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 </math>x\tan{\theta}<math>, so the solution goes off rails here.
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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?
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  <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 </math>3\sec^2{\theta}\csc{\theta}$.
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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

[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 $3\sec^2{\theta}\csc{\theta}$.

1972 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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