AMC10 2015 B

AMC10 2015 B · Q22

AMC10 2015 B · Q22. It mainly tests Quadratic equations, Triangles (properties).

In the figure shown below, $ABCDE$ is a regular pentagon and $AG = 1$. What is $FG + JH + CD$?

如图所示，$ABCDE$是一个正五边形，且$AG=1$。求$FG+JH+CD$的值。

(A) 3 3

(B) $12 - 4\sqrt{5}$ $12 - 4\sqrt{5}$

(D) $1 + \sqrt{5}$ $1 + \sqrt{5}$

(E) $\frac{11 + 11\sqrt{5}}{10}$ $\frac{11 + 11\sqrt{5}}{10}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Triangles $AGB$ and $CHJ$ are isosceles and congruent, so $AG = HC = HJ = 1$. Triangles $AFG$ and $BGH$ are congruent, so $FG = GH$. Triangles $AGF$, $AHJ$, and $ACD$ are similar, so $\frac{a}{b} = \frac{a+b}{c} = \frac{2a+b}{d}$. Because $a = c = 1$, the first equation becomes $\frac{1}{b} = \frac{1+b}{1}$ or $b^2 + b - 1 = 0$, so $b = \frac{-1+\sqrt{5}}{2}$. Substituting this in the second equation gives $d = \frac{1+\sqrt{5}}{2}$, so $b + c + d = 1 + \sqrt{5}$.

答案（D）：三角形 $AGB$ 和 $CHJ$ 是等腰且全等的，所以 $AG = HC = HJ = 1$。三角形 $AFG$ 和 $BGH$ 全等，所以 $FG = GH$。三角形 $AGF$、$AHJ$ 和 $ACD$ 相似，所以 $\frac{a}{b} = \frac{a+b}{c} = \frac{2a+b}{d}$。因为 $a = c = 1$，第一个等式变为 $\frac{1}{b} = \frac{1+b}{1}$，即 $b^2 + b - 1 = 0$，所以 $b = \frac{-1+\sqrt{5}}{2}$。将其代入第二个等式得到 $d = \frac{1+\sqrt{5}}{2}$，因此 $b + c + d = 1 + \sqrt{5}$。

Topics