Difference between revisions of "1990 AHSME Problems/Problem 4"
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MP("16",(8,0),S);MP("10",(18.5,5sqrt(3)/2),E);MP("4",(3,5sqrt(3)),N); | MP("16",(8,0),S);MP("10",(18.5,5sqrt(3)/2),E);MP("4",(3,5sqrt(3)),N); | ||
dot((4,4sqrt(3))); | dot((4,4sqrt(3))); | ||
− | MP("F",(4,4sqrt(3)), | + | MP("F",(4,4sqrt(3)),dir(210)); |
</asy> | </asy> | ||
Line 18: | Line 18: | ||
== Solution == | == Solution == | ||
− | <math>DFE</math> and <math> | + | <math>DFE</math> and <math>AFB</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> | Thus it is one fifth of the length of <math>AD</math>, which means <math>\fbox{B}</math> | ||
Latest revision as of 20:41, 2 July 2021
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.