1991 AHSME Problems/Problem 23

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

Solution by e_power_pi_times_i

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 $\dfrac{1}{2}\cdot\dfrac{14}{15} = \boxed{\textbf{(C) } \dfrac{7}{15}}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1991 AHSME Problems/Problem 23

Problem

Solution 1: Coordinate Geometry

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