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

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 $\fbox{\textbf{(D)}}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 36
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Difference between revisions of "1965 AHSME Problems/Problem 35"

Latest revision as of 15:54, 19 July 2024

Problem

Solution

See Also

@@ Line 19: / Line 19: @@
 segment l;
+// Rectangle ABCD
 draw((0,sqrt(5))--(0,0)--(5,0)--(5,sqrt(5))--(0,sqrt(5)));
 dot((0,sqrt(5)));
@@ Line 29: / Line 30: @@
 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);
-draw(perpendicular(M,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 <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>:
-<math>\fbox{D}</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 ==