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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.

- is 107
- is 33
- is 840
- cannot be determined from the given information

Angles

Algebra

Triangles

But try the problem first...

Answer: is 33.

Source

Suggested Reading

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.

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

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