2021 AIME I Problems/Problem 11
Let be a cyclic quadrilateral with and . Let and be the feet of the perpendiculars from and , respectively, to line and let and be the feet of the perpendiculars from and respectively, to line . The perimeter of is , where and are relatively prime positive integers. Find .
Let be the intersection of and . Let .
Firstly, since , we deduce that is cyclic. This implies that , with a ratio of . This means that . Similarly, . Hence It therefore only remains to find .
From Ptolemy's theorem, we have that . From Brahmagupta's Formula, . But the area is also , so . Then the desired fraction is for an answer of .
The angle between diagonals satisfies (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, or That is, or Thus, or In this context, . Thus, ~y.grace.yu
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