2016 AMC 10A Problem 21

Put the picture on a grid so every center has coordinates, then find the area.

1. Set up. Let line $l$ be the $x$-axis with $Q'$ at the origin. A circle touching the $x$-axis from above has its center straight above the touching point, at a height equal to its radius. So $Q = (0, 2)$, $P$ is at height $1$ to the left, and $R$ is at height $3$ to the right.
2. Find $P$. The circles at $P$ and $Q$ touch from outside, so their centers are $1 + 2 = 3$ apart. The heights differ by $2 - 1 = 1$. Draw a right triangle with the segment $PQ$ as the long side: the vertical leg is $1$, so the horizontal leg is
$$\sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}.$$
So $P = (-2\sqrt2,\ 1)$.
3. Find $R$. The centers $Q$ and $R$ are $2 + 3 = 5$ apart and differ in height by $3 - 2 = 1$. The horizontal leg is $\sqrt{5^2 - 1^2} = \sqrt{24} = 2\sqrt{6}$. So $R = (2\sqrt6,\ 3)$.
4. Area with trapezoids. Drop each center straight down to the $x$-axis. Each region between a side of the triangle and the axis is a trapezoid with vertical parallel sides (the radii):
- Under $PQ$: parallel sides $1$ and $2$, width $2\sqrt2$. Area $\tfrac{1 + 2}{2} \cdot 2\sqrt2 = 3\sqrt2$.
- Under $QR$: parallel sides $2$ and $3$, width $2\sqrt6$. Area $\tfrac{2 + 3}{2} \cdot 2\sqrt6 = 5\sqrt6$.
- Under $PR$: parallel sides $1$ and $3$, width $2\sqrt2 + 2\sqrt6$. Area $\tfrac{1 + 3}{2}(2\sqrt2 + 2\sqrt6) = 4\sqrt2 + 4\sqrt6$.
5. The triangle is the gap between the bent path $P \to Q \to R$ and the straight segment $P \to R$, so its area is the difference of the two totals:
$$(3\sqrt2 + 5\sqrt6) - (4\sqrt2 + 4\sqrt6) = \sqrt6 - \sqrt2.$$
(It is positive, which tells us $Q$ sits a little above the segment $PR$.)

The answer is $\boxed{\textbf{(D)}\ \sqrt{6}-\sqrt{2}}$.

Two circles touching each other and the same line always give a right triangle: the long side is the sum of the radii, and the vertical side is the difference of the radii. That fixes every center. The three centers are not in a straight line (the two slopes $\tfrac{1}{2\sqrt2}$ and $\tfrac{1}{2\sqrt6}$ are different), which is why the area is small but not zero.

With $P(-2\sqrt2, 1)$, $Q(0, 2)$ and $R(2\sqrt6, 3)$, use the shoelace formula:
$$\text{Area} = \tfrac12\left|(-2\sqrt2)(2 - 3) + 0 \cdot (3 - 1) + 2\sqrt6\,(1 - 2)\right| = \tfrac12\left|2\sqrt2 - 2\sqrt6\right| = \sqrt6 - \sqrt2.$$

Let $l$ be the $x$-axis with $Q' = (0,0)$, so $Q = (0, 2)$. Each center sits at height equal to its radius: $P = (p, 1)$ and $R = (r, 3)$ with $p < 0 < r$.

External tangency gives $PQ = 1 + 2 = 3$. The vertical gap between $P$ and $Q$ is $2 - 1 = 1$, so the horizontal gap is $\sqrt{3^2 - 1^2} = \sqrt 8 = 2\sqrt 2$, i.e. $P = (-2\sqrt 2,\, 1)$.

Similarly $QR = 2 + 3 = 5$ with vertical gap $1$, so the horizontal gap is $\sqrt{25 - 1} = 2\sqrt 6$ and $R = (2\sqrt 6,\, 3)$.

Shoelace on $P(-2\sqrt2, 1)$, $Q(0, 2)$, $R(2\sqrt6, 3)$:
$$[PQR] = \tfrac12\left|(-2\sqrt2)(2 - 3) + 0\cdot(3 - 1) + 2\sqrt6\,(1 - 2)\right| = \tfrac12\left|2\sqrt2 - 2\sqrt6\right| = \sqrt6 - \sqrt2.$$

The answer is $\boxed{\textbf{(D)}\ \sqrt{6}-\sqrt{2}}$.