2021 AIME I Problems/Problem 11
Contents
Problem
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
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution 1
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 .
Solution 2 (Finding cos x)
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 [I don't believe this is correct... are the two diagonals of necessarily congruent? -peace09] 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
Solution 4 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)
This solution refers to the Diagram section.
By the Converse of the Inscribed Angle Theorem, if distinct points and lie on the same side of (but not on itself) for which then and are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals and are all cyclic.
Suppose and intersect at and let It follows that and
We obtain the following diagram:
In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have (both supplementary to ) and (both supplementary to ), from which by AA, with the ratio of similitude Similarly, we have (both supplementary to ) and (both supplementary to ), from which by AA, with the ratio of similitude We apply the Transitive Property to and
- We get so by SAS, with the ratio of similitude
- We get so by SAS, with the ratio of similitude
From and the perimeter of is Two solutions follow from here:
Solution 4.1 (Law of Cosines)
Note that holds for all We apply the Law of Cosines to and respectively: We subtract from Finally, substituting this result into gives from which the answer is
~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)
Solution 4.2 (Area Formulas)
Let the brackets denote areas. We find in two different ways:
- Note that holds for all By area addition, we get
- By Brahmagupta's Formula, we get where is the semiperimeter of
Equating the expressions for we have so Since we have It follows that Finally, substituting this result into gives from which the answer is
~MRENTHUSIASM
Remark (Ptolemy's Theorem)
In we have ~MRENTHUSIASM
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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