2018 AMC 8 Problems/Problem 22
Contents
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 .
Sidenote: is a 1"2 ratio because it was given.
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
has half the area of the square. has base equal to half the square side length, and by AA Similarity with , it has 1/(1+2)= 1/3 the height, so has th area of square. 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 .
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=4038
- pi_is_3.14
Video Solutions
- Happytwin
~savannahsolver
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
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.