1977 AMC 12 Problem 30

Question

If

a,

b

and

d

are the lengths of a side, a shortest diagonal and a longest diagonal, respectively, of a regular nonagon (see adjoining figure), then

d=a+b

d^2=a^2+b^2

d^2=a^2+ab+b^2

b=\frac{a+d}{2}

b^2=ad

Write the three chord lengths using the nonagon’s circumradius

Compare

\sin20^\circ+\sin40^\circ

with

\sin80^\circ

using the sum-to-product identity

If the circumradius is

r,

the three chords subtend central angles

40^\circ,

80^\circ,

160^\circ,

respectively. Thus

\begin{aligned} a&=2r\sin20^\circ,\\ b&=2r\sin40^\circ,\\ d&=2r\sin80^\circ. \end{aligned}

The sum-to-product identity gives

\begin{gathered} \sin20^\circ+\sin40^\circ\\ =2\sin30^\circ\cos10^\circ\\ =\cos10^\circ=\sin80^\circ. \end{gathered}

Multiplying by

2r

yields

a+b=d.

Therefore, the correct answer is A.

Problem 30 in other years