Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.
Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA. Find PC.
Angles
Algebra
Triangles
But try the problem first...
Answer: is 33.
AIME I, 1987, Question 9
Geometry Vol I to Vol IV by Hall and Stevens
First hint
Let PC be x, \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA=120 (in degrees)
Second Hint
Applying cosine law \(\Delta\)APB, \(\Delta\)BPC, \(\Delta\)CPA with cos120=\(\frac{-1}{2}\) gives
\(AB^{2}\)=36+100+60=196, \(BC^{2}\)=36+\(x^{2}\)+6x, \(CA^{2}\)=100+\(x^{2}\)+10x
Final Step
By Pathagorus Theorem, \(AB^{2}+BC^{2}=CA^{2}\)
or, \(x^{2}\)+10x+100=\(x^{2}\)+6x+36+196
or, 4x=132
or, x=33.
Math Olympiad Program... taught by Olympians and researchers
From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:
