and $Y$ be the midpoints of $AB$ and $AC$, respectively. On segment $BC$, there is a point $D$, different from its midpoint, such that $\angle{XDY}=\angle{BAC}$. Prove that $AD\perp BC$.
2013 Philippine MO P2
Let $P$ be a point in the interior of triangle $ABC$ . Extend $AP, BP,$ and $CP$ to meet $BC, AC$, and $AB$ at $D, E$, and $F$, respectively. If triangle $APF$, triangle $BPD$ and triangle $CPE$ have equal areas, prove that $P$ is the centroid of triangle $ABC$ .
(16th) 2014 missing
(17th) 2015 missing
2016 Philippine MO P5
Pentagon \(ABCDE\) is inscribed in a circle. Its diagonals \(AC\) and \(BD\) intersect at \(F\). The bisectors of \(\angle BAC\) and \(\angle CDB\) intersect at \(G\). Let \(AG\) intersect \(BD\) at \(H\), let \(DG\) intersect \(AC\) at \(I\), and let \(EG\) intersect \(AD\) at \(J\). If \(FHGI\) is cyclic and \[JA \cdot FC \cdot GH = JD \cdot FB \cdot GI,\]prove that \(G\), \(F\) and \(E\) are collinear.
2017 Philippine MO P4
Circles \(\mathcal{C}_1\) and \(\mathcal{C}_2\) with centers at \(C_1\) and \(C_2\) respectively, intersect at two points \(A\) and \(B\). Points \(P\) and \(Q\) are varying points on \(\mathcal{C}_1\) and \(\mathcal{C}_2\), respectively, such that \(P\), \(Q\) and \(B\) are collinear and \(B\) is always between \(P\) and \(Q\). Let lines \(PC_1\) and \(QC_2\) intersect at \(R\), let \(I\) be the incenter of \(\Delta PQR\), and let \(S\) be the circumcenter of \(\Delta PIQ\). Show that as \(P\) and \(Q\) vary, \(S\) traces the arc of a circle whose center is concyclic with \(A\), \(C_1\) and \(C_2\).
2018 Philippine MO P1
In triangle $ABC$ with $\angle ABC = 60^{\circ}$ and $5AB = 4BC$, points $D$ and $E$ are the feet of the altitudes from $B$ and $C$, respectively. $M$ is the midpoint of $BD$ and the circumcircle of triangle $BMC$ meets line $AC$ again at $N$. Lines $BN$ and $CM$ meet at $P$. Prove that $\angle EDP = 90^{\circ}$.
2019 Philippine MO P4
In acute triangle $ABC $with $\angle BAC > \angle BCA$, let $P$ be the point on side $BC$ such that $\angle PAB = \angle BCA$. The circumcircle of triangle $AP B$ meets side $AC$ again at $Q$. Point $D$ lies on segment $AP$ such that $\angle QDC = \angle CAP$.
Point $E$ lies on line $BD$ such that $CE = CD$. The circumcircle of triangle $CQE$ meets segment $CD$ again at $F$, and line $QF$ meets side $BC$ at $G$. Show that $B, D, F$, and $G$ are concyclic.
2020 Philippine MO P4
Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$ and $D$ the foot of the altitude from $A$. Suppose that $AD=BC$. Point $M$ is the midpoint of $DC$, and the bisector of $\angle ADC$ meets $AC$ at $N$. Point $P$ lies on $\Gamma$ such that lines $BP$ and $AC$ are parallel. Lines $DN$ and $AM$ meet at $F$, and line $PF$ meets $\Gamma$ again at $Q$. Line $AC$ meets the circumcircle of $\triangle PNQ$ again at $E$. Prove that $\angle DQE = 90^{\circ}$.
Segments $AC$ and $BD$ intersect at point $P$ such that $PA = PD$ and $PB = PC$. Let $E$ be the foot of the perpendicular from $P$ to the line $CD$. Prove that the line $PE$ and the perpendicular bisectors of the segments $PA$ and $PB$ are concurrent.
On a cyclic quadrilateral $ABCD$, there is a point $P$ on side $AD$ such that the triangle $CDP$ and the quadrilateral $ABCP$ have equal perimeters and equal areas. Prove that two sides of $ABCD$ have equal lengths.
In triangle $ABC$, let $X$ and $Y$ be the midpoints of $AB$ and $AC$, respectively. On segment $BC$, there is a point $D$, different from its midpoint, such that $\angle{XDY}=\angle{BAC}$. Prove that $AD\perp BC$.
Let $P$ be a point in the interior of triangle $ABC$ . Extend $AP, BP,$ and $CP$ to meet $BC, AC$, and $AB$ at $D, E$, and $F$, respectively. If triangle $APF$, triangle $BPD$ and triangle $CPE$ have equal areas, prove that $P$ is the centroid of triangle $ABC$ .
