Difference between revisions of "1972 AHSME Problems/Problem 30"
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== Problem 30 == | == Problem 30 == | ||
+ | <asy> | ||
+ | real h = 7; | ||
+ | real t = asin(6/h)/2; | ||
+ | real x = 6-h*tan(t); | ||
+ | real y = x*tan(2*t); | ||
+ | draw((0,0)--(0,h)--(6,h)--(x,0)--cycle); | ||
+ | draw((x,0)--(0,y)--(6,h)); | ||
+ | draw((6,h)--(6,0)--(x,0),dotted); | ||
+ | label("L",(3.75,h/2),W); | ||
+ | label("$\theta$",(6,h-1.5),W);draw(arc((6,h),2,270,270-degrees(t)),Arrow(2mm)); | ||
+ | label("6''",(3,0),S); | ||
+ | draw((2.5,-.5)--(0,-.5),Arrow(2mm)); | ||
+ | draw((3.5,-.5)--(6,-.5),Arrow(2mm)); | ||
+ | draw((0,-.25)--(0,-.75));draw((6,-.25)--(6,-.75)); | ||
+ | //Credit to Zimbalono for the diagram | ||
+ | </asy> | ||
+ | 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 <math>\theta</math> is | ||
+ | |||
+ | <math>\textbf{(A) }3\sec ^2\theta\csc\theta\qquad | ||
+ | \textbf{(B) }6\sin\theta\sec\theta\qquad | ||
+ | \textbf{(C) }3\sec\theta\csc\theta\qquad | ||
+ | \textbf{(D) }6\sec\theta\csc^2\theta\qquad | ||
+ | \textbf{(E) }\text{None of these} </math> | ||
+ | |||
== Solution == | == Solution == | ||
+ | <asy> | ||
+ | real h = 7; | ||
+ | real t = asin(6/h)/2; | ||
+ | real x = 6-h*tan(t); | ||
+ | real y = x*tan(2*t); | ||
+ | draw((0,0)--(0,h)--(6,h)--(x,0)--cycle); | ||
+ | draw((x,0)--(0,y)--(6,h)); | ||
+ | draw((6,h)--(6,0)--(x,0),dotted); | ||
+ | label("A",(0,h),NW);label("B",(6,h),NE);label("C",(6,0),SE);label("D",(0,0),SW);label("E",(x,0),N);label("F",(0,y),W); | ||
+ | label("L",(3.75,h/2),W); | ||
+ | label("$\theta$",(6,h-1.5),W);draw(arc((6,h),2,270,270-degrees(t)),Arrow(2mm)); | ||
+ | label("6''",(3,0),S); | ||
+ | draw((2.5,-.5)--(0,-.5),Arrow(2mm)); | ||
+ | draw((3.5,-.5)--(6,-.5),Arrow(2mm)); | ||
+ | draw((0,-.25)--(0,-.75));draw((6,-.25)--(6,-.75)); | ||
+ | //Credit to Zimbalono for the diagram | ||
+ | </asy> | ||
+ | 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>. | ||
− | + | Thanks to xiej for correcting my mistake! | |
− | + | ==See Also== | |
− | + | {{AHSME box|year=1972|num-b=29|num-a=31}} | |
+ | {{MAA Notice}} |
Latest revision as of 19:52, 23 June 2021
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!
See Also
1972 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Problem 31 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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