2022 AMC 10A Problem 5

1. Where are the points? Going around the hexagon we meet $A, P, Q, C, R, S$. Between $A$ and $C$ the hexagon passes $P$ and $Q$, so it cuts off corner $B$: $P$ is on side $AB$ and $Q$ is on side $BC$. In the same way $R$ is on $CD$ and $S$ is on $DA$.
2. Lengths along the square. $AP$ is a side of the hexagon, so $AP = s$. The whole side $AB$ is $1$, so $PB = 1 - s$. In the same way $QC = s$, so $BQ = 1 - s$.
3. The corner triangle. Triangle $PBQ$ has a square corner ($90^\circ$) at $B$, and both legs are $1 - s$. So it is a 45-45-90 triangle, and its slanted side is
$$PQ = (1 - s)\sqrt{2}.$$
4. Use "equilateral". $PQ$ is also a side of the hexagon, so $PQ = s$:
$$(1 - s)\sqrt{2} = s.$$
5. Solve. Multiply out: $\sqrt2 - s\sqrt2 = s$. Move the $s$ terms together: $\sqrt2 = s + s\sqrt2 = s(1 + \sqrt2)$. So
$$s = \frac{\sqrt2}{1 + \sqrt2}.$$
6. Tidy the fraction. Multiply top and bottom by $\sqrt2 - 1$. The bottom becomes $(\sqrt2 + 1)(\sqrt2 - 1) = 2 - 1 = 1$, so
$$s = \sqrt2(\sqrt2 - 1) = 2 - \sqrt2.$$
7. The corner at $D$ is the same picture flipped over the diagonal $AC$, so the same $s$ works there too.

Check with decimals: $s \approx 0.586$, so $1 - s \approx 0.414$, and $0.414 \times 1.414 \approx 0.586$. It matches.

The answer is $\boxed{\textbf{(C)}\ 2 - \sqrt{2}}$.

"Equilateral" hands you an equation: every side equals $s$. The sides lying along the square are easy to write down, and each slanted side cuts off a corner triangle. Here that triangle has two equal legs, so the 45-45-90 rule gives the slanted side, and setting it equal to $s$ finishes the problem.

Test the choices. For (B), $s = \frac12$: the legs are $\frac12$ and the slanted side is $\frac{\sqrt2}{2} \approx 0.71$, which is not $\frac12$. For (E), $s = \frac23$: the legs are $\frac13$ and the slanted side is about $0.47$, not $\frac23$. For (C) the decimal check above shows the slanted side equals $s$.

The hexagon runs $A \to P \to Q \to C \to R \to S \to A$, so $P$ and $Q$ lie on the path from $A$ to $C$ through $B$: $P$ on $\overline{AB}$ and $Q$ on $\overline{BC}$. Likewise $R$ is on $\overline{CD}$ and $S$ on $\overline{DA}$.

Since $AP = s$, we get $PB = 1 - s$; since $QC = s$, we get $BQ = 1 - s$. Triangle $PBQ$ has a right angle at $B$ and equal legs $1 - s$, so its hypotenuse is
$$PQ = (1 - s)\sqrt{2}.$$
The hexagon is equilateral, so $PQ = s$:
$$(1 - s)\sqrt{2} = s \quad\Longrightarrow\quad s(1 + \sqrt{2}) = \sqrt{2} \quad\Longrightarrow\quad s = \frac{\sqrt{2}}{1 + \sqrt{2}} = \sqrt{2}(\sqrt{2} - 1) = 2 - \sqrt{2}.$$
The other half of the hexagon (around corner $D$) is the mirror image, so the same $s$ works there.

The answer is $\boxed{\textbf{(C)}\ 2 - \sqrt{2}}$.