# 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
4 years, 5 months ago

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How could we rewrite this using cyclic sums?

- 4 years, 5 months ago

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 $$(-\pi,\pi]$$: $$-\pi/2,2\arctan(1/3),2\arctan(1/2),\pi$$.

- 4 years, 5 months ago

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

- 4 years, 5 months ago