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.
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.
