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fruit math!

If the image above doesn't show, the equation could be rendered as follows:

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4$$

How would you go about solving this? Trying to do LCD and substituting does not seem to work well.

Dave
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Steve Byer
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    Well, it is apparent that this problem cannot have exactly one solution: if $(a,b,c)$ is a solution, then $(ka,kb,kc)$ with $k\ne0$ is a solution as well. –  Mar 12 '17 at 04:39
  • Finding all values is a little daunting. However, the question asks for any values. If you'd like to just come up with one or two solutions choose any apple and banana (not both zero) you'd like and solve for pineapple. – MattW Mar 12 '17 at 04:48
  • It's only one equation, with three unknowns - the system is likely wildly underdetermined. Try picking anything you like for two and solve for the other. I tried setting apple to 0 and pineapple to 1, and got an irrational number for banana (OK, but probably not ideal). – Reese Johnston Mar 12 '17 at 04:49
  • Oh, and I suppose apple and banana should not be additive inverses either. – MattW Mar 12 '17 at 04:56
  • The usual statement of the problem requires the unknowns to be positive integers. See e.g. https://mathoverflow.net/questions/278747/is-the-set-of-integers-of-the-form-a-bcb-acc-ab-computable – Gerry Myerson Sep 16 '20 at 11:35
  • Related, with some more information on the set of solutions: https://math.stackexchange.com/questions/2779545/ask-for-the-rational-roots-of-fracabc-fracbac-fraccab-4 – Thomas Oct 23 '22 at 13:15

1 Answers1

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Consider the case where $apple = 0$ and $banana = 1$. Denote the pineapple by $p$. Then we have,

$$\frac{0}{1+p}+\frac{1}{0+p}+\frac{p}{1+0} = 4$$ $$\frac{1}{p}+p = 4$$ $$1+p^2 = 4p$$ $$p^2-4p+1 = 0$$ $$p = \frac{4\pm \sqrt{16-4}}{2} = 2\pm \sqrt{3}$$

Thus $$(apple=0, banana=1, pineapple= 2+\sqrt{3})$$ $$(apple=0, banana=1, pineapple= 2-\sqrt{3})$$and are two possible solutions in the real numbers.

benguin
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