drop down menu

Mexico OMMock 2017-21 8p

geometry problems from OMMock =Mexico National Olympiad Mock Exam,  with aops links in the names


collected inside aops here

2017 - 2021

Let $ABC$ be a triangle with circumcenter $O$. Point $D, E, F$ are chosen on sides $AB, BC$ and $AC$, respectively, such that $ADEF$ is a rhombus. The circumcircles of $BDE$ and $CFE$ intersect $AE$ at $P$ and $Q$ respectively. Show that $OP=OQ$.
by Ariel García
Let $ABCD$ be a trapezoid with bases $AD$ and $BC$, and let $M$ be the midpoint of $CD$. The circumcircle of triangle $BCM$ meets $AC$ and $BD$ again at $E$ and $F$, with $E$ and $F$ distinct, and line $EF$ meets the circumcircle of triangle $AEM$ again at $P$. Prove that $CP$ is parallel to $BD$.
by Ariel García
Let $ABC$ be a triangle with circumcirle $\Gamma$, and let $M$ and $N$ be the respective midpoints of the minor arcs $AB$ and $AC$ of $\Gamma$. Let $P$ and $Q$ be points such that $AB=BP$, $AC=CQ$, and $P$, $B$, $C$, $Q$ lie on $BC$ in that order. Prove that $PM$ and $QN$ meet at a point on $\Gamma$.
by Victor Domínguez
Let $C_1$ and $C_2$ be two circles with centers $O_1$ and $O_2$, respectively, intersecting at $A$ and $B$. Let $l_1$ be the line tangent to $C_1$ passing trough $A$, and $l_2$ the line tangent to $C_2$ passing through $B$. Suppose that $l_1$ and $l_2$ intersect at $P$ and $l_1$ intersects $C_2$ again at $Q$. Show that $PO_1B$ and $PO_2Q$ are similar triangles.
by Pablo Valeriano
Let $ABC$ be a scalene triangle with circumcenter $O$, and let $D$ and $E$ be points inside angle $\measuredangle BAC$ such that $A$ lies on line $DE$, and $\angle ADB=\angle CBA$ and $\angle AEC=\angle BCA$. Let $M$ be the midpoint of $BC$ and $K$ be a point such that $OK$ is perpendicular to $AO$ and $\angle BAK=\angle MAC$. Finally, let $P$ be the intersection of the perpendicular bisectors of $BD$ and $CE$. Show that $KO=KP$.
by Victor Domínguez
Let $ABC$ be a triangle. Suppose that the perpendicular bisector of $BC$ meets the circle of diameter $AB$ at a point $D$ at the opposite side of $BC$ with respect to $A$, and meets the circle through $A, C, D$ again at $E$. Prove that $\angle ACE=\angle BCD$.

by José Manuel Guerra and Victor Domínguez
Let $P$ and $Q$ be points in the interior of a triangle $ABC$ such that $\angle APC = \angle AQB = 90^{\circ}$, $\angle ACP = \angle PBC$, and $\angle ABQ = \angle QCB$. Suppose that lines $BP$ and $CQ$ meet at a point $R$. Show that $AR$ is perpendicular to $PQ$.

Let $ABC$ be an obtuse triangle with $AB = AC$, and let $\Gamma$ be the circle that is tangent to $AB$ at $B$ and to $AC$ at $C$. Let $D$ be the point on $\Gamma$ furthest from $A$ such that $AD$ is perpendicular to $BC$. Point $E$ is the intersection of $AB$ and $DC$, and point $F$ lies on line $AB$ such that $BC = BF$ and $B$ lies on segment $AF$. Finally, let $P$ be the intersection of lines $AC$ and $DB$. Show that $PE = PF$.

No comments:

Post a Comment