Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on complex numbers and triangles.

Complex numbers and triangles – AIME I, 2012


Complex numbers a,b and c are zeros of a polynomial P(z)=\(z^{3}+qz+r\) and \(|a|^{2}+|b|^{2}+|c|^{2}\)=250, The points corresponding to a,b,.c in a complex plane are the vertices of right triangle with hypotenuse h, find \(h^{2}\).

  • is 107
  • is 375
  • is 840
  • cannot be determined from the given information

Key Concepts


Complex Numbers

Algebra

Triangles

Check the Answer


But try the problem first…

Answer: is 375.

Source
Suggested Reading

AIME I, 2012, Question 14

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

here q ,r real a real b,c complex and conjugate pair x+iy,x-iy then a+b+c=0 gives a=-2x and by given condition a-x=y then y=-3x

Second Hint

\(|a|^{2}+|b|^{2}+|c|^{2}\)=250 then 24\(x^{2}\)=250

Final Step

h distance between b and c h=2y=-6x then \(h^{2}=36x^{2}\)=36\(\frac{250}{24}\)=375.

Subscribe to Cheenta at Youtube