Difference between revisions of "2019 AIME II Problems/Problem 11"
Firebolt360 (talk | contribs) (→Solution 2) |
|||
Line 28: | Line 28: | ||
-franchester | -franchester | ||
− | ==Solution 2 (Video)== | + | ==Solution 2 (Inversion)== |
+ | Consider an inversion with center <math>A</math> and radius <math>r=AK</math>. Then, we have <math>AB\cdot AB^*=AK^2</math>, or <math>AB^*=\frac{AK^2}{7}</math>. Similarly, <math>AC^*=\frac{AK^2}{9}</math>. Notice that <math>AB^*KC^*</math> is a parallelogram, since <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AC</math> and <math>AB</math>, respectively. Thus, <math>AC^*=B^*K</math>. Now, we get that | ||
+ | <cmath>\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}</cmath> | ||
+ | so by Law of Cosines on <math>\triangle AB^*K</math> we have | ||
+ | <cmath>(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)</cmath> | ||
+ | <cmath>\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}</cmath> | ||
+ | <cmath>\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}</cmath> | ||
+ | <cmath>\Rightarrow AK=\frac{9}{2}</cmath> | ||
+ | Then, our answer is <math>9+2=\boxed{11}</math>. | ||
+ | -brianzjk | ||
+ | |||
+ | ==Solution 3 (Video)== | ||
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI | Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI |
Revision as of 17:53, 7 August 2020
Problem
Triangle has side lengths and Circle passes through and is tangent to line at Circle passes through and is tangent to line at Let be the intersection of circles and not equal to Then where and are relatively prime positive integers. Find
Solution 1
-Diagram by Brendanb4321
Note that from the tangency condition that the supplement of with respects to lines and are equal to and , respectively, so from tangent-chord, Also note that , so . Using similarity ratios, we can easily find However, since and , we can use similarity ratios to get Now we use Law of Cosines on : From reverse Law of Cosines, . This gives us so our answer is .
-franchester
Solution 2 (Inversion)
Consider an inversion with center and radius . Then, we have , or . Similarly, . Notice that is a parallelogram, since and are tangent to and , respectively. Thus, . Now, we get that so by Law of Cosines on we have Then, our answer is . -brianzjk
Solution 3 (Video)
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI
See Also
2019 AIME II (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.