HMMT November

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Let $ABCD$ be a convex quadrilateral with area $202$ , $AB=4$ , and $\angle A=\angle B=90^{\circ}$ such that there is exactly one point $E$ on line $CD$ satisfying $\angle AEB=90^{\circ}$ . Compute the perimeter of $ABCD$ .

Solution

Solution:

The locus of point $E$ such that $\angle AEB=90^{\circ}$ is the circle $\omega$ with diameter $AB$ . Thus, if there exists unique point $E$ , the circle $\omega$ must intersect line $CD$ at exactly one point and hence line $CD$ must be tangent to $\omega$ .

Now, since $\angle DAB=90^{\circ}$ , we get that $AD$ is tangent to $\omega$ , so $DA=DE$ by equal tangents property. Similarly, $CB=CE$ . Thus,

$CD=CE+DE=AD+BC$

However, equating the given area of the quadrilateral gives

$\frac{1}{2}(AD+BC) \cdot AB=202 \Longrightarrow AD+BC=101$

Hence, the final answer is

$CD+(AD+BC)+AB=101+101+4=206$

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