Difference between revisions of "2021 AIME I Problems/Problem 11"
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Firstly, since <math>\angle AA_1D = \angle AD_1D = 90^\circ</math>, we deduce that <math>AA_1D_1D</math> is cyclic. This implies that <math>\triangle A_1OD_1 \sim \triangle AOD</math>, with a ratio of <math>\frac{A_1O}{AO} = \cos \angle A_1OA = \cos \theta</math>. This means that <math>\frac{A_1D_1}{AD} = \cos \theta</math>. Similarly, <math>\frac{A_1B_1}{AB} = \frac{B_1C_1}{BC} = \frac{C_1D_1}{CD} = \cos \theta</math>. Hence <cmath>A_1B_1 + B_1C_1 + C_1D_1 + D_1A_1 = (AB + BC + CD + DA)\cos \theta</cmath> It therefore only remains to find <math>\cos \theta</math>. | Firstly, since <math>\angle AA_1D = \angle AD_1D = 90^\circ</math>, we deduce that <math>AA_1D_1D</math> is cyclic. This implies that <math>\triangle A_1OD_1 \sim \triangle AOD</math>, with a ratio of <math>\frac{A_1O}{AO} = \cos \angle A_1OA = \cos \theta</math>. This means that <math>\frac{A_1D_1}{AD} = \cos \theta</math>. Similarly, <math>\frac{A_1B_1}{AB} = \frac{B_1C_1}{BC} = \frac{C_1D_1}{CD} = \cos \theta</math>. Hence <cmath>A_1B_1 + B_1C_1 + C_1D_1 + D_1A_1 = (AB + BC + CD + DA)\cos \theta</cmath> It therefore only remains to find <math>\cos \theta</math>. | ||
− | From Ptolemy's theorem, we have that <math>(BC)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}( | + | From Ptolemy's theorem, we have that <math>(BC)(AC) = 4\times6+5\times7 = 59</math>. From Brahmagupta's Formula, <math>[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}</math>. But the area is also <math>\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta</math>, so <math>\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}</math>. Then the desired fraction is <math>(4+5+6+7)\cos\theta = \frac{242}{59}</math> for an answer of <math>\boxed{301}</math>. |
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=10|num-a=12}} | {{AIME box|year=2021|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:06, 11 March 2021
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 .
Solution
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 .
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 | ||
All AIME Problems and Solutions |
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