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Equations and Complex numbers | AIME I, 2019 Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

Equations and Complex numbers – AIME 2019


For distinct complex numbers \(z_1,z_2,……,z_{673}\) the polynomial \((x-z_1)^{3}(x-z_2)^{3}…..(x-z_{673})^{3}\) can be expressed as \(x^{2019}+20x^{2018}+19x^{2017}+g(x)\), where g(x) is a polynomial with complex coefficients and with degree at most 2016. The value of \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\) can be expressed in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n

  • is 107
  • is 352
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Complex Numbers

Integers

Check the Answer


Answer: is 352.

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


here \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\)=s=\((z_1z_2+z_1z_3+….z_1z_{673})+(z_2z_3+z_2z_4+…+z_2z_{673})\)

\(+…..+(z_{672}z_{673})\) here

P=\((x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2)…(x-z_{673})(x-z_{673})(x-z_{673})\)

with Vieta’s formula,\(z_1+z_1+z_1+z_2+z_2+z_2+…..+z_{673}+z_{673}+z_{673}\)=-20 then \(z_1+z_2+…..+z_{673}=\frac{-20}{3}\) the first equation and \({z_1}^{2}+{z_1}^{2}+{z_1}^{2}+{z_1z_2}+{z_1z_2}+{z_1z_2}+…..\)=\(3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})\)+\(9({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})\)=\(3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})\)+9s which is second equation

here \((z_1+z_2+…..+z_{673})^{2}=\frac{400}{9}\) from second equation then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2s=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})=\frac{400}{9}\)-2s then with second equation and with vieta s formula \(3(\frac{400}{9}-2s)+9s\)=19 then s=\(\frac{-343}{9}\) then |s|=\(\frac{343}{9}\) where 343 and 9 are relatively prime then 343+9=352.

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