We know the following about the numbers $a$, $b$ and $c$:

$$(a+b)^2=9,\quad(b+c)^2=25, \quad(a+c)^2=81$$

If $a + b + c ≥ 1$, determine the number of possible values for $a + b + c$

  • $\begingroup$ Is there any trick to this, or is it just straightforward counting of possibilities? Also, I assume you meant to specify $a,b,c$ must be integers? $\endgroup$ – Rand al'Thor Mar 27 at 23:04
  • $\begingroup$ There are different possibilities of a, b, and c, for e.g 0+3, and 1+2. $\endgroup$ – HelloWorld1337 Mar 27 at 23:06

We know that:

$$a+b=\pm 3\\b+c=\pm5\\a+c=\pm9$$

It then follows that

by adding them up we get $$ 2(a+b+c) = \pm3\pm5\pm9 $$ Since $a+b+c\ge1$, we need this to be greater or equal to $2$.

So the question boils down to finding

how many choices of the signs make $\pm3\pm5\pm9 \ge2$ true. Clearly $9$ cannot have a minus sign, so we have $9\pm3\pm5 \ge2$. This only fails when both remaining signs are minus signs, so there are three possibilities left.

The solutions are therefore:

$$a+b+c = (9+3+5)/2 = 17/2\\a+b+c = (9+3-5)/2 = 7/2\\a+b+c = (9-3+5)/2 = 11/2$$

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