Difference between revisions of "2021 AIME I Problems/Problem 11"
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From Ptolemy's theorem, we have that <math>(BD)(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>. | From Ptolemy's theorem, we have that <math>(BD)(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>. | ||
+ | |||
+ | ==Finding <math>\cos{x}</math> 2== | ||
+ | The angle <math>\theta</math> between diagonals satisfies <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d}{(s-a)(s-c)}}</cmath> (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). | ||
+ | Thus, <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}</cmath> or <cmath>\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}</cmath> | ||
+ | That is, <math>\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}</math> or <math>\frac{35}{24}</math> | ||
+ | Thus, <math>\cos^2{\frac{\theta}{2}}=\frac{35}{59}</math> or <math>\frac{24}{59}</math> | ||
+ | <cmath>\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\frac{\pm11}{59}</cmath> | ||
+ | In this context, <math>\cos{\theta}>0</math>. Thus, <math>\cos{\theta}=\frac{11}{59}</math> | ||
+ | <cmath>Ans=22*\cos{\theta}=22*\frac{11}{59}=\frac{242}{59}=\frac{m}{n}</cmath> | ||
+ | <cmath>m+n=242+59=\boxed{301}</cmath> | ||
+ | ~y.grace.yu | ||
+ | |||
==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 02:30, 12 March 2021
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 .
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 .
Finding 2
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.