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Problem 569

AMC 10/12, early questions

Geometry Difficulty 3.5 Multiple choice AMC 10 A · United States

Isosceles triangle $ABC$ has $AB = AC = 3\sqrt{6}$ , and a circle with radius $5\sqrt{2}$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ , $B$ , and $C$ ?

Pick one

Official solution

Let $O$ be the center of the circle with radius $5\sqrt{2}$ . Consider the circle with diameter $\overline{AO}$ . Because $\angle ABO$ and $\angle ACO$ are right angles, the opposite angles of quadrilateral $ABOC$ are supplementary, and hence this quadrilateral is cyclic. Thus $O$ is also on the circle that passes through $A$ , $B$ , and $C$ , and by symmetry $\overline{AO}$ is a diameter. By the Pythagorean Theorem,
$AO = \sqrt{(5\sqrt{2})^2 + (3\sqrt{6})^2} = 2\sqrt{26},$
so the circle that passes through $A$ , $B$ , and $C$ has radius $\sqrt{26}$ and area $26\pi$ .

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