2014 AIME I Problems/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 .
Notice that . This means passes through the center of the square.
Draw with on , on such that and intersects at the center of the square which I'll label as .
Let the area of the square be . Then the area of and the area of . This is because is perpendicular to (given in the problem), so is also perpendicular to . These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.
Let the side length of the square be .
Draw and intersects at . .
The area of , so the area of .
Let . Then
Consider the area of .
Solving , we get .
Therefore, the area of
, 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.
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
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