2013 AIME I Problems/Problem 3
Problem
Problem 3
Let be a square, and let and be points on and respectively. The line through parallel to and the line through parallel to divide into two squares and two nonsquare rectangles. The sum of the areas of the two squares is of the area of square Find
Solution
It's important to note that is equivalent to
We define as the length of the side of larger inner square, which is also , as the length of the side of the smaller inner square which is also , and as the side length of . Since we are given that the sum of the areas of the two squares is of the the area of ABCD, we can represent that as . The sum of the two nonsquare rectangles can then be represented as .
Looking back at what we need to find, we can represent as . We have the numerator, and dividing by two gives us the denominator . Dividing gives us an answer of .
Solution 2
Let the side of the square be . Therefore the area of the square is also . We label as and as . Notice that what we need to find is equivalent to: . Since the sum of the two squares () is (as stated in the problem) the area of the whole square, it is clear that the sum of the two rectangles is . Since these two rectangles are congruent, they each have area: . Also note that the area of this is . Plugging this into our equation we get:
Solution 3
Let be , and be . Then we are looking for the value . The areas of the smaller squares add up to of the area of the large square, . Cross multiplying and simplifying we get . Rearranging, we get
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.