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 $C?$

$24\pi$

$25\pi$

$26\pi$

$27\pi$

$28\pi$

Solution:

Let $\odot C_1$ be the circle that is tangent to $AB$ and $AC.$ Then $\angle ABO_1 = \angle ACO_1 = 90^{\circ},$ making the two angles supplementary. This makes $ABO_1C$ cyclic.

Let $\odot O_2$ be the circumcircle of $ABO_1C.$ This makes $\odot O_2$ the circumcircle of $\triangle ABC$ as well.

We also know that $AO_1$ is the diameter of $O_2$ since $AO_1$ bisects $\angle BAC.$

By the Pythagorean theorem, we get that $\begin{align*} AO_1 &= \sqrt{AB^2 + BO_1^2} \\ &= \sqrt{54 + 50} \\ &= 2\sqrt{26}. \end{align*}$ This makes the area of $\odot O_2$ $26 \pi.$

Thus, C is the correct answer.

2021 AMC 10A Fall Problem 15

Solution:

Problem 15 in Other Years