Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Let $ABCD$ be a rectangle, and let $E$ and $F$ be points on segment $AB$ such that $AE = EF = FB$ . If $CE$ intersects the line $AD$ at $P$ , and $PF$ intersects $BC$ at $Q$ , determine the ratio of $BQ$ to $CQ$ .

Solution

Solution:

Answer: $\frac{1}{3}$

Because $\triangle PAE \sim \triangle PDC$ and $AE : DC = 1 : 3$ , we have that $PA : PD = 1 : 3 \Longrightarrow PA : AB = PA : BC = 1 : 2$ . Also, by similar triangles $\triangle PAF \sim \triangle QBF$ , since $AF : BF = 2 : 1$ , $PA : BQ = 2 : 1$ . Then $BQ = \frac{1}{2} PA = \frac{1}{2} \cdot \frac{1}{2} BC = \frac{1}{4} BC$ . Then $BQ : CQ = \frac{1}{3}$ .

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