2008 AMC 10A Problem 25

Let $O$ be the center of the table. Focus on one mat, with outer corners $E$ and $F$ on the table's edge and inner corners $G$ and $H$. So $EF = GH = x$ and $EG = FH = 1$.

1. The inner corners. Each inner corner touches an inner corner of the next mat, so there are $6$ meeting points around the center, evenly spaced. They form a regular hexagon, and $GH$ is one of its sides.
2. A regular hexagon splits into $6$ equilateral triangles meeting at the center. So triangle $OGH$ is equilateral with side $x$.
3. Distance from $O$ to the inner edge. The height of an equilateral triangle with side $x$ is $\frac{\sqrt3}{2}x$. So the inner edge $GH$ is $\frac{\sqrt3}{2}x$ from $O$.
4. Distance from $O$ to the outer edge. The mat is $1$ unit wide, and the outer edge $EF$ is parallel to $GH$ and farther out. So $EF$ is $\frac{\sqrt3}{2}x + 1$ from $O$.
5. One right triangle. Let $M$ be the midpoint of $EF$. Then $OM = \frac{\sqrt3}{2}x + 1$, $ME = \frac{x}{2}$, and $OE = 4$ because $E$ is on the table's edge. The angle at $M$ is a right angle, so by the Pythagorean theorem
$$\left(\frac{\sqrt3}{2}x + 1\right)^2 + \left(\frac{x}{2}\right)^2 = 4^2.$$
6. Solve. Multiply out: $\frac34 x^2 + \sqrt3\,x + 1 + \frac14 x^2 = 16$, which simplifies to
$$x^2 + \sqrt3\,x - 15 = 0.$$
The quadratic formula gives the positive solution
$$x = \frac{-\sqrt3 + \sqrt{3 + 60}}{2} = \frac{-\sqrt3 + 3\sqrt7}{2}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}}$.

When identical pieces are arranged evenly around a center, look for the regular polygon formed by the points where they touch. Here it is a hexagon, and a regular hexagon's side equals its distance from center to corner. That tells you how far one mat sits from the center, and then a single right triangle finishes the job.

Check the answer with estimates. $\sqrt7 \approx 2.646$ and $\sqrt3 \approx 1.732$, so choice (C) is about $\frac{7.94 - 1.73}{2} \approx 3.10$. Put $x = 3.10$ into step 5: $OM \approx 0.866 \times 3.10 + 1 \approx 3.68$ and $ME \approx 1.55$, and $3.68^2 + 1.55^2 \approx 13.5 + 2.4 = 15.9$, very close to $16$. Choice (B), $x = 3$, gives $3.60^2 + 1.5^2 \approx 15.2$, too small, so (C) is the one that fits.