Difference between revisions of "2017 AIME II Problems/Problem 10"
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Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>N</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, or drawing diagonal <math>\overline{BO}</math> and using <math>\frac12bh</math>, we find that its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2} = 1092</math>. Using <math>A = \frac 12 bh</math>, we get <math>2184 = 42h</math>. Simplifying, we get <math>h = 52</math>. This means that the x-coordinate of <math>P = 84 - 52 = 32</math>. Since P lies on <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, you can solve and get that the y-coordinate of <math>P</math> is <math>13</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}</math>. | Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>N</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, or drawing diagonal <math>\overline{BO}</math> and using <math>\frac12bh</math>, we find that its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2} = 1092</math>. Using <math>A = \frac 12 bh</math>, we get <math>2184 = 42h</math>. Simplifying, we get <math>h = 52</math>. This means that the x-coordinate of <math>P = 84 - 52 = 32</math>. Since P lies on <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, you can solve and get that the y-coordinate of <math>P</math> is <math>13</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}</math>. | ||
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==Solution 2 (No Coordinates)== | ==Solution 2 (No Coordinates)== |
Revision as of 13:02, 29 November 2020
Problem
Rectangle has side lengths and . Point is the midpoint of , point is the trisection point of closer to , and point is the intersection of and . Point lies on the quadrilateral , and bisects the area of . Find the area of .
Solution 1
Impose a coordinate system on the diagram where point is the origin. Therefore , , , and . Because is a midpoint and is a trisection point, and . The equation for line is and the equation for line is , so their intersection, point , is . Using the shoelace formula on quadrilateral , or drawing diagonal and using , we find that its area is . Therefore the area of triangle is . Using , we get . Simplifying, we get . This means that the x-coordinate of . Since P lies on , you can solve and get that the y-coordinate of is . Therefore the area of is .
Solution 2 (No Coordinates)
Since the problem tells us that segment bisects the area of quadrilateral , let us compute the area of by subtracting the areas of and from rectangle . To do this, drop altitude onto side and draw a horizontal segment from side to . Since is the midpoint of side , . Denote as . Noting that and are similar, we can write the statement . Using this information, the area of and are and , respectively. Thus, the area of quadrilaterial is . Now, it is clear that point lies on side , so the area of is . Given this, drop altitude (let's call it ) onto . Therefore, . From here, drop an altitude onto . Recognizing that and that and are similar, we write . The area of is given by ~blitzkrieg21
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.