An interesting property of a particular set of triples - multiplying two and adding the other always gives 1 
Find all triples of real numbers such that multiplying any two in a
  triple and adding the third always gives $1$.

When will this be the case? How can we find all such triples?
So far, I've let the numbers be $a$, $b$ and $c$. 
Therefore, $ab+c$ etc must $= 1$ but how can I restrict the possibilities to find all possible triples? 
I seem to think this has something to do with 1s and 0s, eg. $0,0,1$ or $0,1,1$
Many thanks guys!!

You have three equations and three variables to solve for. The three equations are $$ab + c=1$$ $$ac +b=1$$ $$bc + a=1$$
Using the first equation and solving for $c$, we get $$c = 1-ab$$
This now reduces to two equations $$a(1-ab)+b=a+b-a^2b=1$$ $$b(1-ab)+a=a+b-ab^2=1$$
Subtracting the second equation from the first, I get $$-a^2b+ab^2=0$$ This can be factored as $$ab(b-a)=0$$
Which means that there are three possible cases $a=0, b=0,$ and $a=b$.
If $a=0$, it must also be true that $b=c=1$ in order to satisfy the original three equations. Similarly, if $b=0$, it must be true that $a=c=1$. If $a = b \not = 0$, there is the cubic $$2a-a^3=1$$ that must be solved. There are three solutions $$a= \frac{-1 \pm \sqrt{5}}{2}, 1$$
This means that all solutions to the problem are $(0, 1, 1), (1, 0, 1), (1, 1, 0), \left(\frac{-1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}\right)$
