2018 AMC 8 Problems/Problem 22
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Problem 22
Point is the midpoint of side in square and meets diagonal at The area of quadrilateral is What is the area of
Solution 1
Let the area of be . Thus, the area of triangle is and the area of the square is .
By AA similarity, with a 1:2 ratio, so the area of triangle is . Now consider trapezoid . Its area is , which is three-fourths the area of the square. We set up an equation in :
Solving, we get . The area of square is .
Solution 2
We can use analytic geometry for this problem.
Let us start by giving the coordinate , the coordinate , and so forth. and can be represented by the equations and , respectively. Solving for their intersection gives point coordinates .
Now, ’s area is simply or . This means that pentagon ’s area is of the entire square, and it follows that quadrilateral ’s area is of the square.
The area of the square is then .
Solution 3
Note that triangle has half the area of the square and triangle has th. Thus the area of the quadrilateral is th the area of the square. The area of the square is then .
Solution 4
Extend and to meet at . Drop an altitude from to and call it . Also, call . As stated before, we have , so the ratio of their heights is in a ratio, making the altitude from to . Note that this means that the side of the square is . In addition, by AA Similarity in a ratio. This means that the side length of the square is , making .
Now, note that . We have and Subtracting makes We are given that so Therefore, so our answer is - moony_eyed
Solution 5
Solution with Cartesian and Barycentric Coordinates:
We start with the following
Claim: Given a square , let be the midpoint of and let . Then .
Proof. We use Cartesian coordinates. Let be the origin, . We have that and are governed by the equations and , respectively. Solving, . The result follows.
Now we apply Barycentric Coordinates w.r.t. . We let . Then .
In the barycentric coordinate system, the area formula is where is a random triangle and is the reference triangle. Using this, we find that Let so that . Then we have so the answer is .
-franzliszt
Video Solution
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
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Followed by Problem 23 | |
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Set s to be the bottom left triangle. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.