160. A very nice "slicing" problem.

by Virgil Nicula, Oct 19, 2010, 7:58 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=372309

Let $ABC$ be an acute triangle with $A=2B$ . Denote $D\in (BC)$ so that $AD\perp BC$ and $E\in (AD)$ such that $3\cdot m(\widehat {BCE})=C$ . Ascertain $m(\widehat{BEC})$ .

Proof 1 (trigonometric). Denote $B=\phi$ and $m(\widehat {ABE})=x$ . Show easily that $\left\|\begin{array}{ccc} m(\widehat{EAB})=90^{\circ}-\phi & \wedge & m(\widehat{EAC})=3\phi-90^{\circ}\\\\ m(\widehat{EBD})=\phi -x & \wedge & m(\widehat{EBA})=x\\\\ m(\widehat{ECA})=120^{\circ}-2\phi & \wedge & m(\widehat{ECB})=60^{\circ}-\phi\end{array}\right\|$ .

Since $\phi\not\in\left\{60^{\circ},90^{\circ}\right\}$ , i.e. $0\not\in \left\{\cos\phi ,\cos (30^{\circ}+\phi )\right\}$ , we can simplify by $\cos\phi$ and $\cos (30^{\circ}+\phi )$ lower down $\Downarrow$ . Apply the trigonometric form of Ceva's theorem :

$\frac {\sin (90^{\circ}-\phi )} {\sin (3\phi-90^{\circ})}\cdot\frac {\sin (\phi -x)}{\sin x}\cdot\frac {\sin (120^{\circ}-2\phi )}{\sin (60^{\circ}-\phi )}=1$ $\iff$ $\frac {\sin (\phi -x)}{\sin x}=-\frac {\cos 3\phi } {\cos \phi}\cdot\frac {\cos (30^{\circ}+\phi )}{\sin (60^{\circ}+2\phi )}$ $\iff$ $\sin\phi\cdot \cot x-\cos \phi =\frac {3-4\cos^2\phi }{2\sin (30^{\circ}+\phi )}$ $\iff$

$\sin\phi\cdot \cot x=\frac {\cos\phi (\cos\phi+\sqrt 3\sin\phi )+3-4\cos^2\phi}{\cos\phi +\sqrt 3\sin\phi}$ $\iff$ $\sin\phi\cdot \cot x=\frac {3\sin^2\phi +\sqrt 3\sin\phi \cos\phi}{\cos\phi +\sqrt 3\sin\phi}$ $\iff$ $\cot x=\frac {3\sin\phi +\sqrt 3\cos\phi}{\cos\phi +\sqrt 3\sin\phi}$ $\iff$

$\cot x=\sqrt 3$ $\iff$ $x=30^{\circ}$ . Denote $F\in CE\cap AB$ . Since $m(\widehat{AFE})=60^{\circ}$ obtain $m(\widehat{BEF})=m(\widehat{EBA})=x=30^{\circ}$ $\implies$ $\boxed{\begin{array}{c} FE=FB\\\ m(\widehat{BEC})=150^{\circ}\end{array}}$ .

Remark. If can show synthetically that $FE=FB$ , then all is O.K. Here is a synthetical proof.

Proof 2 (synthetic).
This post has been edited 23 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:47 PM

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