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, | + | draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,dot); |
− | MP("B",(0,0),SW);MP("A",(0, | + | draw((2,2)--(0,0)--(0,1)--cycle,dot); |
− | MP("E",(0, | + | draw((0,2)--(1,0),dot); |
− | 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> | </asy> | ||
− | If <math>ABCD</math> is a <math> | + | 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 | ||
Line 16: | Line 18: | ||
\text{(E) } \frac{3}{5}</math> | \text{(E) } \frac{3}{5}</math> | ||
− | == Solution == | + | == Solution 1: Coordinate Geometry== |
− | <math>\ | + | Solution by e_power_pi_times_i |
+ | |||
+ | |||
+ | 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 13:25, 14 December 2016
Problem
If is a square, is the midpoint of , is the midpoint of , and intersect at , and and intersect at , then the area of quadrilateral is
Solution 1: Coordinate Geometry
Solution by e_power_pi_times_i
First, we find out the coordinates of the vertices of quadrilateral , then use the Shoelace Theorem to solve for the area. Denote as . Then . Since I is the intersection between lines and , and since the equations of those lines are and , . Using the same method, the equation of line is , so . Using the Shoelace Theorem, the area of is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.