1972 AHSME Problems/Problem 30
Problem 30
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 is
Solution
Let the rectangle have crease with on , and let be on such that is a reflection of over . Notice that triangles and are similar, so by setting with , giving we have that . Noticing that
Note from xiej: x\tan{\theta}BC = BF = x\tan{\theta}\frac{(18-3x)^2}{3x-9}+36 = x^2\tan^2{\theta} \Rightarrow \frac{3(x-6)^2}{x-3}+36 = \frac{3x^2}{x-3} = x^2\tan^2{\theta}x = \frac{3\tan^2{\theta}+3}{\tan^2{\theta}} = \frac{3\sec^2{\theta}}{\tan^2{\theta}} = 3\csc^2{\theta}BE = x\sqrt{\tan^2{\theta}+1} = x\sec{\theta}3\csc^2{\theta}\sec{\theta}$which is not in the answer list?
If anyone could correct my solution, please help and thanks!
P.S correct answer is$ (Error compiling LaTeX. ! Missing $ inserted.)3\sec^2{\theta}\csc{\theta}$.
1972 AHSME (Problems • Answer Key • Resources) | ||
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