Difference between revisions of "2024 AIME I Problems/Problem 10"
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==Problem== | ==Problem== | ||
− | Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. | + | Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. If <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>, <math>AP</math> can be written as the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m + n</math>. |
==Solution 1== | ==Solution 1== |
Revision as of 20:03, 6 February 2024
Contents
[hide]Problem
Let be a triangle inscribed in circle . Let the tangents to at and intersect at point , and let intersect at . If , , and , can be written as the form , where and are relatively prime integers. Find .
Solution 1
From the tangency condition we have . With LoC we have and . Then, . Using LoC we can find : . Thus, . By Power of a Point, so which gives . Finally, we have .
~angie.
Solution 2
We know is the symmedian, which implies where is the midpoint of . By Appolonius theorem, . Thus, we have
~Bluesoul
Solution 3
Extend sides and to points and , respectively, such that and are the feet of the altitudes in . Denote the feet of the altitude from to as , and let denote the orthocenter of . Call the midpoint of segment . By the Three Tangents Lemma, we have that and are both tangents to , and since is the midpoint of , . Additionally, by angle chasing, we get that: Also, Furthermore, From this, we see that with a scale factor of . By the Law of Cosines, Thus, we can find that the side lengths of are . Then, by Stewart's theorem, . By Power of a Point, Thus, Therefore, the answer is .
~mathwiz_1207
Solution 4 (LoC spam)
Connect lines and . From the angle by tanget formula, we have . Therefore by AA similarity, . Let . Using ratios, we have Similarly, using angle by tangent, we have , and by AA similarity, . By ratios, we have However, because , we have so Now using Law of Cosines on in triangle , we have Solving, we find . Now we can solve for . Using Law of Cosines on we have
Solving, we get Now we have a system of equations using Law of Cosines on and ,
Solving, we find , so our desired answer is .
~evanhliu2009
Solution 5
Following from the law of cosines, we can easily get , , .
Hence, , , . Thus, .
Denote by the circumradius of . In , following from the law of sines, we have .
Because and are tangents to the circumcircle , and . Thus, .
In , we have and . Thus, following from the law of cosines, we have
Following from the law of cosines,
Therefore,
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn.org
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.