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Difference between revisions of "1991 AHSME Problems/Problem 23"

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If <math>ABCD</math> is a <math>2X2</math> square, <math>E</math> is the midpoint of <math>\overline{AB}</math>,<math>F</math> is the midpoint of <math>\overline{BC}</math>,<math>\overline{AF}</math> and <math>\overline{DE}</math> intersect at <math>I</math>, and <math>\overline{BD}</math> and <math>\overline{AF}</math> intersect at <math>H</math>, then the area of quadrilateral <math>BEIH</math> is
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If <math>ABCD</math> is a <math>2\times2</math> square, <math>E</math> is the midpoint of <math>\overline{AB}</math>,<math>F</math> is the midpoint of <math>\overline{BC}</math>,<math>\overline{AF}</math> and <math>\overline{DE}</math> intersect at <math>I</math>, and <math>\overline{BD}</math> and <math>\overline{AF}</math> intersect at <math>H</math>, then the area of quadrilateral <math>BEIH</math> is
  
 
<math>\text{(A) } \frac{1}{3}\quad
 
<math>\text{(A) } \frac{1}{3}\quad

Revision as of 19:09, 11 October 2016

Problem

[asy] draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,dot); draw((2,2)--(0,0)--(0,1)--cycle,dot); draw((0,2)--(1,0),dot); MP("B",(0,0),SW);MP("A",(0,2),NW);MP("D",(2,2),NE);MP("C",(2,0),SE); MP("E",(0,1),W);MP("F",(1,0),S);MP("H",(2/3,2/3),E);MP("I",(2/5,6/5),N); dot((1,0));dot((0,1));dot((2/3,2/3));dot((2/5,6/5)); [/asy]


If $ABCD$ is a $2\times2$ square, $E$ is the midpoint of $\overline{AB}$,$F$ is the midpoint of $\overline{BC}$,$\overline{AF}$ and $\overline{DE}$ intersect at $I$, and $\overline{BD}$ and $\overline{AF}$ intersect at $H$, then the area of quadrilateral $BEIH$ is

$\text{(A) } \frac{1}{3}\quad \text{(B) } \frac{2}{5}\quad \text{(C) } \frac{7}{15}\quad \text{(D) } \frac{8}{15}\quad \text{(E) } \frac{3}{5}$

Solution

$\fbox{C}$

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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