2014 AIME I Problems/Problem 13
Contents
Problem 13
On square , points , and lie on sides and respectively, so that and . Segments and intersect at a point , and the areas of the quadrilaterals and are in the ratio Find the area of square .
Solution (Official Solution, MAA)
Let be the side length of , let , and be the midpoints of and , respectively, let be the foot of the perpendicular from to , let be the foot of the perpendicular from to . The fraction of the area of the square which is occupied by trapezoid is so is the center of . Thus , , are collinear, and . Similarly, the fraction of the area occupied by trapezoid is , so and .
Because , the area of is the sum Rectangle has area . If , then has area Therefore .
Because these areas are in the ratio , it follows that from which we get . Note that , so and . Then
Solution 1
Let be the side length of , let . Let and be the midpoints of and , respectively; because , is also the center of the square. Draw through , with on , on . Segments and divide the square into four congruent quadrilaterals, each of area . Then The fraction of the total area occupied by parallelogram is , so .
Because , with , we get . Now and because , with , we get . By Pythagoras' Theorem on , we get with roots or . The former leads to a square with diagonal less than , which can't be, since ; therefore and .
Solution 2 (Lazy)
, a multiple of . In addition, , which is . Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to and must be a multiple of . All of these triples are primitive:
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting :
Thus, is the only valid answer.
Solution 3
Continue in the same way as solution 1 to get that has area , and . You can then find has length .
Then, if we drop a perpendicular from to at , We get .
Thus, , and we know , and . Thus, we can set up an equation in terms of using the Pythagorean theorem.
is extraneous, so . Since the area is , we have it is equal to
-Alexlikemath
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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