Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on complex numbers and triangles.
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 \(h^{2}\).
Complex Numbers
Algebra
Triangles
Answer: is 375.
AIME I, 2012, Question 14
Complex Numbers from A to Z by Titu Andreescue
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
\(|a|^{2}+|b|^{2}+|c|^{2}\)=250 then 24\(x^{2}\)=250
h distance between b and c h=2y=-6x then \(h^{2}=36x^{2}\)=36\(\frac{250}{24}\)=375.
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