AMC10 2002 A

AMC10 2002 A · Q23

AMC10 2002 A · Q23. It mainly tests Triangles (properties), Pythagorean theorem.

Points A, B, C, and D lie on a line, in that order, with AB = CD and BC = 12. Point E is not on the line, and BE = CE = 10. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find AB.

点 A、B、C、D 在一条直线上，按此顺序，AB = CD，BC = 12。点 E 不在直线上，且 BE = CE = 10。$ riangle AED$ 的周长是 $\triangle BEC$ 周长的两倍。求 AB。

(A) 15/2 15/2

(B) 8 8

(D) 9 9

(E) 19/2 19/2

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Let $H$ be the midpoint of $\overline{BC}$. Then $\overline{EH}$ is the perpendicular bisector of $\overline{AD}$, and $\triangle AED$ is isosceles. Segment $\overline{EH}$ is the common altitude of the two isosceles triangles $\triangle AED$ and $\triangle BEC$, and $$EH=\sqrt{10^2-6^2}=8.$$ Let $AB=CD=x$ and $AE=ED=y$. Then $2x+2y+12=2(32)$, so $y=26-x$. Thus, $$8^2+(x+6)^2=y^2=(26-x)^2 \text{ and } x=9.$$

（D）设 $H$ 为 $\overline{BC}$ 的中点，则 $\overline{EH}$ 是 $\overline{AD}$ 的垂直平分线，且 $\triangle AED$ 为等腰三角形。线段 $\overline{EH}$ 是两个等腰三角形 $\triangle AED$ 与 $\triangle BEC$ 的公共高，因此 $$EH=\sqrt{10^2-6^2}=8.$$ 令 $AB=CD=x$ 且 $AE=ED=y$。则 $2x+2y+12=2(32)$，所以 $y=26-x$。因此， $$8^2+(x+6)^2=y^2=(26-x)^2 \text{，且 } x=9.$$

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