2026 AMC 10A Problem 17

Everything in the picture is arranged evenly around the center of the big hexagon, so measure distances from that center. Call the center $O$.

1. How far $O$ is from each side. The big hexagon is made of $6$ equilateral triangles with side $1$ that meet at $O$. The distance from $O$ to a side is the height of one of these triangles. Half of an equilateral triangle with side $1$ is a $30$-$60$-$90$ triangle with hypotenuse $1$ and short leg $\frac12$, so the height is
$$\sqrt{1 - \tfrac14} = \frac{\sqrt3}{2} \approx 0.866.$$

2. How far each square reaches. Turn the picture so that one side of the big hexagon is at the bottom. The square on that side goes straight up a distance $1$. The center $O$ is only $\frac{\sqrt3}{2}$ above the bottom side, so the top edge of this square is above $O$ by
$$h = 1 - \frac{\sqrt3}{2} \approx 0.134.$$

1. Pair it with the opposite square. The top side of the big hexagon also has a square, hanging straight down, and by the same reasoning its bottom edge is a distance $h$ below $O$. A point inside both squares must be below the first square's top edge and above the second square's bottom edge. So it lies in a narrow band between two parallel lines, each a distance $h$ from $O$. (In the figure, these are the two horizontal lines running along the top and the bottom of the shaded region.)

1. All three pairs. The hexagon has three pairs of opposite sides, each pair turned $60^\circ$ from the last. Each pair gives a band like the one in Step 3. The region inside all three bands is a regular hexagon centered at $O$, and the distance from $O$ to each of its sides is $h$. (This center-to-side distance of a regular polygon is called its apothem.) The other two sides of each square are $\frac12$ away from $O$ on either side, far outside this tiny region, so they do not cut into it. This small hexagon is the shaded one.

5. Area of a regular hexagon from its apothem. Split the small hexagon into $6$ equilateral triangles meeting at $O$. Each has height $h$. In an equilateral triangle the height is $\frac{\sqrt3}{2}$ times the side, so each side is $\frac{2h}{\sqrt3}$. One triangle has area
$$\frac12 \cdot \frac{2h}{\sqrt3} \cdot h = \frac{h^2}{\sqrt3},$$
and six of them give
$$\text{area} = \frac{6h^2}{\sqrt3} = 2\sqrt3\, h^2.$$

6. Compute $h^2$. Square $1 - \frac{\sqrt3}{2}$ like any difference:
$$h^2 = 1 - 2 \cdot \frac{\sqrt3}{2} + \frac{3}{4} = \frac{7}{4} - \sqrt3.$$

7. The area.
$$2\sqrt3 \left(\frac74 - \sqrt3\right) = \frac{7\sqrt3}{2} - 6.$$
As a check, this is about $6.062 - 6 = 0.062$, a tiny area, just like the small shaded region in the picture.

8. Read off the numbers. $a = 7$, $b = 3$, $c = 2$, $d = 6$. Here $7$ and $2$ are relatively prime and $3$ is not divisible by any square, so the form is correct, and
$$a + b + c + d = 7 + 3 + 2 + 6 = 18.$$

The answer is $\boxed{\textbf{(C)}\ 18}$.

When a figure has a center and everything is spread evenly around it, measure from the center. Each square only matters through one number, how far its far edge reaches past the center, and that single distance describes the whole shaded hexagon. The region inside several strips between parallel lines is a polygon, and when the strips are evenly turned, the polygon is regular. Finally, any regular polygon splits into identical triangles at its center, which gives its area from either the side or the apothem.

Find the side length of the small hexagon first and use the usual formula $\frac{3\sqrt3}{2} s^2$ for a regular hexagon with side $s$.

From Steps 1 to 4, the small hexagon's apothem is $h = 1 - \frac{\sqrt3}{2}$, and the apothem of a regular hexagon is $\frac{\sqrt3}{2}$ times its side. So
$$s = \frac{2h}{\sqrt3} = \frac{2}{\sqrt3} - 1 = \frac{2\sqrt3}{3} - 1, \qquad s^2 = \frac{4}{3} - \frac{4\sqrt3}{3} + 1 = \frac{7 - 4\sqrt3}{3}.$$
Then
$$\frac{3\sqrt3}{2} \cdot \frac{7 - 4\sqrt3}{3} = \frac{\sqrt3\,(7 - 4\sqrt3)}{2} = \frac{7\sqrt3}{2} - 6,$$
the same area, so $a + b + c + d = 18$.