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# Department of Science, Math, & Technology

E = mcï¿½
Fruit Math
Posted: Posted September 1st
Edited September 1st by Xhin
 In response to some of the dumb fruit math things that pop up on facebook occasionally, here's one that's a bit more challenging: I'm particularly interested in your means of solving this probelm!
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I'm not good in mathematics but what is fruit math?

Edited September 1st by Big One 100
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You have to find numbers to substitute in for the fruit such that the expression makes sense mathematically.

So, for example, to start out, I used the first letter of each fruit name as an algebraic variable: (A/(B+C)) + (B/(A+C)) + (C/(A+B)) = 4. From there, though, the paths I've been taking to try to find this answer are ridiculously convoluted and none of them are leading me anywhere, so I'll have to leave it up to the eggheads to figure this one out.

Posted September 1st by Black Yoshi
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I'm in the same boat honestly.

I ran across this problem elsewhere and fruitized it, but it's much older.

I'll share my own notes soon.

Posted September 1st by Xhin
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Xhin
Ground's what's around

(A/(B+C)) + (B/(C+A)) + (C/(A+B)) = 4

A/(B+C) = 4 - (B/(C+A)) + (C/(A+B))
A = (4 - (B/C+A) + (C/(A+B)))*(B+C)
A = (4B - (B/C+A)B + (C/(A+B))B) + (4C - (B/C+A)C + (C/(A+B))C)
A = (4B - ((B^2)/(CB+AB)) + (CB/(AB+B^2)) + (4C - ((BC/C^2+AC) + (C^2/(AC+BC))
A = 4B + 4C - BC(AB+B^2) + BC/(C^2+AC) + B^2/(CB+AB) + C^2/(AC+BC)

I'll continue this when less drunk maybe

Posted September 1st by Ghowilo
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You have one equation and three variables, so there aren't enough constraints to give a unique solution.

For example, I let A=0, B=1, and then solved the resulting quadratic to get C = 2+sqrt(3) as one solution set.

Posted September 2nd by EN
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I tried a similar approach: set C = 0, then solve for A and B. I wound up with three solutions for B, of which one was exactly 4, so I used that to avoid endless decimals and came up with A = 8 +- 4*sqrt(3).

Posted September 2nd by Black Yoshi
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And then a thought occurs: no one said we can't make all three variables equal to the same number, right? So, in that case, you'd basically have (X/2X) + (X/2X) + (X/2X) = 4, or (3/2)x = 4, which is far, far simpler than trying to solve for three different variables in one equations.

In this case, (3/2)x = 4 --> x = 4*(2/3) = 2.66666666... = 8/3

Unless we have to have unique numbers for each variable, in which case I'm right back at square one.

Posted September 2nd by Black Yoshi
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(X/2X) + (X/2X) + (X/2X) does not give (3/2)X though. In each of those fractions the X's would cancel out, and you would have 3/2 = 4, which is a contradiction. So all of the variables can't be equal.

Posted September 2nd by EN
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Ah damn, thanks for catching that. I guess I was getting a bit ahead of myself, or maybe my algebra's a bit rustier than I thought.

Posted September 3rd by Black Yoshi
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