Printable

2017 AMC 10A Problem 22

2017 AMC 10A Problem 22AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 4 of 5 · about 8 min · position P21-25

Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

A)
$\frac{4\sqrt{3}\pi}{27}-\frac{1}{3}$
B)
$\frac{\sqrt{3}}{2}-\frac{\pi}{8}$
C)
$\frac{1}{2}$
D)
$\sqrt{3}-\frac{2\sqrt{3}\pi}{9}$
E)
$\frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$

Answer

Key insight

Radii to the tangent points make angle BOC = 120 degrees, so the triangle's overlap with the circle is a 120-degree segment; compare it to the triangle's area.

Solution

Let $O$ be the center and take the radius to be $1$ . Since $AB$ is tangent at $B$ , $OB \perp AB$ , so $\angle ABO = 90^\circ$ ; likewise $\angle ACO = 90^\circ$ . In quadrilateral $ABOC$ the angle at $A$ is $60^\circ$ , so $\angle BOC = 360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ$ . Because $\angle ABC = 60^\circ$ and $\angle ABO = 90^\circ$ , the center lies on the opposite side of $BC$ from $A$ , with $\angle OBC = 30^\circ$ .

Chord $BC$ in the isosceles triangle $OBC$ with legs $1$ and apex $120^\circ$ has length $2\cos 30^\circ = \sqrt3$ . So the equilateral triangle has side $\sqrt3$ and area
$\frac{\sqrt3}{4} \cdot (\sqrt3)^2 = \frac{3\sqrt3}{4}.$

The circle stays inside $\angle BAC$ (it is tangent to both sides), so the part of the triangle inside the circle is exactly the circular segment cut off by chord $BC$ on $A$ 's side. Its area is the $120^\circ$ sector minus triangle $OBC$ :
$\frac{120}{360}\pi - \frac12 \cdot 1 \cdot 1 \cdot \sin 120^\circ = \frac{\pi}{3} - \frac{\sqrt3}{4}.$

The fraction of the triangle inside the circle is
$\frac{\frac{\pi}{3} - \frac{\sqrt3}{4}}{\frac{3\sqrt3}{4}} = \frac{4\pi}{9\sqrt3} - \frac13 = \frac{4\sqrt3\pi}{27} - \frac13,$
so the fraction outside is
$1 - \left(\frac{4\sqrt3\pi}{27} - \frac13\right) = \frac43 - \frac{4\sqrt3\pi}{27}.$

The answer is $\boxed{\textbf{(E)}\ \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}}$ .

Why this works

Tangency gives right angles at the points of contact, and those right angles pin down the central angle $\angle BOC$ and hence the radius-to-side ratio. Once the geometry is fixed, the overlap of a circle and a polygon bounded by a single chord is a circular segment, computed as sector minus triangle. Choosing radius $1$ keeps the arithmetic clean; the answer is a ratio and does not depend on scale.

Alternative approach

Estimate: $\frac{4\sqrt3\pi}{27} \approx 0.806$ , so choice (E) is about $0.53$ , choice (A) about $0.47$ , and (D) about $0.52$ . A sketch shows the segment covers a bit under half the triangle, so the outside fraction is a bit over $\tfrac12$ ; among (C), (D), (E) only an exact computation separates them, but (A) and (B) can be discarded quickly.

The trap

Placing the circle's center inside the triangle (as if it were the incircle or circumcircle) instead of on the far side of BC.

Common mistakes

Placing the circle's center inside the triangle (as if it were the incircle or circumcircle) instead of on the far side of BC.
Reporting the fraction inside the circle, choice (A), instead of the fraction outside.

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Cut the figure into known shapes (triangles, rectangles, sectors)