Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it Ukraine

A parallelogram $ABCD$ with $AC > BD$ is given. Let $k_1$ be the circle with diameter $AC$ and $k_2$ the circle with diameter $DC$ . $k_1$ meets line $AB$ at the point $E$ , $k_2$ meets the line $AC$ at the points $C$ and $O$ , and line $AD$ at the point $F$ . Suppose that $AO = a$ , $FO = b$ and $\angle BAC = 45^\circ$ . Find the ratio of the areas of the triangles $AOE$ and $COF$ .

Answer: $\left(\frac{a}{b}\right)^2$ .

Solution

Finally, $EF$ and $AC$ are chords in the circle $k_1$ , hence $EO \cdot OF = AO \cdot OC$ , which implies:
$\frac{S(AOE)}{S(COF)} = \frac{AO \cdot GE}{FO \cdot CO} = \left(\frac{AO}{FO}\right)^2 = \left(\frac{a}{b}\right)^2$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.