Difference between revisions of "1990 AHSME Problems/Problem 4"
m (→Problem) |
|||
| Line 9: | Line 9: | ||
</asy> | </asy> | ||
| − | 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 15:05, 10 November 2017
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",(21,5sqrt(3)),NE);MP("D",(5,5sqrt(3)),N);MP("E",(1,5sqrt(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))); MP("F",(4,4sqrt(3)),W); [/asy]](http://latex.artofproblemsolving.com/4/8/c/48c62bae9fa86f0f007e8fddf092b32bf0d8aa6f.png)
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. 
