2021 AMC Fall 10A Problem 15

Let $O$ be the center of the given circle. A radius drawn to a point of tangency is perpendicular to the tangent line, so $OB \perp AB$ and $OC \perp AC$: the angles $\angle ABO$ and $\angle ACO$ are both right angles.

A right angle subtended by segment $AO$ means that $B$ and $C$ both lie on the circle with diameter $AO$ (Thales' theorem in reverse). Since $A$ is trivially on that circle too, the circle with diameter $AO$ is the circle through $A$, $B$, $C$.

Compute $AO$ from right triangle $ABO$:
$$AO^2 = AB^2 + OB^2 = (3\sqrt6)^2 + (5\sqrt2)^2 = 54 + 50 = 104.$$
The circumradius is $R = \tfrac12 AO$, so $R^2 = \tfrac{104}{4} = 26$ and the area is $\pi R^2 = 26\pi$.

The answer is $\boxed{\textbf{(C)}\ 26\pi}$.

Two tangent segments from an external point $A$ to a circle form right angles with the radii, and any two right angles on the same segment $AO$ put their vertices on one circle with that segment as diameter. Recognizing this hidden circle makes the circumradius a one-line Pythagorean computation; no trigonometry or angle chasing is required. Whenever a figure has several right angles sharing a hypotenuse, look for a common circle.

Without the diameter trick: let $\theta = \angle BAO$, so $\tan\theta = \dfrac{5\sqrt2}{3\sqrt6} = \dfrac{5}{3\sqrt3}$, and $\angle BAC = 2\theta$. Then $BC = 2 \cdot AB \sin\theta$ and by the extended law of sines $R = \dfrac{BC}{2\sin 2\theta} = \dfrac{AB}{2\cos\theta} = \dfrac{AO}{2}$, the same conclusion with more work.

• Assuming the given circle of radius 5 sqrt 2 is the circle through A, B, C and answering 25 pi, choice (B).
• Taking $AO$ itself as the radius, giving $104\pi$, or dropping the factor of $\tfrac12$ somewhere and choosing (E).