2015 AMC 10B Problem 22

Attempt Problem 22 of the 2015 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10B solutions, or check the answer key.

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22.

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

3 3

1245 12-4\sqrt{5}

5+253 \dfrac{5+2\sqrt{5}}{3}

1+5 1+\sqrt{5}

11+11510 \dfrac{11+11\sqrt{5}}{10}

Answer: D
Concepts:regular polygonsimilarityquadratic
Difficulty rating: 1880
Solution:

By symmetry, AG=HC=HJ=1AG=HC=HJ=1, and triangles AFGAFG and BGHBGH are congruent, so FG=GHFG=GH. Let FG=bFG=b, and let CD=dCD=d.

The similar triangles in the pentagon give 1b=1+b\frac1b=1+b and 1+b1=d.\frac{1+b}{1}=d. The first equation is b2+b1=0b^2+b-1=0, so b=1+52b=\frac{-1+\sqrt{5}}{2}. Then d=1+b=1+52d=1+b=\frac{1+\sqrt{5}}{2}.

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

Thus, the correct answer is D.

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