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Harvard-MIT Math Tournament

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

$A$ is the center of a semicircle, with radius $A D$ lying on the base. $B$ lies on the base between $A$ and $D$ , and $E$ is on the circular portion of the semicircle such that $E B A$ is a right angle. Extend $E A$ through $A$ to $C$ , and put $F$ on line $C D$ such that $E B F$ is a line. Now $E A = 1$ , $A C = \sqrt{2}$ , $B F = \frac{2 - \sqrt{2}}{4}$ , $C F = \frac{2 \sqrt{5} + \sqrt{10}}{4}$ , and $D F = \frac{2 \sqrt{5} - \sqrt{10}}{4}$ . Find $D E$ .

Solution

Solution:

Let $\theta = \angle A E D$ and $x = D E$ . By the law of cosines on triangle $A D E$ , we have
$1 = 1 + x^{2} - 2x \cos \theta \implies 2x \cos \theta = x^{2}.$
Then by the law of cosines on triangle $C D E$ (note that $C D = \sqrt{5}$ ), we have
$5 = (1 + \sqrt{2})^{2} + x^{2} - 2(1 + \sqrt{2}) x \cos \theta = (1 + \sqrt{2})^{2} + x^{2} - (1 + \sqrt{2}) x^{2}.$
Solving the quadratic equation gives $x = \sqrt{2 - \sqrt{2}}$ .

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