Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Philippines

Problem:

Given triangle $ABC$ , let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ . Let $F$ be the intersection of $BE$ and $CD$ . If $\triangle DBF$ has an area of $4$ , $\triangle BFC$ has an area of $6$ , and $\triangle FCE$ has an area of $5$ , find the area of quadrilateral $ADFE$ .

Solution

Solution:

Let the area of quadrilateral $ADFE$ be $x$ . By Menelaus' Theorem, $\frac{AD}{DB} \cdot \frac{BF}{FE} \cdot \frac{EC}{CA} = 1$ . Since $\frac{AD}{DB} = \frac{x+5}{10}$ , $\frac{BF}{FE} = \frac{6}{5}$ , and $\frac{EC}{CA} = \frac{11}{x+15}$ , we have $\frac{66(x+5)}{50(x+15)} = 1$ , or $x = \frac{105}{4}$ or $26.25$ .

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