Page 1 of 1
IGO 2016 Advanced/2
Posted: Tue Jan 10, 2017 4:20 pm
by Thamim Zahin
2. In acute-angled triangle $ABC$, altitude of $A$ meets $BC$ at $D$, and $M$ is midpoint of $AC$. Suppose that $X$ is a point such that $\angle AXB = \angle DXM = 90^o$ (assume that $X$ and $C$ lie on opposite sides of the line $BM$). Show that $ \angle XMB = 2\angle MBC$.
Re: IGO 2016 Advanced/2
Posted: Wed Jan 11, 2017 6:46 pm
by Thanic Nur Samin
Cute problem.
Let $E$ be the midpoint of $AB$. Let $Y$ be the intersection of $AD$ and $EM$. Clearly, $\angle DXM=\angle DYM=90^{\circ}$, and so $DYXM$ is cyclic.
Again, $\angle ADB=\angle AXB=90^{\circ}$. So, $ABDX$ is cyclic.
Now, $\angle XME=\angle XMY=\angle XDY=\angle XDA=\angle XBA=\angle XBE$. So, $BMXE$ is cyclic.
Also, since $\triangle AXB$ is a right traingle, $E$ is the circumcenter. So, $EB=EX$. But since $BMXE$ is cyclic, this implies $\angle XME=\angle EMB=\angle MBC$. So, $\angle XMB=\angle XME+\angle EMB=2\angle MBC$.