1972 AMC 12 Problem 30

Question

A rectangular piece of paper

6

inches wide is folded as in the diagram so that one corner touches the opposite side. The length in inches of the crease

L

in terms of angle

\theta

is:

3\sec^2\theta\csc\theta

6\sin\theta\sec\theta

3\sec\theta\csc\theta

6\sec\theta\csc^2\theta

None of these

Let

h

be the height of the rectangular sheet and use the equal lengths created by the fold

The diagram gives

\frac{6}{h}=\sin(2\theta)

and

\frac{L}{h}=\sec\theta

Let

h

be the sheet’s height. The congruent right triangles created by reflecting the folded corner across the crease give

\frac6h=\sin(2\theta)=2\sin\theta\cos\theta,

so

h=\frac{3}{\sin\theta\cos\theta}.

They also give

\frac{L}{h}=\sec\theta.

Hence

\begin{aligned} L&=h\sec\theta\\ &=\frac{3\sec\theta} {\sin\theta\cos\theta}\\ &=3\sec^2\theta\csc\theta. \end{aligned}

Therefore, the correct answer is A.

Problem 30 in other years