2016 AMC 10A Problem 21

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

Two circles tangent to the same line and to each other form a right trapezoid whose slanted side is the sum of the radii and whose vertical leg is the difference of the radii; the Pythagorean theorem gives the horizontal offset of the tangency points. That converts the whole configuration into explicit coordinates, after which any triangle area is mechanical. The centers are not collinear because the slopes $\frac{1}{2\sqrt2}$ and $\frac{1}{2\sqrt6}$ differ, which is why the area is small but nonzero.

Trapezoids: $[PQR] = [P'PRR'] - [P'PQQ'] - [Q'QRR']$, where each is a right trapezoid with the radii as parallel sides. This gives $\frac{1+3}{2}(2\sqrt2 + 2\sqrt6) - \frac{1+2}{2}\cdot 2\sqrt2 - \frac{2+3}{2}\cdot 2\sqrt6 = \sqrt2 - \sqrt6$ in signed form; the area is its absolute value $\sqrt6 - \sqrt2$.

• Assuming the three centers are collinear (answer 0), or using the sum of radii as the horizontal distance between tangency points.
• Placing $P$ and $R$ on the same side of $Q$, which ignores the condition that $Q'$ lies between $P'$ and $R'$ and changes the area.