AMC12 2018 A

AMC12 2018 A · Q11

AMC12 2018 A · Q11. It mainly tests Triangles (properties), Circle theorems.

A paper triangle with sides of lengths 3, 4, and 5 inches, as shown, is folded so that point A falls on point B. What is the length in inches of the crease?

一个纸质三角形，边长分别为3、4和5英寸，如图所示，将其折叠使得点A落在点B上。折痕的长度是多少英寸？

(A) $1 + \frac{1}{2}\sqrt{2}$ $1 + \frac{1}{2}\sqrt{2}$

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

(D) $\frac{15}{8}$ $\frac{15}{8}$

(E) 2 2

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The paper’s long edge $\overline{AB}$ is the hypotenuse of right triangle $ACB$, and the crease lies along the perpendicular bisector of $\overline{AB}$. Because $AC>BC$, the crease hits $\overline{AC}$ rather than $\overline{BC}$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the intersection of $\overline{AC}$ and the line through $D$ perpendicular to $\overline{AB}$. Then the crease in the paper is $\overline{DE}$. Because $\triangle ADE \sim \triangle ACB$, it follows that $\dfrac{DE}{AD}=\dfrac{CB}{AC}=\dfrac{3}{4}$. Thus $$ DE=AD\cdot \frac{CB}{AC}=\frac{5}{2}\cdot \frac{3}{4}=\frac{15}{8}. $$

答案（D）：纸张的长边 $\overline{AB}$ 是直角三角形 $ACB$ 的斜边，折痕位于 $\overline{AB}$ 的垂直平分线上。由于 $AC>BC$，折痕会与 $\overline{AC}$ 相交而不是与 $\overline{BC}$ 相交。设 $D$ 为 $\overline{AB}$ 的中点，设 $E$ 为 $\overline{AC}$ 与过 $D$ 且垂直于 $\overline{AB}$ 的直线的交点，则纸上的折痕为 $\overline{DE}$。因为 $\triangle ADE \sim \triangle ACB$，所以 $\dfrac{DE}{AD}=\dfrac{CB}{AC}=\dfrac{3}{4}$。因此 $$ DE=AD\cdot \frac{CB}{AC}=\frac{5}{2}\cdot \frac{3}{4}=\frac{15}{8}. $$

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