Difference between revisions of "1972 AHSME Problems/Problem 30"
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− | Let the rectangle <math>ABCD</math> have crease <math>BE</math> with <math>E</math> on <math>CD</math>, and let <math>F</math> be on <math>AD</math> such that <math>F</math> is a reflection of <math>C</math> over <math>BE</math>. Notice that triangles <math>ABF</math> and <math>DEF</math> are similar, so by setting <math>CE = EF = x</math> with <math>DE = 6-x</math>, giving <math>DF = 2\sqrt{3x-9}</math> we have that <math>AF = \frac{18-3x}{\sqrt{3x-9}}</math>. Noticing that | + | Let the rectangle <math>ABCD</math> have crease <math>BE</math> with <math>E</math> on <math>CD</math>, and let <math>F</math> be on <math>AD</math> such that <math>F</math> is a reflection of <math>C</math> over <math>BE</math>. Notice that triangles <math>ABF</math> and <math>DEF</math> are similar, so by setting <math>CE = EF = x</math> with <math>DE = 6-x</math>, giving <math>DF = 2\sqrt{3x-9}</math> we have that <math>AF = \frac{18-3x}{\sqrt{3x-9}}</math>. Noticing that <math>BC = BF = x\cot{\theta}</math> gives <math>\frac{(18-3x)^2}{3x-9}+36 = x^2\cot^2{\theta} \Rightarrow \frac{3(x-6)^2}{x-3}+36 = \frac{3x^2}{x-3} = x^2\cot^2{\theta}</math>. <math>x = \frac{3\cot^2{\theta}+3}{\cot^2{\theta}} = \frac{3\csc^2{\theta}}{\cot^2{\theta}} = 3\sec^2{\theta}</math>. Noticing that <math>BE = x\sqrt{\cot^2{\theta}+1} = x\csc{\theta}</math> gives the answer to be <math>3\sec^2{\theta}\csc{\theta}</math>. |
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Revision as of 13:38, 23 August 2017
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 gives . . Noticing that gives the answer to be .
Thanks to xiej for correcting my mistake!
1972 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Problem 31 | |
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