How can we get the approximate values $a,b,c$?

Question

The condition and relation are the followings :

$0 < a,b,c < 1$
$a + b + c = 1$
$(1-a)^2 + b^2 + c^2 =1 $

It seems that are infinitely many solutions. This can be osberved by substituting 2. in 3. which leads to $b^2+bc+c^2=1$ and the positive solutions of this equation lies on a kind of ellipse centered in the origin. Do you want them all or just one? — Surb, Mar 30 '15 at 13:32
The complete conditions are equivalent to $b^2+bc+c^2=0.5$ with $0<b,c<1$. — user26486, Mar 30 '15 at 13:35
@user31415 I think they can be reduced further to $b^2+bc+c^2=1/2$ with $0<b,c$. — Nathanson, Mar 30 '15 at 13:39
@user31415 The points on the ellipse already satisfy the inequality $<1$. The ellipse condition makes that inequality superfluous. — Nathanson, Mar 30 '15 at 13:43
That's true - b and c will both be < $\frac{1}{\sqrt{2}}$ anyway. — J Richard Snape, Mar 30 '15 at 13:45

score 0 · Answer 1 · answered Mar 30 '15 at 13:39

You don't need to approximate. You can simplify 2 and 3 together to give

$b^2 + bc + c^2 = \frac{1}{2}$

Combine that with 1 and you get an infinite set of solutions, one that has a completely defined relation between $b$ and $c$, with actually tighter limits on $b$ and $c$. $a$ is always just $1 - (b+c)$.

How can we get the approximate values $a,b,c$?

1 Answers1