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 endpoints 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:

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.

2008 AMC 12A Problem 22

Solution:

Problem 22 in Other Years