Difference between revisions of "1965 AHSME Problems/Problem 35"
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== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | |
+ | <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 <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 inches and its width is less than inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is , then the width is:
Solution
Let the rectangle be with and , as in the diagram. We desire a line such that reflecting point across that line yields point . For this to happen, the line must be perpendicular to the diagonal , and it must go through the midpoint of (let it be point ). Let the intersection of this line with be point and with be point . From the problem, we know that . By HL congruence, , so , where is some number. Furthermore, . By AA similarity, , so . , , , and , so we can rewrite this proportion to solve for in terms of :
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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