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Mischievous Equality - Problem 5

Find all triplets of real numbers \((a,b,c)\) with \(a^2+b^2=c^2\) satisfying

\[2(a^3+b^3+c^3)=ab(3a+3b-4c)+bc(3b+3c-4a)+ca(3c+3a-4b)\]

Note by Cody Johnson
3 years, 9 months ago

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How could we rewrite this using cyclic sums?

Cody Johnson - 3 years, 9 months ago

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We can assume \(a=c\sin\theta,b=c\cos\theta\) then solve: \[3(\sin^3\theta+\cos^3\theta+1)+18(\sin\theta\cos\theta)=(\sin\theta+\cos\theta+1)^3,\] that leads to: \[ \cos(\theta/2)\cdot\sin(\pi/4+\theta/2)\cdot (\cos(\theta/2)-3\sin(\theta/2))\cdot(\cos(\theta/2)-2\sin(\theta/2))\cdot (\sin \theta+\cos \theta-2)=0,\] hence \(\theta\) can take four values in \((-\pi,\pi]\): \(-\pi/2,2\arctan(1/3),2\arctan(1/2),\pi\).

Jack D'Aurizio - 3 years, 9 months ago

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So what are our possible values of \((a,b,c)\)?

Cody Johnson - 3 years, 9 months ago

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