2022 AMC 10A Problem 5

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

"Equilateral" supplies equations: each side of the hexagon that is not along the square is the hypotenuse of a corner triangle, and the corner triangle is isosceles right because the two cut-off lengths are both $1 - s$. The symmetry across diagonal $AC$ means one corner determines everything.

Check numerically: $2 - \sqrt{2} \approx 0.586$, so $1 - s \approx 0.414$ and $(1 - s)\sqrt{2} \approx 0.586$, matching $s$. Choice (B) fails because $s = \frac12$ would give $PQ = \frac{\sqrt{2}}{2} \ne \frac12$.

• Mis-simplifying sqrt(2)/(1 + sqrt(2)); multiply top and bottom by sqrt(2) - 1 to get 2 - sqrt(2).
• Assuming $P$ and $Q$ are midpoints (giving $s = \frac12$) without checking that $PQ$ would then have a different length.