2015 AMC 10B Problem 22

Question

In the figure shown below,

ABCDE

is a regular pentagon and

AG=1.

What is

FG + JH + CD?

3

12-4\sqrt{5}

\dfrac{5+2\sqrt{5}}{3}

1+\sqrt{5}

\dfrac{11+11\sqrt{5}}{10}

Use similar triangles with side lengths

AG=1

,

FG=b

, and

CD=d

The first similarity gives

b^2+b-1=0

By symmetry,

AG=HC=HJ=1

, and triangles

AFG

and

BGH

are congruent, so

FG=GH

. Let

FG=b

, and let

CD=d

.

The similar triangles in the pentagon give

\frac1b=1+b

and

\frac{1+b}{1}=d.

The first equation is

b^2+b-1=0

, so

b=\frac{-1+\sqrt{5}}{2}

. Then

d=1+b=\frac{1+\sqrt{5}}{2}

.

Therefore

\begin{aligned} FG+JH+CD &= b+1+d \\ &= 1+\sqrt{5}. \end{aligned}

Thus, the correct answer is D.

Problem 22 in other years