2008 AMC 10A Problem 25

Pick a mat with outer corners $P$ and $Q,$ and let $R$ be the point on the circle diametrically opposite $P.$ Then $\triangle PQR$ is right-angled at $Q$ with hypotenuse $PR = 8.$ The inner corners of adjacent mats meet in isosceles triangles with vertex angle $120^\circ$ and sides $x,$ whose base is $\sqrt{3}\,x.$ Together with the two mat widths, $QR = \sqrt{3}\,x + 2.$ By the Pythagorean theorem, $$\left(\sqrt{3}\,x + 2\right)^2 + x^2 = 64,$$ which simplifies to $x^2 + \sqrt{3}\,x - 15 = 0.$ Taking the positive root, $$x = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$$ Thus, the correct answer is C.