2011 AMC 10A Problem 18

Put the tangency point of circles $A$ and $B$ at the origin $T = (0, 0)$, with centers $A = (-1, 0)$ and $B = (1, 0)$. Circle $C$ touches segment $\overline{AB}$ at its midpoint $T$, so its center is one unit straight up: $C = (0, 1)$. In particular $T$ lies on circle $C$, so circle $C$ is not tangent to $A$ or $B$; it cuts into both.

Circles $A$ and $C$: the centers are $\sqrt{1^2 + 1^2} = \sqrt2$ apart. The point $P = (-1, 1)$ is at distance $1$ from both $A$ and $C$, so the two circles meet at $T$ and $P$. Quadrilateral $ATCP$ has four sides of length $1$ and diagonal $AC = \sqrt2$, so it is a unit square: the common chord $TP$ subtends a right angle at each center.

The overlap of the two circles is therefore two $90^\circ$ circular segments, each a quarter-disk minus a right isosceles triangle with legs $1$:
$$\text{lens} = 2\left(\frac{\pi}{4} - \frac12\right) = \frac{\pi}{2} - 1.$$
By the mirror symmetry across the $y$-axis, circle $C$ overlaps circle $B$ in a congruent lens, and the two lenses share only the point $T$.

Removing both lenses from circle $C$:
$$\pi - 2\left(\frac{\pi}{2} - 1\right) = 2.$$

The answer is $\boxed{\textbf{(C)}\ 2}$.

Two unit circles whose centers are $\sqrt2$ apart cross at right angles, and $90^\circ$ segments have the tidy area $\frac{\pi}{4} - \frac12$. Coordinates make the distances and the intersection points visible immediately. Whenever a problem mentions tangency and midpoints, locate every center and compute center-to-center distances before doing any area work.

The points $T$, $P = (-1, 1)$, $Q = (1, 1)$ and the top point $(0, 2)$ form a square inscribed in circle $C$ with diagonals of length $2$, so its area is $2$. Circle $C$ outside this square consists of four congruent $90^\circ$ segments. The lens shared with circle $A$ is the lower-left segment of circle $C$ plus one congruent segment of circle $A$ bulging past chord $TP$. So the desired region is the square plus the two upper segments minus the two bulging segments, and the segments cancel exactly: the area is that of the square, $2$.

• Assuming circle C is tangent to circles A and B, when it actually passes through their point of tangency and overlaps both.
• Treating each lens as a single $90^\circ$ segment ($\frac{\pi}{4} - \frac12$) instead of two, which produces $\frac{\pi}{2} + 1$, choice (E).
• Taking the sector angle as $60^\circ$ or $120^\circ$ from a hasty sketch rather than checking that $AT = TC = CP = PA = 1$ with $AC = \sqrt2$.