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Difference between revisions of "1965 AHSME Problems/Problem 35"

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== Solution ==
 
== Solution ==
  
 +
<asy>
  
 +
import geometry;
  
−
<math>\fbox{D}</math>
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point M;
 +
segment l;
 +
 
 +
// Rectangle ABCD
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draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5)));
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dot((0,sqrt(5)));
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label("A", (0,sqrt(5)), NW);
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dot((0,0));
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label("B", (0,0), SW);
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dot((5,0));
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label("C", (5,0), SE);
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dot((5,sqrt(5)));
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label("D", (5, sqrt(5)), NE);
 +
 
 +
// Segment AC and point M
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M=(2.5,sqrt(5)/2);
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l=line((0,sqrt(5)),(5,0));
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draw(l);
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dot(M);
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label("M",M,W);
 +
 
 +
// Segments AX, CY, and XY
 +
pair[] x=intersectionpoints(perpendicular(M,l),(0,0)--(5,0));
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pair[] y=intersectionpoints(perpendicular(M,l),(0,sqrt(5))--(5,sqrt(5)));
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dot(x[0]);
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label("X",x[0],SW);
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dot(y[0]);
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label("Y",y[0],NE);
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draw((0,sqrt(5))--x[0]);
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draw((5,0)--y[0]);
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draw(x[0]--y[0]);
 +
 
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// Right Angle Markers
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markscalefactor=0.025;
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draw(rightanglemark((0,sqrt(5)),M,y[0])); // Angle AMY
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draw(rightanglemark((5,0),M,x[0])); // Angle CMX
 +
draw(rightanglemark((0,sqrt(5)),(0,0),(5,0))); // Angle ABC
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draw(rightanglemark((0,sqrt(5)), (5,sqrt(5)),(5,0))); // Angle ADC
 +
 
 +
// Length Labels
 +
label("$5$",(2.5,0),S);
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label("$w$",(0,sqrt(5)/2),W);
 +
 
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</asy>
 +
 
 +
Let the rectangle be <math>ABCD</math> with <math>AB=CD=w</math> and <math>AD=BC=5</math>, as in the diagram. We desire a line such that reflecting point <math>C</math> across that line yields point <math>A</math>. For this to happen, the line must be perpendicular to the diagonal <math>\overline{AC}</math>, and it must go through the midpoint of <math>\overline{AC}</math> (let it be point <math>M</math>). Let the intersection of this line with <math>\overline{BC}</math> be point <math>X</math> and with <math>\overline{AD}</math> be point <math>Y</math>. From the problem, we know that <math>XY=\sqrt{6}</math>. By [[Congruent (geometry)#HL Congruence|HL congruence]], <math>\triangle AMY \cong \triangle CMX</math>, so <math>AM=CM=x</math>, where <math>x</math> is some number. Furthermore, <math>XM=MY=\frac{\sqrt{6}}{2}</math>. By [[AA similarity]], <math>\triangle MXC \sim \triangle BAC</math>, so <math>\frac{MX}{MC}=\frac{BA}{BC}</math>. <math>MX=\frac{\sqrt{6}}{2}</math>,  <math>MC=x</math>, <math>BA=w</math>, and <math>BC=5</math>, so we can rewrite this proportion to solve for <math>x</math> in terms of <math>w</math>:
 +
\begin{align*}
 +
\frac{\sqrt{6}/2}{x}&=\frac{w}{5} \\
 +
\frac{5\sqrt{6}}{2}&=xw \\
 +
x&=\frac{5\sqrt{6}}{2w}
 +
\end{align*}
 +
By the [[Pythagorean Theorem]] on <math>\triangle ABC</math>, we know that <math>w^2+25=4x^2</math>, and we can plug in our new expression for <math>x</math> into this equation to solve for <math>w</math>:
 +
\begin{align*}
 +
w^2+25&=4(\frac{5\sqrt{6}}{2w})^2 \\
 +
w^2+25&=\frac{25*6}{w^2} \\
 +
w^4+25w^2-150&=0 \\
 +
(w^2-5)(w^2+30)&=0
 +
\end{align*}
 +
Because <math>w>0</math>, <math>w^2=5</math>, and so <math>w=\boxed{\sqrt{5}}</math>, which corresponds to answer choice <math>\fbox{\textbf{(D)}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:54, 19 July 2024