(17th) 2015 missing
2016 Philippine MO P5
Pentagon \(ABCDE\) is inscribed in a circle. Its diagonals \(AC\) and \(BD\) intersect at \(F\). The bisectors of \(\angle BAC\) and \(\angle CDB\) intersect at \(G\). Let \(AG\) intersect \(BD\) at \(H\), let \(DG\) intersect \(AC\) at \(I\), and let \(EG\) intersect \(AD\) at \(J\). If \(FHGI\) is cyclic and \[JA \cdot FC \cdot GH = JD \cdot FB \cdot GI,\]prove that \(G\), \(F\) and \(E\) are collinear.
2017 Philippine MO P4
Circles \(\mathcal{C}_1\) and \(\mathcal{C}_2\) with centers at \(C_1\) and \(C_2\) respectively, intersect at two points \(A\) and \(B\). Points \(P\) and \(Q\) are varying points on \(\mathcal{C}_1\) and \(\mathcal{C}_2\), respectively, such that \(P\), \(Q\) and \(B\) are collinear and \(B\) is always between \(P\) and \(Q\). Let lines \(PC_1\) and \(QC_2\) intersect at \(R\), let \(I\) be the incenter of \(\Delta PQR\), and let \(S\) be the circumcenter of \(\Delta PIQ\). Show that as \(P\) and \(Q\) vary, \(S\) traces the arc of a circle whose center is concyclic with \(A\), \(C_1\) and \(C_2\).
In triangle $ABC$ with $\angle ABC = 60^{\circ}$ and $5AB = 4BC$, points $D$ and $E$ are the feet of the altitudes from $B$ and $C$, respectively. $M$ is the midpoint of $BD$ and the circumcircle of triangle $BMC$ meets line $AC$ again at $N$. Lines $BN$ and $CM$ meet at $P$. Prove that $\angle EDP = 90^{\circ}$.
2019 Philippine MO P4
In acute triangle $ABC $with $\angle BAC > \angle BCA$, let $P$ be the point on side $BC$ such that $\angle PAB = \angle BCA$. The circumcircle of triangle $AP B$ meets side $AC$ again at $Q$. Point $D$ lies on segment $AP$ such that $\angle QDC = \angle CAP$.
Point $E$ lies on line $BD$ such that $CE = CD$. The circumcircle of triangle $CQE$ meets segment $CD$ again at $F$, and line $QF$ meets side $BC$ at $G$. Show that $B, D, F$, and $G$ are concyclic.
2020 Philippine MO P4
Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$ and $D$ the foot of the altitude from $A$. Suppose that $AD=BC$. Point $M$ is the midpoint of $DC$, and the bisector of $\angle ADC$ meets $AC$ at $N$. Point $P$ lies on $\Gamma$ such that lines $BP$ and $AC$ are parallel. Lines $DN$ and $AM$ meet at $F$, and line $PF$ meets $\Gamma$ again at $Q$. Line $AC$ meets the circumcircle of $\triangle PNQ$ again at $E$. Prove that $\angle DQE = 90^{\circ}$.
In convex quadrilateral $ABCD$, $\angle CAB = \angle BCD$. $P$ lies on line $BC$ such that
$AP = PC$, $Q$ lies on line $AP$ such that $AC$ and $DQ$ are parallel, $R$ is the point of
intersection of lines $AB$ and $CD$, and $S$ is the point of intersection of lines $AC$ and $QR$.
Line $AD$ meets the circumcircle of $AQS$ again at $T$. Prove that $AB$ and $QT$ are parallel.
2021 Philippine MO P8
In right triangle $ABC$, $\angle ACB = 90^{\circ}$ and $\tan A > \sqrt{2}$. $M$ is the midpoint of
$AB$, $P$ is the foot of the altitude from $C$, and $N$ is the midpoint of $CP$. Line $AB$ meets the
circumcircle of $CNB$ again at $Q$. $R$ lies on line $BC$ such that $QR$ and $CP$ are parallel, $S$
lies on ray $CA$ past $A$ such that $BR = RS$, and $V$ lies on segment $SP$ such that $AV = VP$.
Line $SP$ meets the circumcircle of $CPB$ again at $T$. $W$ lies on ray $VA$ past $A$ such that
$2AW = ST$, and $O$ is the circumcenter of $SPM$. Prove that lines $OM$ and $BW$ are
perpendicular.
Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.
In $\triangle ABC$, let $D$ be the point on side $BC$ such that $AB+BD=DC+CA.$ The line $AD$ intersects the circumcircle of $\triangle ABC$ again at point $X \neq A$. Prove that one of the common tangents of the circumcircles of $\triangle BDX$ and $\triangle CDX$ is parallel to $BC$.
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