IGO 2016 Advanced/2
- Thamim Zahin
- Posts:98
- Joined:Wed Aug 03, 2016 5:42 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$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: IGO 2016 Advanced/2
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, $\angle XMB=\angle XME+\angle EMB=2\angle MBC$.
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.
Because destroying everything mindlessly isn't cool enough.