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Difference between revisions of "1972 AHSME Problems/Problem 30"

(Created page with "== Problem 30 == == Solution == 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>A...")
 
(found error in solution, fix later)
Line 1: Line 1:
 
== Problem 30 ==
 
== Problem 30 ==
  
 +
<asy>
 +
real h = 7;
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real t = asin(6/h)/2;
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real x = 6-h*tan(t);
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real y = x*tan(2*t);
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draw((0,0)--(0,h)--(6,h)--(x,0)--cycle);
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draw((x,0)--(0,y)--(6,h));
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draw((6,h)--(6,0)--(x,0),dotted);
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label("L",(3.75,h/2),W);
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label("$\theta$",(6,h-1.5),W);draw(arc((6,h),2,270,270-degrees(t)),Arrow(2mm));
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label("6''",(3,0),S);
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draw((2.5,-.5)--(0,-.5),Arrow(2mm));
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draw((3.5,-.5)--(6,-.5),Arrow(2mm));
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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
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\textbf{(B) }6\sin\theta\sec\theta\qquad
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\textbf{(C) }3\sec\theta\csc\theta\qquad
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\textbf{(D) }6\sec\theta\csc^2\theta\qquad
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\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
 +
 +
Note from xiej: <math>BC=BF=x\cot{\theta}, 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?
  
−
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\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}</math>.
+
P.S correct answer is </math>3\sec^2{\theta}\csc{\theta}$.
 +
 
 +
{{AHSME box|year=1972|num-b=29|num-a=31}}
 +
{{MAA Notice}}

Revision as of 20:50, 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$ (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
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