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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

But try the problem first…

Source

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.