Difference between revisions of "2019 AIME II Problems/Problem 11"
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So median <cmath>AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot \frac{AK^2}{7\cdot 9} \implies AK = \frac{9}{2}.</cmath> | So median <cmath>AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot \frac{AK^2}{7\cdot 9} \implies AK = \frac{9}{2}.</cmath> | ||
− | ''' | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
==See Also== | ==See Also== |
Revision as of 04:17, 31 August 2022
Contents
[hide]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
- The motivation for using the Law of Cosines ("LoC") is after finding the similar triangles it's hard to figure out what to do with and yet we know which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealing with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.
-First
11111111:L)xiexie
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 (Death By Trig Bash)
14. Let the centers of the circles be and where the has the side length contained in the circle. Now let This implies by the angle by by tangent. Then we also know that Now we first find We use law of cosines on to obtain Then applying law of sines on we obtain Using similar logic we obtain
Now we know that Thus using law of cosines on yields While this does look daunting we can write the above expression as Then factoring yields The area Now is twice the length of the altitude of so we let the altitude be and we have Thus our desired length is
Solution 4 (Video)
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI
Solution 5 (Olympiad Geometry)
By the definition of , it is the spiral center mapping , which means that it is the midpoint of the -symmedian chord. In particular, if is the midpoint of and is the reflection of across , we have . By Stewart's Theorem, it then follows that
Solution 6 (Inversion simplified)
The median of is
Consider an inversion with center and radius (inversion with respect the red circle). Let and be inverse points for and respectively.
Image of line is line lies on this line.
Image of is line (circle passes through K, C and is tangent to the line at point Diagram shows circle and its image using same color).
Similarly, is the image of the circle ).
Therefore is a parallelogram, is median of and Then, we have . with coefficient
So median vladimir.shelomovskii@gmail.com, vvsss
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.