Difference between revisions of "2019 AIME II Problems/Problem 11"
Brendanb4321 (talk | contribs) m (→Problem) |
|||
Line 27: | Line 27: | ||
Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math>. This gives us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>. | Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math>. This gives us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>. | ||
-franchester | -franchester | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=nJydO5CLuuI | ||
+ | |||
+ | Please like, share, and subscribe! | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=10|num-a=12}} | {{AIME box|year=2019|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:58, 6 August 2019
Contents
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
On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=nJydO5CLuuI
Please like, share, and subscribe!
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.