2013 AMC 12A Problem 12

Question

The angles in a particular triangle are in arithmetic progression, and the side lengths are

4,

5,

and

x.

The sum of the possible values of

x

equals

a + \sqrt{b} + \sqrt{c},

where

a,

b,

and

c

are positive integers. What is

a + b + c?

36

38

40

42

44

Angles in arithmetic progression summing to

180^\circ

force the middle angle to be

60^\circ

Apply the Law of Cosines in each case for which side is opposite the

60^\circ

angle

If the angles are

\alpha - \delta, \alpha, \alpha + \delta,

their sum

3\alpha = 180^\circ

gives

\alpha = 60^\circ,

so one angle is

60^\circ.

If

x

is opposite the

60^\circ

angle, the Law of Cosines gives

\begin{gathered} x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ \\ = 21, \end{gathered}

so

x = \sqrt{21}.

If

5

is opposite the

60^\circ

angle, then

25 = x^2 - 4x + 16,

whose positive solution is

x = 2 + \sqrt{13}.

If

4

is opposite, then

16 = x^2 - 5x + 25

has no real solution.

The sum of the possible values is

2 + \sqrt{13} + \sqrt{21},

so

a + b + c = 2 + 13 + 21 = 36.

Thus, the correct answer is A.

Problem 12 in other years