2021 AIME II Problems/Problem 12
A convex quadrilateral has area and side lengths and in that order. Denote by the measure of the acute angle formed by the diagonals of the quadrilateral. Then can be written in the form , where and are relatively prime positive integers. Find .
Solution 1 (Sines and Cosines)
Since we are asked to find , we can find and separately and use their values to get . We can start by drawing a diagram. Let the vertices of the quadrilateral be , , , and . Let , , , and . Let , , , and . We know that is the acute angle formed between the intersection of the diagonals and . We are given that the area of quadrilateral is . We can express this area using the areas of triangles , , , and . Since we want to find and , we can represent these areas using as follows: We know that . Therefore it follows that: From here we see that . Now we need to find . Using the Law of Cosines on each of the four smaller triangles, we get following equations: We know that for all . We can substitute this value into our equations to get: If we subtract from , the squared terms cancel, leaving us with: From here we see that .
Since we have figured out and , we can calculate : Therefore our answer is .
~ Steven Chen (www.professorchenedu.com)
Solution 2 (Right Triangles)
In convex quadrilateral let and Let and be the feet of the perpendiculars from and respectively, to We obtain the following diagram:
Let and We apply the Pythagorean Theorem to right triangles and respectively: Let the brackets denote areas. We get We subtract from From right triangles and we have It follows that Finally, we divide by from which the answer is
Let , , , be the vertices of the quadrilateral, , , , be the lengths of the sides of , and and be the lengths of the diagonals of . By Bretschneider's formula, . Thus, . Also, ; solving for yields . Since is acute, is positive, so . Solving for yields , for a final answer of .
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