2023 AMC 10A Problem 24

Only the frame's side length $s$ is needed: the answer is $\frac{3\sqrt3}{2}s^2 - 6 \cdot \frac{3\sqrt3}{2}$.

By rotational symmetry every frame edge looks the same: its block sits at distance $p$ from one endpoint and $q$ from the other, so $s = p + 1 + q$, and at each corner one adjacent block is $p$ away and the other is $q$ away. The given $\frac37$ is the smaller of $p, q$.

Place a corner $V$ at the origin with one frame edge along the positive $x$-axis and the other along direction $120^\circ$. The block on the $x$-axis has base from $(p, 0)$ to $(p+1, 0)$; its upper-left side runs from $\left(p - \tfrac12, \tfrac{\sqrt3}{2}\right)$ to $(p, \sqrt3)$, on the line $y = \sqrt3(x - p + 1)$.

The other block's nearest vertex is $W = \left(-\tfrac q2, \tfrac{q\sqrt3}{2}\right)$. From $W$ its next side runs $1$ unit parallel to the $x$-axis to $W' = \left(1 - \tfrac q2, \tfrac{q\sqrt3}{2}\right)$, and the following side climbs at $60^\circ$, along the line $y = \sqrt3(x + q - 1)$.

"Aligned" means these two slanted sides lie on a common line (the blocks touch along it), so $-p + 1 = q - 1$, i.e. $p + q = 2$. Hence $s = 3$ regardless of the value $\frac37$.

The empty area is
$$\frac{3\sqrt3}{2}\cdot 9 - 9\sqrt3 = \frac{27\sqrt3}{2} - \frac{18\sqrt3}{2} = \frac{9\sqrt3}{2}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{9 \sqrt{3}}{2}}$.

The blocks' area is fixed, so everything hinges on the frame's side, and the true constraint is the alignment of neighboring blocks, not the $\frac37$. Translating "sides are collinear" into equal line equations at a single corner uses the $120^\circ$ corner and the hexagon's $60^\circ$ slopes. Checking which given numbers actually matter is a valuable habit on hard problems.

With $p = \frac37$ and $q = \frac{11}{7}$ known, verify: the two aligned sides overlap along the common line for a length of $\frac37$, matching the picture where neighboring blocks touch. Numerically, $\frac{9\sqrt3}{2} \approx 7.79$; the decoys $\frac{216\sqrt3}{49} \approx 7.64$ and $\frac{243\sqrt3}{49} \approx 8.59$ come from misusing $\frac37$.

• Building the frame side from 3/7 alone (e.g. 1 + 6/7 or similar), which produces the decoy answers with 49 in the denominator.
• Using area $\frac{\sqrt3}{4}s^2$ (a triangle) or $\sqrt3 s^2$ for a regular hexagon instead of $\frac{3\sqrt3}{2}s^2$.