If one observes the two solutions for a and b, they are appear similar to a = e^(i*pi/3) and b = exp(-i*pi/3) (or vice versa) in complex coordinates. Knowing that cos(t) = (e^(i*t)+e^(-i*t))/2 and DeMoivre's Theorem, the answer just deduces to: 2*cos(11*pi/3) = 2(1/2) = 1.
I did a different approach, i solved for the values of a and b and this is what i got: a^11+b^11=(1+sqrt(3))/2)^11+(1-sqrt(3))÷2)^11 and it is 30 something .
Tbh, I just solved for a and b and used that to get the numerical solution on my calculator. I don't have paper to write on at the moment. They're (1+-sqrt(3))/2, interchangeably. It's nice that the sums of their powers reliably cancel out the irrational part.
What a stupid way to solve a simple problem? Russian math Olympiad? Ya sure! a+b=1 and ab=-1/2. Solve the quadratic equation for a and b. Then evaluate the result.
@@vlad_bizin It's not 1950s. I use a calculator. In Olympiad there is no questions like that if you can use a calculator to evaluate an expression. This is for elementary schools.
I believe you don’t have to go to a quadratic equation because ab = -1/2 => 2ab = -1, subtract from second equation, a^2 - 2ab + b^2 = 3, (a-b)^2 = 3, a-b = sqrt(3) or - sqrt(3). Now add to the first equation: 2a = 1 + sqrt(3), 1 - sqrt(3), a = (1 + sqrt(3))/2, (1 - sqrt(3))/2. The rest is trivial.
If one observes the two solutions for a and b, they are appear similar to a = e^(i*pi/3) and b = exp(-i*pi/3) (or vice versa) in complex coordinates. Knowing that cos(t) = (e^(i*t)+e^(-i*t))/2 and DeMoivre's Theorem, the answer just deduces to:
2*cos(11*pi/3) = 2(1/2) = 1.
Russian are among the top mathematicians specially back to Soviet Union era
to be fair, it was either do Math, plow fields, or Gulag.
@@UZPvNUCaaQdFWell gulag didn't happen after 1960s !
I did a different approach, i solved for the values of a and b and this is what i got:
a^11+b^11=(1+sqrt(3))/2)^11+(1-sqrt(3))÷2)^11 and it is 30 something .
Tbh, I just solved for a and b and used that to get the numerical solution on my calculator. I don't have paper to write on at the moment. They're (1+-sqrt(3))/2, interchangeably. It's nice that the sums of their powers reliably cancel out the irrational part.
You can't just give a numerical solution in olympiads.
What a stupid way to solve a simple problem? Russian math Olympiad? Ya sure! a+b=1 and ab=-1/2. Solve the quadratic equation for a and b. Then evaluate the result.
Then come on, do it. I will watch you raise the roots of the system to power 11 😊
@@vlad_bizin It's not 1950s. I use a calculator. In Olympiad there is no questions like that if you can use a calculator to evaluate an expression. This is for elementary schools.
I believe you don’t have to go to a quadratic equation because ab = -1/2 => 2ab = -1, subtract from second equation, a^2 - 2ab + b^2 = 3, (a-b)^2 = 3, a-b = sqrt(3) or - sqrt(3). Now add to the first equation: 2a = 1 + sqrt(3), 1 - sqrt(3), a = (1 + sqrt(3))/2, (1 - sqrt(3))/2. The rest is trivial.
@@bayareapianist If you think you are allowed to use a calculator, then you know nothing about olympiads.
@@Tommy_007 haha I was one of the finalist! There are ppl here in YT who just want to trash other.
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BOYKOTT RUSSIA