Difference between revisions of "1990 AHSME Problems/Problem 4"

m (Problem)
(Problem)
 
Line 6: Line 6:
 
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)),W);
+
MP("F",(4,4sqrt(3)),dir(210));
 
</asy>
 
</asy>
  

Latest revision as of 00:35, 19 January 2019

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)),dir(210)); [/asy]

Let $ABCD$ be a parallelogram with $\angle{ABC}=120^\circ, AB=16$ and $BC=10.$ Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4.$ If $\overline{BE}$ intersects $\overline{AD}$ at $F$, then $FD$ is closest to

$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

$DFE$ and $ADB$ are similar triangles, so $FD$ is one quarter the length of the corresponding side $AF$. Thus it is one fifth of the length of $AD$, which means $\fbox{B}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS