Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Cyclic quadrilateral $ABCD$ has side lengths $AB = 1$ , $BC = 2$ , $CD = 3$ , and $AD = 4$ . Determine $AC / BD$ .

Solution

Solution:

Answer: $\dfrac{5}{7}$ . Let the diagonals intersect at $P$ . Note that triangles $ABP$ and $DCP$ are similar, so that $3AP = DP$ and $3BP = CP$ . Additionally, triangles $BCP$ and $ADP$ are similar, so that $2BP = AP$ . It follows that
$\frac{AC}{BD} = \frac{AP + PC}{BP + PD} = \frac{2BP + 3BP}{BP + 6BP} = \frac{5}{7}$

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