Question

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}.$ Three of the sides of this quadrilateral have length $200.$ What is the length of the fourth side?

$200$

$200\sqrt{2}$

$200\sqrt{3}$

$300\sqrt{2}$

$500$

Solution:

Let $O$ be the circle's center, and let $AD$ meet $OB$ and $OC$ at $E$ and $F$ . Equal chords $AB,BC,CD$ give equal central angles.

From the equal-angle relationships, $\triangle OAB\sim\triangle ABE$ . Since $OA=OB=200\sqrt{2}$ and $AB=200$ , the similarity gives $AE=AB=200$ and $BE=100\sqrt{2}$ . Similarly, $FD=CD=200$ .

In triangle $OBC$ , the same scaling gives $EF=\frac12BC=100$ . Therefore $AD=AE+EF+FD=200+100+200=500.$

Thus, the correct answer is E.

2016 AMC 10A Problem 24

Solution:

Problem 24 in Other Years