2011 AMC 10A Problem 18

Everything depends on where the centers are, so place the circles on a grid first.

1. Set up coordinates. Circles $A$ and $B$ touch at one point; call it $T$ and put it at $(0, 0)$. Their centers are then $A = (-1, 0)$ and $B = (1, 0)$, and $T$ is the midpoint of $\overline{AB}$.
2. Place circle $C$. It touches segment $\overline{AB}$ at $T$, so its center is $1$ unit straight above $T$: $C = (0, 1)$. Notice that $T$ is on circle $C$. So circle $C$ does not just touch $A$ and $B$; it cuts into both of them.
3. Where circles $A$ and $C$ cross. The point $P = (-1, 1)$ is $1$ unit from $A$ (straight up) and $1$ unit from $C$ (straight left). So both circles pass through $T$ and $P$.
4. Spot a square. The four points $A = (-1,0)$, $T = (0,0)$, $C = (0,1)$, $P = (-1,1)$ are the corners of a $1 \times 1$ square. So at center $A$, the segments to $T$ and $P$ make a right angle, and the same is true at center $C$.
5. Area of the overlap (a lens). The overlap of circles $A$ and $C$ is split by the line $TP$ into two identical pieces. Each piece is a quarter of a circle of radius $1$ (area $\frac{\pi}{4}$) minus a right triangle with legs $1$ and $1$ (area $\frac{1}{2}$). So
$$\text{lens} = 2\left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{2} - 1.$$
6. The same on the right. The picture is symmetric, so circle $C$ overlaps circle $B$ in another lens of area $\frac{\pi}{2} - 1$. The two lenses meet only at the single point $T$, so they don't overlap each other.
7. Subtract. Circle $C$ has area $\pi$. Take away the two lenses:
$$\pi - 2\left(\frac{\pi}{2} - 1\right) = \pi - \pi + 2 = 2.$$

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

Two circles of radius $1$ whose centers are $\sqrt2$ apart (the diagonal of a unit square) overlap in a lens made of two quarter-circle-minus-triangle pieces, each of area $\frac{\pi}{4} - \frac{1}{2}$. The lesson: whenever a problem talks about circles touching, first find every center and the distances between them. The picture you imagine at first (circles just touching) can be wrong.

A cut-and-move view. The points $T = (0,0)$, $P = (-1,1)$, $Q = (1,1)$ and $(0,2)$ are all on circle $C$ and form a square whose diagonals have length $2$. Its area is $\frac{2 \times 2}{2} = 2$.

Circle $C$ is this square plus four identical curved slivers, one on each side. The lower-left sliver lies inside circle $A$, and circle $A$ also pokes into the square with a sliver of the same size and shape across the line $TP$. The same happens on the right with circle $B$. So the region we want is the square, plus the two upper slivers of circle $C$, minus the two slivers where circles $A$ and $B$ poke into the square. Those four slivers are all the same size, so they cancel, leaving the square's area, $2$.

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 $\boxed{\textbf{(C)}\ 2}$.