Difference between revisions of "1990 AHSME Problems/Problem 4"
(Created page with "== Problem == <asy> draw((0,0)--(16,0)--(21,5*sqrt(3))--(5,5*sqrt(3))--cycle,dot); draw((5,5*sqrt(3))--(1,5*sqrt(3))--(16,0),dot); MP("A",(0,0),S);MP("B",(16,0),S);MP("C",(21,5sq...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | <math>DFE</math> and <math>ADB</math> are similar triangles, so <math>FD</math> is one quarter the length of the corresponding side <math>AF</math>. |
+ | Thus it is one fifth of the length of <math>AD</math>, which means <math>\fbox{B}</math> | ||
== See also == | == See also == |
Revision as of 02:51, 4 February 2016
Problem
Let be a parallelogram with and Extend through to so that If intersects at , then is closest to
Solution
and are similar triangles, so is one quarter the length of the corresponding side . Thus it is one fifth of the length of , which means
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.