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> | + | 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>. Find <math>AP</math>, if <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>. |
==Solution 1== | ==Solution 1== |
Revision as of 14:09, 3 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 . Find , if , , and .
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
Well 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
Video Solution 1 by OmegaLearn.org
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 |
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