Difference between revisions of "2019 AIME II Problems/Problem 11"
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− | ==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 18:53, 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.