Difference between revisions of "1991 AHSME Problems/Problem 23"
(Created page with "== Problem == <asy> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle),dots); MP("B",(0,0),SW);MP("A",(0,1),NW);MP("D",(1,1),NE);MP("C",(1,0),SE); MP("E",(0,.5),W);MP("F",(.5,0),S); dot((.5...") |
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== Problem == | == Problem == | ||
<asy> | <asy> | ||
− | draw((0,0)--(0, | + | draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,dot); |
− | MP("B",(0,0),SW);MP("A",(0, | + | draw((2,2)--(0,0)--(0,1)--cycle,dot); |
− | MP("E",(0, | + | draw((0,2)--(1,0),dot); |
− | dot(( | + | MP("B",(0,0),SW);MP("A",(0,2),NW);MP("D",(2,2),NE);MP("C",(2,0),SE); |
+ | MP("E",(0,1),W);MP("F",(1,0),S);MP("H",(2/3,2/3),E);MP("I",(2/5,6/5),N); | ||
+ | dot((1,0));dot((0,1));dot((2/3,2/3));dot((2/5,6/5)); | ||
</asy> | </asy> | ||
Line 17: | Line 19: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | <math>\fbox{C}</math> |
== See also == | == See also == |
Revision as of 14:29, 28 September 2014
Problem
If is a square, is the midpoint of , is the midpoint of , and intersect at , and and intersect at , then the area of quadrilateral is
Solution
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.