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# Complex Numbers and Triangles | AIME I, 2012 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Complex Numbers and Triangles.

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

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.

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