2023 AMC 10A Problem 24

Plan. The six blocks always cover the same area, so we only need the side length of the frame. Then
$$\text{empty area} = (\text{frame area}) - 6 \times (\text{block area}).$$
A regular hexagon with side $s$ is made of $6$ equilateral triangles, so its area is $6 \times \frac{\sqrt3}{4}s^2 = \frac{3\sqrt3}{2}s^2$.

Step 1: What one frame edge looks like. Look at the figure. Each frame edge has one block lying along it. Between the block and one end of the edge there is a short gap, and between the block and the other end there is a long gap. The problem says the short gap is $\frac37$. Call the long gap $g$. Then
$$\text{frame side} = \tfrac37 + 1 + g.$$
At every corner of the frame, one block is $\frac37$ away and the neighboring block is $g$ away.

Step 2: Use the fact that neighbors touch. Pick a corner $V$ of the frame. Call $A$ the block that is $\frac37$ from $V$ and $B$ the block that is $g$ from $V$ (on the other edge at $V$). In the figure, $A$ and $B$ touch along a shared side. Call the straight line through that shared side $\ell$.

1. Extend $\ell$ past $B$ until it hits the frame edge that $B$ sits on, at a point $X$. The corner of $B$ nearest $V$, the next corner of $B$, and $X$ form a small triangle with two $60^\circ$ angles (each hexagon angle is $120^\circ$, so the angle next to it on a straight line is $60^\circ$). So it is an equilateral triangle with side $1$, and $X$ is $1$ unit closer to $V$ than $B$ is. That gives
$$VX = g - 1.$$
2. Now extend $\ell$ past $A$ the other way, and extend $A$'s frame edge past $V$ beyond the corner. They meet at a point $Z$ outside the frame. The corner of $A$ nearest $V$, the next corner of $A$ (the one where $A$'s side on $\ell$ begins), and $Z$ again form an equilateral triangle with side $1$. So $Z$ is $1$ unit from $A$'s corner, which is $\frac37$ from $V$. That makes
$$ZV = 1 - \tfrac37 = \tfrac47.$$
3. Now look at triangle $ZVX$. Its angle at $V$ is $180^\circ - 120^\circ = 60^\circ$ (the frame corner is $120^\circ$), and its angle at $Z$ is $60^\circ$ (from the equilateral triangle in step 2). So triangle $ZVX$ is equilateral too, and
$$VX = ZV = \tfrac47.$$
4. Put steps 1 and 3 together: $g - 1 = \frac47$, so $g = \frac{11}{7}$.

Step 3: The frame side.
$$\text{frame side} = \tfrac37 + 1 + \tfrac{11}{7} = \tfrac{3 + 7 + 11}{7} = 3.$$

Step 4: The area.
$$\text{frame area} = \tfrac{3\sqrt3}{2} \times 9 = \tfrac{27\sqrt3}{2}, \qquad \text{blocks} = 6 \times \tfrac{3\sqrt3}{2} = 9\sqrt3 = \tfrac{18\sqrt3}{2}.$$
$$\text{empty area} = \tfrac{27\sqrt3}{2} - \tfrac{18\sqrt3}{2} = \tfrac{9\sqrt3}{2}.$$

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

Every angle in this picture is $60^\circ$ or $120^\circ$, so extending sides keeps producing equilateral triangles, and those turn angle facts into length facts. The real rule that fixes the frame is "neighboring blocks share a side line." In fact the two gaps always add up to $2$, so the frame side is $3$ no matter what the short gap is. On hard problems, check which given numbers really matter.

Use coordinates. Put $V$ at the origin with one frame edge along the positive $x$-axis and the other pointing up-left at $120^\circ$. Let block $A$ sit on the $x$-axis at distance $p$ from $V$ and block $B$ on the other edge at distance $q$ from $V$. Working out the corners of the two hexagons, $A$'s upper-left side lies on the line $y = \sqrt3\,(x - p + 1)$ and $B$'s matching side lies on $y = \sqrt3\,(x + q - 1)$. The two blocks share this line, so $-p + 1 = q - 1$, which means $p + q = 2$ and the frame side is $p + 1 + q = 3$.