Prove that EI bisects <BEC

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Prove that EI bisects <BEC

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Source: Saudi Arabia IMO TST Day II Problem 1

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TheMaskedMagician

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TheMaskedMagician

#1 h Jul 22, 2014, 6:59 PM • 2 Y

Y by Adventure10, Mango247

Let $\Gamma$ be a circle with center $O$ and $AE$ be a diameter. Point $D$ lies on segment $OE$ and point $B$ is the midpoint of one of the arcs $\widehat{AE}$ of $\Gamma$ . Construct point $C$ such that $ABCD$ is a parallelogram. Lines $EB$ and $CD$ meet at $F$ . Line $OF$ meets the minor arc $\widehat{EB}$ at $I$ . Prove that $EI$ bisects $\angle BEC$ .

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#2 h Jul 23, 2014, 12:34 AM • 2 Y

Y by Adventure10, Mango247

Hmmm this is kind of sad.

Trivially $BCED, BFDO$ are cyclic. Let $\angle OBD =x$ . Then we can easily get $\angle BEC =\tfrac{\pi}{4}+x$ and $\angle EFI =\tfrac{\pi}{2}-x$ . Since $AOI$ is isosceles $\angle AIO= \tfrac{\pi}{8}-\tfrac{x}{2}$ so $\angle FIE = \tfrac{3\pi}{8}+\tfrac{x}{2}$ so $\angle BEI =\tfrac{\pi}{8}+\tfrac{x}{2}=\tfrac{1}{2}\angle BEC$ , as desired.

One can simply cram the angle measures into the diagram and be done with no effort.

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#3 h Jul 23, 2014, 8:40 PM • 1 Y

Y by Adventure10

By simple angle chasing we get $BCED$ and $BFDO$ cyclic and then we just angle chase again.I think the problems in Saudi Arabia TST are too easy.

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