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Difference between revisions of "1988 AHSME Problems/Problem 20"

(Created page with "==Problem== In one of the adjoining figures a square of side <math>2</math> is dissected into four pieces so that <math>E</math> and <math>F</math> are the midpoints of opposite...")
 
(Added a solution with explanation)
 
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==Solution==
 
==Solution==
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Within <math>WXYZ</math>, the parallelogram piece has vertical side <math>BF = \sqrt{1^2 + 2^2} = \sqrt{5}</math>, and diagonal side <math>FD = 1</math>. Thus the triangle in the bottom-right hand corner (the one with horizontal side <math>YZ</math>) must have hypotenuse <math>1</math>, and the only such triangle in the original figure is <math>\triangle AFG</math>, so we deduce <math>YZ = AG = \frac{\frac{1}{2} \times 1 \times 2}{\frac{1}{2} \times \sqrt{5}} = \frac{2}{\sqrt{5}}.</math> Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is <math>2^2 = 4</math>, and thus the vertical side <math>WZ</math> is <math>\frac{4}{\frac{2}{\sqrt{5}}} = 2\sqrt{5}</math>, so the required ratio is <math>\frac{2\sqrt{5}}{\frac{2}{\sqrt{5}}} = \sqrt{5} \times \sqrt{5} = 5</math>, which is <math>\boxed{\text{E}}</math>.
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== See also ==
 
== See also ==

Latest revision as of 13:11, 27 February 2018

Problem

In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label("A", A, NW); label("B", B, SW); label("C", C, SE); label("D", D, NE); label("E", E, S); label("F", F, N); label("G", G, E); label("X", X, NW); label("Y", Y, SW); label("Z", Z, SE); label("W", W, NE); [/asy]

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 1+2\sqrt{3}\qquad \textbf{(C)}\ 2\sqrt{5}\qquad \textbf{(D)}\ \frac{8+4\sqrt{3}}{3}\qquad \textbf{(E)}\ 5$


Solution

Within $WXYZ$, the parallelogram piece has vertical side $BF = \sqrt{1^2 + 2^2} = \sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\triangle AFG$, so we deduce $YZ = AG = \frac{\frac{1}{2} \times 1 \times 2}{\frac{1}{2} \times \sqrt{5}} = \frac{2}{\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is $2^2 = 4$, and thus the vertical side $WZ$ is $\frac{4}{\frac{2}{\sqrt{5}}} = 2\sqrt{5}$, so the required ratio is $\frac{2\sqrt{5}}{\frac{2}{\sqrt{5}}} = \sqrt{5} \times \sqrt{5} = 5$, which is $\boxed{\text{E}}$.

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
All AHSME Problems and Solutions

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