Difference between revisions of "1988 AHSME Problems/Problem 20"
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Revision as of 02:03, 23 October 2014
Problem
In one of the adjoining figures a square of side 
is dissected into four pieces so that 
and 
are the midpoints
of opposite sides and 
is perpendicular to 
. These four pieces can then be reassembled into a rectangle as shown
in the second figure. The ratio of height to base, 
, 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]](http://latex.artofproblemsolving.com/b/a/9/ba97279261a01408420684879845a8987e0a013b.png)
Solution
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
