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

(Created page with "== Problem == <asy> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle),dots); MP("B",(0,0),SW);MP("A",(0,1),NW);MP("D",(1,1),NE);MP("C",(1,0),SE); MP("E",(0,.5),W);MP("F",(.5,0),S); dot((.5...")
 
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== Problem ==
 
== Problem ==
 
<asy>
 
<asy>
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle),dots);
+
draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,dot);
MP("B",(0,0),SW);MP("A",(0,1),NW);MP("D",(1,1),NE);MP("C",(1,0),SE);
+
draw((2,2)--(0,0)--(0,1)--cycle,dot);
MP("E",(0,.5),W);MP("F",(.5,0),S);
+
draw((0,2)--(1,0),dot);
dot((.5,0));dot((0,.5));
+
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>
 
</asy>
  
Line 17: Line 19:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{}</math>
+
<math>\fbox{C}</math>
  
 
== See also ==
 
== See also ==

Revision as of 15:29, 28 September 2014

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 $2X2$ 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 (ProblemsAnswer KeyResources)
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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