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

Answer: is 375.

AIME I, 2012, Question 14

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

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.

## Other useful links

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