Difference between revisions of "1990 AHSME Problems/Problem 4"
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− | Let <math>ABCD</math> be a parallelogram with <math>\angle{ABC}=120^\circ, AB= | + | Let <math>ABCD</math> be a parallelogram with <math>\angle{ABC}=120^\circ, AB=16</math> and <math>BC=10.</math> Extend <math>\overline{CD}</math> through <math>D</math> to <math>E</math> so that <math>DE=4.</math> If <math>\overline{BE}</math> intersects <math>\overline{AD}</math> at <math>F</math>, then <math>FD</math> is closest to |
<math>\text{(A) } 1\quad | <math>\text{(A) } 1\quad |
Revision as of 16:05, 10 November 2017
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.