2008 AMC 12A Problem 22

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}

For one mat, let

P, Q

be its outer corners and

R

the point diametrically opposite

P;

then

\triangle PQR

is right-angled with

PR = 8

The inner corners form isosceles triangles with vertex angle

120^\circ,

contributing a segment of length

\sqrt{3}\,x

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.

Problem 22 in other years