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Find all rational number $a,b,c$ satisfy:

$$a+b+c=abc$$

I try to change this in different forms like $(ab-1)c = a+b$, $(ac-1)b = a+c$, $(cb-1)a = b+c$ etc but it won't help...

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The original draft of this answer seems not to be enough:

If the ratios tangents of half-angles are rational, does that mean the ratios of tangents of the angles not divided by $2$ are rational. It would seem not. Then double the angles, and get tangents $$ \frac{2w}{1-w^2}\text{ and }\frac{2x}{1-x^2}. $$ Is the ratio of these rational? That ratio is $\dfrac{1-w^2}{1-x^2}$. Is that rational if $\dfrac{w}{x}$ is rational? E.g. is it rational if $w/x=3$? In other words, must $(1-9x^2)/(1-x^2)$ be rational regardless of the value of $x$? It would seem not, so there's more work to do on this.

The original draft of this answer appears below:

If $\alpha+\beta+\gamma=\pi$ then $\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$, and I think you should be able to show that if $a,b,c$ satisfy the proposed identity, then suitable $\alpha,\beta,\gamma$ exist so that $a=\tan\alpha$ etc. According to the law of tangents, the ratios of sums and differences of the sides of a triangle are rational only if the same is true of the tangents of half the sums and differences of the opposite angles. Via trigonometric identities I think you should be able to show that the ratios of sides are rational iff the ratios of tangents of opposite angles are rational. In that way, you should be able to reduce the problem to that of finding non-right-angled triangles in which the ratios of lengths of sides are rational. Then just remember the triangle inequality. This gives you lots of solutions.

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