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 | ||
− | Note from xiej: <math>BC=BF=x\cot{\theta}, not < | + | Note from xiej: <math>BC=BF=x\cot{\theta}</math>, not <math>x\tan{\theta}</math>, so the solution goes off rails here. |
− | < | + | <math>BC = BF = x\tan{\theta}</math> gives <math>\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}</math>. <math>x = \frac{3\tan^2{\theta}+3}{\tan^2{\theta}} = \frac{3\sec^2{\theta}}{\tan^2{\theta}} = 3\csc^2{\theta}</math>. Noticing that <math>BE = x\sqrt{\tan^2{\theta}+1} = x\sec{\theta}</math> gives the answer to be <math>3\csc^2{\theta}\sec{\theta}</math> which is not in the answer list? |
If anyone could correct my solution, please help and thanks! | If anyone could correct my solution, please help and thanks! | ||
− | P.S correct answer is < | + | P.S correct answer is <math>3\sec^2{\theta}\csc{\theta}</math>. |
{{AHSME box|year=1972|num-b=29|num-a=31}} | {{AHSME box|year=1972|num-b=29|num-a=31}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:50, 22 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
Note from xiej: , not , so the solution goes off rails here.
gives . . Noticing that gives the answer to be which is not in the answer list?
If anyone could correct my solution, please help and thanks!
P.S correct answer is .
1972 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Problem 31 | |
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