Question

A round table has radius $4.$ Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x.$ Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x?$

$2\sqrt{5} - \sqrt{3}$

$3$

$\dfrac{3\sqrt{7} - \sqrt{3}}{2}$

$2\sqrt{3}$

$\dfrac{5 + 2\sqrt{3}}{2}$

Solution:

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.

2008 AMC 10A Problem 25

Solution:

Problem 25 in Other Years