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### APMO 2017 P2

Posted: Sat May 27, 2017 1:11 pm
Let $\bigtriangleup ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of $\angle BAC$ and the circumcircle of $\bigtriangleup ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of $\angle BAC$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$.

### Re: APMO 2017 P2

Posted: Mon May 29, 2017 12:34 am
Let the reflection of $DM$ over $M$ be $D'M$.Thus we concurr that $AD'BD$ is a parallelogram as $M$ is designed to be the midpoint of $AB$.Let the midpoint of $AC$ be $X$.Thus we have $AX$=$CX$; $ZX$ as a common side of triangles $ZXA$ and $ZXC$;$\angle ZXA$=$\angle ZXC$.So,by $SAS$,we have $\bigtriangleup ZXA$ $\cong$ $\bigtriangleup ZXC$ which follows $ZA$=$ZC$.Now let $\angle BAD$ be $\alpha$ and $\angle D'AB$ to be $\gamma$.Now, trivial angle chasing shows that $\angle D'AZ=270^{\circ}-\alpha$-$\gamma$.Again $\angle DCZ$=$\angle BCD$+$\angle ACB$+$\angle ACZ=90^{\circ}+\angle ACB$.Since $\angle ACB$=$\angle ADB$=$\angle AD'B$,it's clear that $\angle DCZ$=$\angle D'AZ$=270-$\alpha$-$\gamma$.So,by applying $SAS$ again,we get $\bigtriangleup D'AZ$ $\cong$ $\bigtriangleup DCZ$,hence $DZ$=$D'Z$.Therefore, $\bigtriangleup DD'Z$ is isosceles showing that $ZM$ $\perp$ $DD'$.So,$\angle ZMD=90^{\circ}$,which was our intended goal leaving $AMDZ$ cyclic.So,we are done.

### Re: APMO 2017 P2

Posted: Tue Sep 12, 2017 7:41 pm
Let $M$ and $N$ be the midpoints of $AB$ and $AC$.$MN$ meets $DC$ at $X$.Then $\angle AMX=\angle ABC=\angle ADX$.So $A$,$M$,$D$,$X$ are cyclic.Then $\angle NXC=\angle MXD=\angle MAD=\angle DAC=\angle AZN=\angle CZN$ implies $Z$,$N$,$C$,$X$ are cyclic so $\angle CXZ=\angle DXZ=90$,$\angle DAZ=90$.
So $A$,$M$,$D$,$X$,$Z$ are cyclic.