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Difference between revisions of "1990 AHSME Problems/Problem 4"
m (→‎Solution: Small point naming mistake.)
 
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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)),W);
+
MP("F",(4,4sqrt(3)),dir(210));
 
</asy>
 
</asy>
  
−
Let <math>ABCD</math> be a parallelogram with <math>\angle{ABC}=120^\circ, AB=6</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
+
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
Line 18: Line 18:
  
 
== Solution ==
 
== Solution ==
−
<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>.
+
<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 19:41, 2 July 2021

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 $AFB$ 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 (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

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