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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Triangle and Trigonometry.

Point P is located inside triangle ABC so that angles PAB,PBC and PCA are all congruent. The sides of the triangle have lengths AB=13, BC=14, CA=15, and the tangent of angle PAB is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

- is 107
- is 463
- is 840
- cannot be determined from the given information

Triangles

Angles

Trigonometry

But try the problem first...

Answer: is 463.

Source

Suggested Reading

AIME, 1999, Question 14

Geometry Revisited by Coxeter

First hint

Let y be the angleOAB=angleOBC=angleOCA then from three triangles within triangleABC we have \(b^{2}=a^{2}+169-26acosy\) \(c^{2}=b^{2}+196-28bcosy\) \(a^{2}=c^{2}+225-30ccosy\) adding these gives cosy(13a+14b+15c)=295

Second Hint

[ABC]=[AOB]+[BOC]+[COA]=\(\frac{siny(13a+14b+15c)}{2}\)=84 then (13a+14b+15c)siny=168

Final Step

tany=\(\frac{168}{295}\) then 168+295=463.

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- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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