Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

4. As shown in Figure 2, given that $\odot O$ is the incircle of rhombus $A B C D$ with side length 16, points $E$ and $F$ are on sides $A B$ and $B C$ respectively, and $E F$ is tangent to $\odot O$ at point $M$ . If $B E=4, B F=13$ , then the length of $E F$ is

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

4. 10. 5 .

As shown in Figure 6, let
$\odot O$ be tangent to $AB$ at
point $N$ , and connect $AC$ ,
$BD$ , $OE$ , $OF$ ,
$ON$ .
From the given conditions, we have
$\begin{aligned} & \angle COF \\ = & 180^{\circ}-\angle OFC-\angle OCF \\ = & 90^{\circ}-\frac{1}{2} \angle EFC+\frac{1}{2} \angle EBF \\ = & \frac{1}{2}(\angle BFE+\angle EBF)=\angle AEO . \end{aligned}$

Thus, $\triangle AOE \sim \triangle CFO$ .
Then $\frac{OA}{CF}=\frac{AE}{OC} \Rightarrow \frac{OA}{16-13}=\frac{16-4}{OA}$
$\Rightarrow OA=6 \text {. }$

In the right triangle $\triangle OAB$ , by the projection theorem, we get
$AN=\frac{OA^2}{AB}=\frac{36}{16}=\frac{9}{4} \text {. }$

Then $BN=AB-AN=16-\frac{9}{4}=\frac{55}{4}$ .
Therefore, $EF=2BN-BE-BF=10.5$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.