Difference between revisions of "2017 AIME II Problems/Problem 10"
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pair A,B,C,D,M,n,O,P; | pair A,B,C,D,M,n,O,P; | ||
A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); | A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); | ||
− | fill(C--D--P--cycle, | + | fill(C--D--P--cycle,blue); |
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(C--M); | draw(C--M); |
Revision as of 21:23, 4 September 2021
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 , we can write the statement Using this information, the area of and are and respectively. Thus, the area of quadrilateral 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 and jdong2006
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.