Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.3 AMC 10/12 Find the answer

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?

Pick one

Solution

Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$

$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$

From there, $\triangle{OAB} \sim \triangle{ABE}$ , thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$ . Since $\frac{OA}{AB} = \frac{OB}{AE}$ , we have that $AB = AE$ . Similarly, $CD = DF$ .
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$

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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.