AMC12 2021 B

AMC12 2021 B · Q11

AMC12 2021 B · Q11. It mainly tests Pythagorean theorem, Coordinate geometry.

Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$. Let $P$ be the point on $\overline{AC}$ such that $PC=10$. There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

三角形 $ABC$ 有 $AB=13,BC=14$ 和 $AC=15$。让 $P$ 为 $\overline{AC}$ 上的点，使得 $PC=10$。线 $BP$ 上恰有两点 $D$ 和 $E$，使得四边形 $ABCD$ 和 $ABCE$ 是梯形。$DE$ 的距离是多少？

(A) \frac{42}5 \frac{42}5

(B) 6\sqrt2 6\sqrt2

(D) 12\sqrt2 12\sqrt2

(E) 18 18

Answer

Correct choice: (D)

正确答案：（D）

Solution

Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$. Then $\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$ by the trapezoid condition, where $D, E\in\overline{BP}$. Since $PC=10$, point $P$ is $\tfrac{10}{15}=\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$. Thus line $BP$ has equation $y=x$. Since $\overline{AD}\parallel\overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$-coordinate as $A$, except it'll also lie on the line $y=x$. Therefore, $D=(12, 12). \, \blacksquare$ To find the location of point $E$, we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\overline{AB}$, or $\tfrac{12}{5}$, and has equation $y=\tfrac{12}{5}x-\tfrac{168}{5}$. The intersection of this line with $y=x$ is $(24, 24)$. Therefore point $E$ is located at $(24, 24). \, \blacksquare$ The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$, which is $\boxed{\textbf{(D) }12\sqrt2}$.

将三角形置于坐标平面上，令 $A=(5, 12), B=(0, 0),$ 和 $C=(14, 0)$。由梯形条件，$\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$，其中 $D, E\in\overline{BP}$。由于 $PC=10$，点 $P$ 从 $C$ 到 $A$ 的 $\tfrac{10}{15}=\tfrac{2}{3}$ 处，位于 $(8, 8)$。因此线 $BP$ 的方程为 $y=x$。由于 $\overline{AD}\parallel\overline{BC}$ 且 $\overline{BC}$ 平行于地面，$D$ 与 $A$ 有相同 $y$ 坐标，且位于直线 $y=x$ 上。因此，$D=(12, 12)。\, \blacksquare$ 要找到点 $E$ 的位置，需要找到 $y=x$ 与通过 $C$ 且平行于 $\overline{AB}$ 的直线的交点。该直线的斜率与 $\overline{AB}$ 相同，为 $\tfrac{12}{5}$，方程为 $y=\tfrac{12}{5}x-\tfrac{168}{5}$。该直线与 $y=x$ 的交点为 $(24, 24)$。因此点 $E$ 位于 $(24, 24)。\, \blacksquare$ 距离 $DE$ 等于 $(12, 12)$ 和 $(24, 24)$ 之间的距离，为 $\boxed{\textbf{(D) }12\sqrt2}$。

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