2015 AMC 10B Problem 22

Each interior angle is $108^\circ$ and each diagonal makes a $36^\circ$ angle with the adjacent sides, so every triangle in the figure has angles from $\{36^\circ, 72^\circ, 108^\circ\}$. Let $\varphi = \frac{1+\sqrt5}{2}$, so $\varphi^2 = \varphi + 1$ and $\frac{1}{\varphi} = \varphi - 1$. Standard fact: in a $36$-$72$-$72$ triangle, base $= \text{leg}/\varphi$ (bisect a base angle to get a similar triangle: $x/1 = (1-x)/x$).

$GB$ and $HC$. In $\triangle ABG$, $\angle GAB = \angle GBA = 36^\circ$, so $GB = GA = 1$. The reflection through $B$'s axis of symmetry swaps $A \leftrightarrow C$ and $E \leftrightarrow D$, hence $G \leftrightarrow H$, so $HC = 1$ as well.

$FG$. In $\triangle AFG$, $\angle FAG = 36^\circ$ and $\angle AGF = 180^\circ - 108^\circ = 72^\circ$, so it is $36$-$72$-$72$ with legs $AF = AG = 1$: $FG = \frac{1}{\varphi}$. By symmetry the inner pentagon $FGHIJ$ is regular with side $\frac{1}{\varphi}$, so $GH = \frac1\varphi$ too.

$CD$. $AC = AG + GH + HC = 2 + \frac1\varphi = 1 + \varphi = \varphi^2$. Triangle $ACD$ is $36$-$72$-$72$ with legs $AC = AD = \varphi^2$, so $CD = \varphi^2/\varphi = \varphi$.

$JH$. $J$ and $H$ are non-adjacent vertices of the inner pentagon, so $JH$ is its diagonal: $JH = \varphi \cdot \frac1\varphi = 1$.

Therefore
$$FG + JH + CD = (\varphi - 1) + 1 + \varphi = 2\varphi = 1 + \sqrt5.$$

The answer is $\boxed{\textbf{(D)}\ 1+\sqrt5}$.

A regular pentagon with its diagonals is built entirely from golden triangles, so every segment is the unit times a power of $\varphi$; once one segment is known, the whole figure is known. The identity $\varphi^2 = \varphi + 1$ (and $1/\varphi = \varphi - 1$) turns what looks like radical arithmetic into bookkeeping. Memorize: side $\cdot \varphi =$ diagonal, and the diagonals cut each other into pieces of length side$/\varphi$ and side$/\varphi^2$.

• Assuming G is the midpoint of AC, or misreading which points F, G, H, J are, e.g. treating JH as a side of the inner pentagon rather than a diagonal.
• Using the wrong ratio for the $36$-$72$-$72$ triangle (base $= \varphi \cdot$ leg instead of leg$/\varphi$), which scales every length the wrong way.