AMC12 2024 A

AMC12 2024 A · Q20

AMC12 2024 A · Q20. It mainly tests Probability (basic), Counting & probability misc.

Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline {AB}$ and $\overline{AC},$ respectively, of equilateral triangle $\triangle ABC.$ Which of the following intervals contains the probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC?$

在等边三角形 $\triangle ABC$ 的边 $\overline{AB}$ 和 $\overline{AC}$ 上，分别均匀独立随机选择点 $P$ 和 $Q$。以下哪个区间包含 $\triangle APQ$ 的面积小于 $\triangle ABC$ 面积一半的概率？

(A) \left[\frac 38, \frac 12\right] \left[\frac 38, \frac 12\right]

(B) \left(\frac 12, \frac 23\right] \left(\frac 12, \frac 23\right]

(D) \left(\frac 34, \frac 78\right] \left(\frac 34, \frac 78\right]

(E) \left(\frac 78, 1\right] \left(\frac 78, 1\right]

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $\overline{AP}=x$ and $\overline{AQ}=y$. Applying the sine formula for a triangle's area, we see that \[[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\] Without loss of generality, we let $AB=BC=CA=1$, and thus $[\Delta ABC]=\dfrac{\sqrt3}4$; we therefore require $\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12$ for $0\le x,y\le1$. (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.) A quick rough sketch of $y=\dfrac1{2x}$ on the square given by $x,y\in[0,1]$ reveals that the curve intersects the boundaries at $(0.5,1)$ and $(1,0.5)$, and it is actually quite (very) obvious that the area bounded by the inequality $xy\le0.5$ and the aforementioned unit square is more than $\dfrac34$ but less than $\dfrac78$ (cf. the diagram below). Thus, our answer is $\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$.

设 $\overline{AP}=x$ 和 $\overline{AQ}=y$。应用三角形面积的正弦公式，我们有 \[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\] 不失一般性，令 $AB=BC=CA=1$，则 $[\Delta ABC]=\dfrac{\sqrt3}4$；因此要求 $\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12$，其中 $0\le x,y\le1$。（注：可以使用三角形面积比率跳过大部分内容 ~A_MatheMagician。）在由 $x,y\in[0,1]$ 给定的正方形上快速粗略绘制 $y=\dfrac1{2x}$，曲线在 $(0.5,1)$ 和 $(1,0.5)$ 与边界相交，很明显不等式 $xy\le0.5$ 与上述单位正方形围成的面积大于 $\dfrac34$ 但小于 $\dfrac78$（参见下图）。因此答案为 $\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$。

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