AMC12 2016 A

AMC12 2016 A · Q15

AMC12 2016 A · Q15. It mainly tests Circle theorems, Area & perimeter.

Circles with centers $P$, $Q$, and $R$, having radii $1$, $2$, and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P'$, $Q'$, and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of $\triangle PQR$?

半径分别为 $1$、$2$、$3$ 的三个圆，其圆心分别为 $P$、$Q$、$R$，位于直线 $l$ 的同一侧，并分别在 $P'$、$Q'$、$R'$ 处与直线 $l$ 相切，且 $Q'$ 位于 $P'$ 与 $R'$ 之间。以 $Q$ 为圆心的圆与另外两个圆都外切。求 $\triangle PQR$ 的面积。

(A) 0 0

(B) $\sqrt{\frac{2}{3}}$ $\sqrt{\frac{2}{3}}$

(D) $\sqrt{6} - \sqrt{2}$ $\sqrt{6} - \sqrt{2}$

(E) $\sqrt{\frac{3}{2}}$ $\sqrt{\frac{3}{2}}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $X$ be the foot of the perpendicular from $P$ to $\overline{QQ'}$, and let $Y$ be the foot of the perpendicular from $Q$ to $\overline{RR'}$. By the Pythagorean Theorem, $P'Q' = PX = \sqrt{(2+1)^2 - (2-1)^2} = \sqrt{8}$ and $Q'R' = QY = \sqrt{(3+2)^2 - (3-2)^2} = \sqrt{24}.$ The required area can be computed as the sum of the areas of the two smaller trapezoids, $PQQ'P'$ and $QRR'Q'$, minus the area of the large trapezoid, $PRR'P'$: $\frac{1+2}{2}\sqrt{8} + \frac{2+3}{2}\sqrt{24} - \frac{1+3}{2}\left(\sqrt{8}+\sqrt{24}\right)=\sqrt{6}-\sqrt{2}.$

答案（D）：令 $X$ 为从 $P$ 到 $\overline{QQ'}$ 的垂足，令 $Y$ 为从 $Q$ 到 $\overline{RR'}$ 的垂足。由勾股定理， $P'Q' = PX = \sqrt{(2+1)^2 - (2-1)^2} = \sqrt{8}$ 以及 $Q'R' = QY = \sqrt{(3+2)^2 - (3-2)^2} = \sqrt{24}.$ 所求面积可由两个较小梯形 $PQQ'P'$ 与 $QRR'Q'$ 的面积之和，减去大梯形 $PRR'P'$ 的面积得到： $\frac{1+2}{2}\sqrt{8} + \frac{2+3}{2}\sqrt{24} - \frac{1+3}{2}\left(\sqrt{8}+\sqrt{24}\right)=\sqrt{6}-\sqrt{2}.$

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