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China Western Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it China

In a given trapezium $ABCD$ , $AD \parallel BC$ . Suppose $E$ is a variable point on $AB$ , $O_1$ and $O_2$ are circumcenters of $\triangle AED$ and $\triangle BEC$ respectively. Prove that the length of $O_1O_2$ is a fixed value. (posed by Leng Gangsong)

Solution

Proof As shown in the figure, we join $EO_1$ and $EO_2$ , then $\angle AEO_1 = 90^\circ - \angle ADE$ , $\angle BEO_2 = 90^\circ - \angle BCE$ . Hence
$\angle O_1EO_2 = \angle ADE + \angle ECB$ .

Since $AD \parallel BC$ , through $E$ constructing a line parallel to $AD$ , we can prove $\angle DEC = \angle ADE + \angle BCE$ , so $\angle O_1EO_2 = \angle DEC$ .
Further, by the sine rule, we can show
$\frac{DE}{EC} = \frac{2O_1E \sin A}{2O_2E \sin B} = \frac{O_1E}{O_2E}$
Thus $\triangle DEC \sim \triangle O_1EO_2$ . Therefore
$\frac{O_1O_2}{DC} = \frac{O_1E}{DE} = \frac{O_1E}{2O_1E \sin A} = \frac{1}{2 \sin A}$
So $O_1O_2 = \frac{DC}{2\sin A}$ , which is a fixed value. The proposition is proven.

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