Problem

The length of a rectangle is $5$ inches and its width is less than $4$ inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is $\sqrt {6}$, then the width is:

$\textbf{(A)}\ \sqrt {2} \qquad \textbf{(B) }\ \sqrt {3} \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ \sqrt{5}\qquad \textbf{(E) }\ \sqrt{\frac{11}{2}}$

Solution

[asy] import geometry; point M; segment l; // Rectangle ABCD draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5))); dot((0,sqrt(5))); label("A", (0,sqrt(5)), NW); dot((0,0)); label("B", (0,0), SW); dot((5,0)); label("C", (5,0), SE); dot((5,sqrt(5))); label("D", (5, sqrt(5)), NE); // Segment AC and point M M=(2.5,sqrt(5)/2); l=line((0,sqrt(5)),(5,0)); draw(l); dot(M); label("M",M,W); // Segments AX, CY, and XY pair[] x=intersectionpoints(perpendicular(M,l),(0,0)--(5,0)); pair[] y=intersectionpoints(perpendicular(M,l),(0,sqrt(5))--(5,sqrt(5))); dot(x[0]); label("X",x[0],SW); dot(y[0]); label("Y",y[0],NE); draw((0,sqrt(5))--x[0]); draw((5,0)--y[0]); draw(x[0]--y[0]); // Right Angle Markers markscalefactor=0.025; draw(rightanglemark((0,sqrt(5)),M,y[0])); // Angle AMY draw(rightanglemark((5,0),M,x[0])); // Angle CMX draw(rightanglemark((0,sqrt(5)),(0,0),(5,0))); // Angle ABC draw(rightanglemark((0,sqrt(5)), (5,sqrt(5)),(5,0))); // Angle ADC // Length Labels label("$5$",(2.5,0),S); label("$w$",(0,sqrt(5)/2),W); [/asy]

Let the rectangle be $ABCD$ with $AB=CD=w$ and $AD=BC=5$, as in the diagram. We desire a line such that reflecting point $C$ across that line yields point $A$. For this to happen, the line must be perpendicular to the diagonal $\overline{AC}$, and it must go through the midpoint of $\overline{AC}$ (let it be point $M$). Let the intersection of this line with $\overline{BC}$ be point $X$ and with $\overline{AD}$ be point $Y$. From the problem, we know that $XY=\sqrt{6}$. By HL congruence, $\triangle AMY \cong \triangle CMX$, so $AM=CM=x$, where $x$ is some number. Furthermore, $XM=MY=\frac{\sqrt{6}}{2}$. By AA similarity, $\triangle MXC \sim \triangle BAC$, so $\frac{MX}{MC}=\frac{BA}{BC}$. $MX=\frac{\sqrt{6}}{2}$, $MC=x$, $BA=w$, and $BC=5$, so we can rewrite this proportion to solve for $x$ in terms of $w$: \begin{align*} \frac{\sqrt{6}/2}{x}&=\frac{w}{5} \\ \frac{5\sqrt{6}}{2}&=xw \\ x&=\frac{5\sqrt{6}}{2w} \end{align*} By the Pythagorean Theorem on $\triangle ABC$, we know that $w^2+25=4x^2$, and we can plug in our new expression for $x$ into this equation to solve for $w$: \begin{align*} w^2+25&=4(\frac{5\sqrt{6}}{2w})^2 \\ w^2+25&=\frac{25*6}{w^2} \\ w^4+25w^2-150&=0 \\ (w^2-5)(w^2+30)&=0 \end{align*} Because $w>0$, $w^2=5$, and so $w=\boxed{\sqrt{5}}$, which corresponds to answer choice $\fbox{\textbf{(D)}}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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