AMC12 2002 A

AMC12 2002 A · Q22

AMC12 2002 A · Q22. It mainly tests Probability (basic), Triangles (properties).

Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$ , and $AB = 10$. Let $P$ be randomly chosen inside $ABC$ , and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. What is the probability that $BD > 5\sqrt2$?

三角形 $ABC$ 是直角三角形，且 $\angle ACB$ 为直角，$m\angle ABC = 60^\circ$，并且 $AB = 10$。在 $ABC$ 内随机选取点 $P$，将 $\overline{BP}$ 延长与 $\overline{AC}$ 相交于 $D$。$BD > 5\sqrt2$ 的概率是多少？

(A) $\frac{2 - \sqrt{2}}{2}$ $\frac{2 - \sqrt{2}}{2}$

(B) $\frac{1}{3}$ $\frac{1}{3}$

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) $\frac{5 - \sqrt{5}}{5}$ $\frac{5 - \sqrt{5}}{5}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Clearly $BC=5$ and $AC=5\sqrt{3}$. Choose a $P'$ and get a corresponding $D'$ such that $BD'= 5\sqrt{2}$ and $CD'=5$. For $BD > 5\sqrt2$ we need $CD>5$, creating an isosceles right triangle with hypotenuse $5\sqrt {2}$ . Thus the point $P$ may only lie in the triangle $ABD'$. The probability of it doing so is the ratio of areas of $ABD'$ to $ABC$, or equivalently, the ratio of $AD'$ to $AC$ because the triangles have identical altitudes when taking $AD'$ and $AC$ as bases. This ratio is equal to $\frac{AC-CD'}{AC}=1-\frac{CD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}$. Thus the answer is $\boxed{C}$.

显然 $BC=5$ 且 $AC=5\sqrt{3}$。选取一点 $P'$ 并得到对应的 $D'$，使得 $BD'= 5\sqrt{2}$ 且 $CD'=5$。要使 $BD > 5\sqrt2$，需要 $CD>5$，从而形成一条斜边为 $5\sqrt {2}$ 的等腰直角三角形。因此点 $P$ 只能落在三角形 $ABD'$ 内。其概率为 $ABD'$ 与 $ABC$ 的面积之比；等价地，由于以 $AD'$ 和 $AC$ 为底时两三角形的高相同，该比值也等于 $AD'$ 与 $AC$ 的比值。这个比值为 $\frac{AC-CD'}{AC}=1-\frac{CD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}$。因此答案是 $\boxed{C}$。

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