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

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\text{(E) } \frac{3}{5}</math>
 
\text{(E) } \frac{3}{5}</math>
  
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== Solution ==
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== Solution 1: Coordinate Geometry==
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<math>\fbox{C}</math>
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First, we find out the coordinates of the vertices of quadrilateral <math>BEIH</math>, then use the Shoelace Theorem to solve for the area. Denote <math>B</math> as <math>(0,0)</math>. Then <math>E (0,1)</math>. Since I is the intersection between lines <math>DE</math> and <math>AF</math>, and since the equations of those lines are <math>y = \dfrac{1}{2}x + 1</math> and <math>y = -2x + 2</math>, <math>I (\dfrac{2}{5}, \dfrac{6}{5})</math>. Using the same method, the equation of line <math>BD</math> is <math>y = x</math>, so <math>H (\dfrac{2}{3}, \dfrac{2}{3})</math>. Using the Shoelace Theorem, the area of <math>BEIH</math> is <math>\dfrac{1}{2}\cdot\dfrac{14}{15} = \boxed{\textbf{(C) } \dfrac{7}{15}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:24, 14 December 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 1: Coordinate Geometry

First, we find out the coordinates of the vertices of quadrilateral $BEIH$, then use the Shoelace Theorem to solve for the area. Denote $B$ as $(0,0)$. Then $E (0,1)$. Since I is the intersection between lines $DE$ and $AF$, and since the equations of those lines are $y = \dfrac{1}{2}x + 1$ and $y = -2x + 2$, $I (\dfrac{2}{5}, \dfrac{6}{5})$. Using the same method, the equation of line $BD$ is $y = x$, so $H (\dfrac{2}{3}, \dfrac{2}{3})$. Using the Shoelace Theorem, the area of $BEIH$ is $\dfrac{1}{2}\cdot\dfrac{14}{15} = \boxed{\textbf{(C) } \dfrac{7}{15}}$.

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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