AMC12 2019 A

AMC12 2019 A · Q22

AMC12 2019 A · Q22. It mainly tests Triangles (properties), Pythagorean theorem.

Circles $\omega$ and $\gamma$, both centered at $O$, have radii 20 and 17, respectively. Equilateral triangle $ABC$, whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$, has vertex $A$ on $\omega$, and the line containing side $\overline{BC}$ is tangent to $\gamma$. Segments $\overline{AO}$ and $\overline{BC}$ intersect at $P$, and $\frac{BP}{CP} = 3$. Then $AB$ can be written in the form $\frac{m}{\sqrt{n}} - \frac{p}{\sqrt{q}}$ for positive integers $m, n, p, q$ with $\gcd(m, n) = \gcd(p, q) = 1$. What is $m + n + p + q$?

圆 $\omega$ 和 $\gamma$ 都以 $O$ 为圆心，半径分别为 20 和 17。正三角形 $ABC$ 的内部位于 $\omega$ 的内部但位于 $\gamma$ 的外部，顶点 $A$ 在 $\omega$ 上，边 $\overline{BC}$ 所在直线与 $\gamma$ 相切。线段 $\overline{AO}$ 和 $\overline{BC}$ 相交于 $P$，且 $\frac{BP}{CP} = 3$。则 $AB$ 可写成 $\frac{m}{\sqrt{n}} - \frac{p}{\sqrt{q}}$ 的形式，其中 $m, n, p, q$ 为正整数且 $\gcd(m, n) = \gcd(p, q) = 1$。求 $m + n + p + q$。

(A) 42 42

(B) 86 86

(D) 114 114

(E) 130 130

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $D$ and $Q$ be the projections of points $A$ and $O$ onto line $BC$, respectively, and let $s=AB$. It follows that $AD=\frac{s\sqrt3}{2}$. Because $\frac{BP}{CP}=3$ and $D$ is the midpoint of $BC$, it follows that $P$ is the midpoint of $CD$ and $DP=\frac{s}{4}$. It then follows from the Pythagorean Theorem that $$ AP=\sqrt{AD^2+DP^2}=\frac{s\sqrt{13}}{4}. $$ Note that $\triangle QOP\sim\triangle DAP$; therefore $\frac{OP}{OQ}=\frac{AP}{AD}$, so $$ OP=\frac{\frac{s\sqrt{13}}{4}}{\frac{s\sqrt3}{2}}\cdot 17=\frac{17\sqrt{13}}{2\sqrt3}. $$ Then $$ AP+OP=\frac{s\sqrt{13}}{4}+\frac{17\sqrt{13}}{2\sqrt3}=20, $$ which yields $$ s=\frac{80}{\sqrt{13}}-\frac{34}{\sqrt3}. $$ Therefore the requested answer is $80+13+34+3=130$.

答案（E）：设 $D$ 和 $Q$ 分别为点 $A$ 和点 $O$ 在直线 $BC$ 上的垂足，并令 $s=AB$。可得 $AD=\frac{s\sqrt3}{2}$。因为 $\frac{BP}{CP}=3$ 且 $D$ 是 $BC$ 的中点，所以 $P$ 是 $CD$ 的中点，且 $DP=\frac{s}{4}$。由勾股定理可得 $$ AP=\sqrt{AD^2+DP^2}=\frac{s\sqrt{13}}{4}. $$ 注意到 $\triangle QOP\sim\triangle DAP$；因此 $\frac{OP}{OQ}=\frac{AP}{AD}$，所以 $$ OP=\frac{\frac{s\sqrt{13}}{4}}{\frac{s\sqrt3}{2}}\cdot 17=\frac{17\sqrt{13}}{2\sqrt3}. $$ 于是 $$ AP+OP=\frac{s\sqrt{13}}{4}+\frac{17\sqrt{13}}{2\sqrt3}=20, $$ 从而得到 $$ s=\frac{80}{\sqrt{13}}-\frac{34}{\sqrt3}. $$ 因此所求答案为 $80+13+34+3=130$。

Topics