Randomly generated title

by math_explorer, Sep 10, 2010, 1:48 PM

math_explorer wrote:
Let $AD$, $BE$, $CF$ be the altitudes of acute $\triangle ABC$. P, Q are on segments $DF$ and $EF$. Prove that $\angle PAQ = \angle DAC$ iff AP bisects $\angle QPF$. (29th IMO Canada practice problem)

So apparently the key was the thing about the sides of a triangle externally bisecting the angles of the feet of the altitudes of the triangle. Fail fail fail fail.

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So, the next problem, which was randomly selected from my book by the brain of the person sitting behind me.

Suppose $\triangle ABC$ is a triangle and $P$ is a point inside it such that $\angle PAC = \angle PCB = \angle PBA$. If $BP$ bisects $AC$, prove that $AB = AC$. (No source quoted... I think? Also there might be a typo or something. What's that called, the Brocard point or something?)
This post has been edited 2 times. Last edited by math_explorer, Sep 11, 2010, 12:09 PM
geo problem
geometry

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