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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

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.

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