228. OJM - Romania, 2010 problem 3.

by Virgil Nicula, Feb 23, 2011, 10:11 AM

Quote:
OJM-Romania. Let $ABC$ be an $A$-isosceles triangle with $A=40^{\circ}$ and the points $S\in (AB)$ , $T\in (BC)$

so that $m(\angle BAT)=m(\angle BCS)=10^{\circ}$ . Denote $P\in AT\cap TS$ . Prove that $BT=2\cdot PT$ .
Similar problem. Let $ABC$ be an $A$-isosceles triangle with $A=80^{\circ}$ and the points $S\in (AB)$ , $T\in (BC)$

so that $m(\angle BAT)=m(\angle BCS)=20^{\circ}$ . Denote $P\in AT\cap TS$ . Prove that $BT=2\cdot PS$ .

An easy extension. Let $ABC$ be an $A$-isosceles triangle with $A=4\phi<120^{\circ}$ and the points $S\in (AB)$ , $T\in (BC)$ such

that $m(\angle BAT)=m(\angle BCS)=\phi$ . Denote $P\in AT\cap TS$ . Prove that $BT\cdot \sin 3\phi =PT$ or $BT\cdot \cos 3\phi =PS$ .

Proof. Show easily that $AT\perp CS$ and the quadrilateral $ACTS$ is cyclically. Therefore, $m(CST)=3\phi$ and

$m(\angle BST)=B$ , i.e. $TB=TS$ . From the $P$-right-angled triangle $SPT$ obtain easily the relations from conclusion.

Remark. For $\phi :=10^{\circ}$ obtain the first problem and for $\phi =20^{\circ}$ obtain the second problem.
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:50 PM

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