Math Puzzle: Totaling up A,B, C [closed]

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 $$a + b + c$$

• 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? Mar 27 '20 at 23:04
• There are different possibilities of a, b, and c, for e.g 0+3, and 1+2. Mar 27 '20 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:

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