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Memorial "Aleksandar Blazhevski-Cane" 2020-23 (N. Macedonia) 7p

geometry problems from Memorial Mathematical Contest "Aleksandar Blazhevski-Cane", a North Macedonian contest with aops links


aops collection here

geometry collected inside aops here

                                             2020 - 2023

Seniors

A convex quadrilateral ABCD is given in which the bisectors of the interior angles \angle ABC and \angle ADC have a common point on the diagonal AC. Prove that the bisectors of the interior angles \angle BAD and \angle BCD have a common point on the diagonal BD.

Let ABCD be a cyclic quadrilateral such that AB=AD. Let E and F be points on the sides BC and CD, respectively, such that BE+DF=EF. Prove that \angle BAD = 2 \angle EAF.

Let \triangle ABC be a triangle with circumcenter O. The perpendicular bisectors of the segments OA,OB and OC intersect the lines BC,CA and AB at D,E and F, respectively. Prove that D,E,F are collinear.

Let ABC be an acute triangle with incircle \omega, incenter I, and A-excircle \omega_{a}.
Let \omega and \omega_{a} meet BC at X and Y, respectively. Let Z be the intersection
point of AY and \omega which is closer to A. The point H is the foot of the altitude from A.
Show that HZ, IY and AX are concurrent.

Let ABCD be a cyclic quadrilateral inscribed in a circle \omega with center O. The lines AD and BC meet at E, while the lines AB and CD meet at F. Let P be a point on the segment EF such that OP \perp EF. The circle \Gamma_{1} passes through A and E and is tangent to \omega at A, while \Gamma_{2} passes through C and F and is tangent to \omega at C. If \Gamma_{1} and \Gamma_{2} meet at X and Y, prove that PO is the bisector of \angle XPY.

 by Nikola Velov
Let ABCD be a cyclic quadrilateral such that AB = AD + BC and CD < AB. The diagonals AC and BD intersect at P, while the lines AD and BC intersect at Q. The angle bisector of \angle APB meets AB at T. Show that the circumcenter of the triangle CTD lies on the circumcircle of the triangle CQD.
by Nikola Velov

Juniors

Let ABC be an acute triangle with altitude AD (D \in BC). The line through C parallel to
AB meets the perpendicular bisector of AD at G. Show that AC = BC if and only if
\angle AGC = 90^{\circ}.

Let ABC be an acute triangle with incircle \omega, incenter I, and A-excircle \omega_{a}.
Let \omega and \omega_{a} meet BC at X and Y, respectively. Let Z be the intersection
point of AY and \omega which is closer to A. The point H is the foot of the altitude from A.
Show that HZ, IY and AX are concurrent.

Let ABCD be a cyclic quadrilateral such that AB = AD + BC and CD < AB. The diagonals AC and BD intersect at P, while the lines AD and BC intersect at Q. The angle bisector of \angle APB meets AB at T. Show that the circumcenter of the triangle CTD lies on the circumcircle of the triangle CQD.
by Nikola Velov


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