1959 AMC 12 Problem 36

Question

The base of a triangle is

80,

and one of the base angles is

60^\circ.

The sum of the lengths of the other two sides is

90.

The shortest side is:

45

40

36

17

12

Let the side adjacent to the

60^\circ

angle be

x

, so the third side is

90-x

Apply the law of cosines with included sides

80

and

x

Let the side adjacent to the

60^\circ

base angle be

x,

and let the opposite side be

90-x.

The law of cosines gives

\begin{aligned} (90-x)^2 &=80^2+x^2\\ &\quad{}-2(80)(x)\cos60^\circ. \end{aligned}

Simplifying yields

8100-180x=6400-80x,

so

x=17.

The side lengths are

17,\ 73,

and

80,

and the shortest is

17.

Thus, the correct answer is D.

Problem 36 in other years