2015 AMC 10B Problems/Problem 22

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

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2015 AMC 10B Problems/Problem 22