2008 AMC 12A Problem 22

Take one mat with outer corners $P$ and $Q,$ and let $R$ be the point of the table’s edge diametrically opposite $P.$ Then $PR = 8$ is a diameter, so $\triangle PQR$ has a right angle at $Q,$ with $PQ = x.$ Along $QR,$ the inner corners of neighboring mats meet in an isosceles triangle with two sides of length $x$ and vertex angle $120^\circ,$ whose base is $\sqrt{3}\,x.$ Hence $QR = \sqrt{3}\,x + 2.$ The Pythagorean Theorem gives $$x^2 + \left(\sqrt{3}\,x + 2\right)^2 = 64,$$ which simplifies to $x^2 + \sqrt{3}\,x - 15 = 0.$ Taking the positive root, $$x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$$ Thus, C is the correct answer.