drop down menu

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


No comments:

Post a Comment