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
Solution 3 (Pythagorean Theorem)
We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length WLOG we focus on diagonal To find the diagonal of the inner quadrilateral, we drop the altitude from and and calculate the length of Let be (Thus By Pythagorean theorem, we have Now let be (thus making ). Similarly, we have We see that , the scaled down diagonal is just which is times our original diagonal implying a scale factor of Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply making our answer -fidgetboss_4000
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