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