2015 AMC 10B Problem 22

Step 1: the angles. Each corner of a regular pentagon is $108^\circ$. The two lines from a corner to the far corners split that $108^\circ$ into three equal angles of $36^\circ$. So every triangle in the picture has angles made from $36^\circ$, $72^\circ$ and $108^\circ$.

Step 2: the one number we need. Let $\varphi = \dfrac{1 + \sqrt5}{2} \approx 1.618$, the golden ratio. Two facts about it:
- $\varphi^2 = \varphi + 1$ and $\dfrac{1}{\varphi} = \varphi - 1$.
- In a triangle with angles $36^\circ$, $72^\circ$, $72^\circ$, each long side is $\varphi$ times the short side. (If you cut one $72^\circ$ angle in half, you get a smaller triangle with the same angles, and matching up the sides gives exactly this ratio.)

Step 3: $GB$ and $HC$. In triangle $ABG$, the angles at $A$ and at $B$ are both $36^\circ$. Two equal angles mean two equal sides, so $GB = GA = 1$. The pentagon looks the same when flipped left-to-right across the line through $B$ and the middle of the opposite side; that flip swaps $G$ and $H$ and swaps $A$ and $C$. So $HC = GA = 1$.

Step 4: $FG$. In triangle $AFG$, the angle at $A$ is $36^\circ$ and $AF = AG = 1$, so the other two angles are $72^\circ$ each. The long sides are $1$, so the short side is
$$FG = \frac{1}{\varphi}.$$
The small pentagon $FGHIJ$ in the middle is also regular, so all its sides are $\frac1\varphi$, including $GH$.

Step 5: $CD$. Along the diagonal $AC$:
$$AC = AG + GH + HC = 1 + \frac{1}{\varphi} + 1 = 2 + (\varphi - 1) = \varphi + 1 = \varphi^2.$$
Triangle $ACD$ has angles $36^\circ$, $72^\circ$, $72^\circ$, with long sides $AC = AD = \varphi^2$. So its short side is
$$CD = \frac{\varphi^2}{\varphi} = \varphi.$$
This matches the pattern in the big pentagon: a diagonal ($\varphi^2$) is $\varphi$ times a side ($\varphi$).

Step 6: $JH$. In the small pentagon, $J$ and $H$ are not next to each other (the corner $I$ is between them), so $JH$ is a diagonal of the small pentagon. By the same pattern, a diagonal is $\varphi$ times a side:
$$JH = \varphi \cdot \frac{1}{\varphi} = 1.$$

Step 7: add.
$$FG + JH + CD = (\varphi - 1) + 1 + \varphi = 2\varphi = 1 + \sqrt5.$$

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

A regular pentagon with all its diagonals is built from triangles with angles $36^\circ$, $72^\circ$ and $108^\circ$, and in all of them the long side is $\varphi$ times the short side. So once one length is known, every length is that number times a power of $\varphi$. The rule $\varphi^2 = \varphi + 1$ turns messy square-root arithmetic into simple adding. Worth remembering: in a regular pentagon, a diagonal is $\varphi$ times a side.

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