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1972 AHSME Problems/Problem 30

Revision as of 20:50, 22 August 2017 by Xiej (talk | contribs) (found error in solution, fix later)

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 $\theta$ is

$\textbf{(A) }3\sec ^2\theta\csc\theta\qquad \textbf{(B) }6\sin\theta\sec\theta\qquad \textbf{(C) }3\sec\theta\csc\theta\qquad \textbf{(D) }6\sec\theta\csc^2\theta\qquad \textbf{(E) }\text{None of these}$

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 $ABCD$ have crease $BE$ with $E$ on $CD$, and let $F$ be on $AD$ such that $F$ is a reflection of $C$ over $BE$. Notice that triangles $ABF$ and $DEF$ are similar, so by setting $CE = EF = x$ with $DE = 6-x$, giving $DF = 2\sqrt{3x-9}$ we have that $AF = \frac{18-3x}{\sqrt{3x-9}}$. Noticing that

Note from xiej: $BC=BF=x\cot{\theta}, not$x\tan{\theta}$, so the solution goes off rails here.$BC = BF = x\tan{\theta}$gives$\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}$. Noticing that$BE = x\sqrt{\tan^2{\theta}+1} = x\sec{\theta}$gives the answer to be$3\csc^2{\theta}\sec{\theta}$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. Unknown error_msg)3\sec^2{\theta}\csc{\theta}$.

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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