IGO 2016 Advanced/2

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IGO 2016 Advanced/2

Post Number:#1  Unread postby Thamim Zahin » Tue Jan 10, 2017 4:20 pm

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$.
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Re: IGO 2016 Advanced/2

Post Number:#2  Unread postby Thanic Nur Samin » Wed Jan 11, 2017 6:46 pm

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, $\angle XMB=\angle XME+\angle EMB=2\angle MBC$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.
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