Question

Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6,$ and a circle with radius $5\sqrt2$ 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 $C?$

$24\pi$

$25\pi$

$26\pi$

$27\pi$

$28\pi$

Answer: C

Concepts:circumcircle, circumcenter, and circumradius cyclic quadrilateral tangent line Pythagorean Theorem

Difficulty rating: 1820

Solution:

Let $O_1$ be the center of the circle tangent to $AB$ and $AC.$ Then $\angle ABO_1=\angle ACO_1=90^\circ,$ so $A,B,O_1,C$ are concyclic.

Because the right angles at $B$ and $C$ subtend $AO_1,$ the segment $AO_1$ is a diameter of this circle. Let $O_2$ be its center. The same circle passes through $A,B,$ and $C,$ so it is the desired circumcircle.

By the Pythagorean Theorem in $\triangle ABO_1,$ $\begin{aligned} AO_1&=\sqrt{AB^2+BO_1^2}\\ &=\sqrt{54+50}=2\sqrt{26}. \end{aligned}$ Therefore, the circumradius is $\sqrt{26},$ and the requested area is $26\pi.$

Thus, C is the correct answer.

2021 AMC 10A Fall Problem 15

Answer: C

Difficulty rating: 1820

Solution:

Problem 15 in Other Years