Difference between revisions of "2019 AIME II Problems/Problem 11"
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| − | ==Solution 2== | + | ==Solution 2 (Video)== |
| − | + | Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI | |
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==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 17:19, 31 May 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
![[asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("$A$",A,dir(105)); label("$B$",B,dir(-135)); label("$C$",C,dir(-75)); dot((2.68,2.25)); label("$K$",(2.68,2.25),dir(-150)); label("$\omega_1$",(-6,1)); label("$\omega_2$",(14,6)); label("$7$",(A+B)/2,dir(140)); label("$8$",(B+C)/2,dir(-90)); label("$9$",(A+C)/2,dir(60)); [/asy]](http://latex.artofproblemsolving.com/0/a/4/0a43806f045704f060a925ba2a72fcf6b8be42c7.png)
-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, ![\[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\]](http://latex.artofproblemsolving.com/8/8/6/8867ab7875bbe6e6f649ea78a84f8317d641e110.png)
Also note that 
, so 
. Using similarity ratios, we can easily find ![\[AK^2=BK*KC\]](http://latex.artofproblemsolving.com/b/7/e/b7e51130904658596ee1ff6506f1d4db24711d75.png)
However, since 
and 
, we can use similarity ratios to get ![\[BK=\frac{7}{9}AK, CK=\frac{9}{7}AK\]](http://latex.artofproblemsolving.com/2/7/0/270c4befa6737344711dd2c3bc280ddb1146d852.png)
Now we use Law of Cosines on 
: From reverse Law of Cosines, 
. This gives us ![\[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\]](http://latex.artofproblemsolving.com/6/1/6/61697df0eb91e3f51d7716219b36fe134ea86702.png)
![\[\implies \frac{196}{81}AK^2=49\]](http://latex.artofproblemsolving.com/f/0/6/f06b39eb1136cb0f7857a6633b223882a2df86c4.png)
![\[AK=\frac{9}{2}\]](http://latex.artofproblemsolving.com/8/d/2/8d25e0f67f32c7dcc9c674a12a21fd74c66e1b83.png)
so our answer is 
.
-franchester
Solution 2 (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. 
