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.
Triangles
Angles
Trigonometry
Answer: is 463.
AIME, 1999, Question 14
Geometry Revisited by Coxeter
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
[ABC]=[AOB]+[BOC]+[COA]=\(\frac{siny(13a+14b+15c)}{2}\)=84 then (13a+14b+15c)siny=168
tany=\(\frac{168}{295}\) then 168+295=463.
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