# 2015 AMC 10B Problems/Problem 22

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Solution

Triangle $AFG$ is isosceles so $AG$= $AF$= $1$. Using the symmetry of pentagon $FGHIJ$, notice that $\angle{FAG}$ is congruent to $\angle{FBI}$, so triangles $JHG$ and $DIJ$ are congruent to our original triangle $AFG$. Therefore, $JH$ = $1$. Now, we still need to find the length of $FG$ and $DC$. Also, we know that $FJ$ = $FG$ since pentagon $FGHIJ$ is regular. Let's call the length of $FG$ and $FJ$ $x$. Now we can solve for $x$. Triangles $AFG$ and $AJH$ are similar. So, $$\frac{1}{x+1} = \frac{x}{1}$$

From this, we get $x = \frac{\sqrt{5} -1}{2}$.

Now, we just have to find the length of $DC$ which we'll call $y$. We already know that $DJ$ = $AF$ = $1$. Triangles $AFG$ and $ADC$ are similar so we have, $$\frac{1}{2+x} = \frac{x}{y}$$

Solving for $y$ we get $y = \frac{\sqrt{5} +1}{2}$

Adding up $FG$ , $JH$, and $CD$ gives us $1 + x + y$ which is $\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

Solution by arebei2