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($a,b,c$) are not necessarily integers. Find the difference between the maximum possible and the minimum possible value of $c$.

I have seen a solution to the problem using the Cauchy-Schwarz inequality, but is there any other way to do this? And please keep it to a level where I can understand? Thanks.

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A very bad solution: The solution set of the system is the intersection of a plane and a sphere, hence is a circle. So you can parametrize it using only one parameter t. Then study c(t). – Taladris May 6 '13 at 2:09
up vote 2 down vote accepted

another way is:

$(a+b)^2=a^2+b^2+2ab \le a^2+b^2+a^2+b^2 \to a^2+b^2 \ge \dfrac{(a+b)^2}{2}$

$12=a^2+b^2+c^2 \ge \dfrac{(a+b)^2}{2}+c^2=\dfrac{(2-c)^2}{2}+c^2 \to 3c^2-4c+4 \le 24 \to 3c^2-4c-20 \le 0 \to (3c-10)(c+2) \le 0 \to -2 \le c \le \dfrac{10}{3}$

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Thank you very much! – Ovi May 6 '13 at 1:33

We are looking at where a plane that has symmetry in the $3$ variables meets a sphere.

The highest and lowest points have equal $x$ and $y$ coordinates. So we get the equations $2a+c=2$, $2a^2+c^2=12$. Eliminate $a$. We have $4a^2=(2-c)^2$ and $4a^2=24-2c^2$, giving $3c^2-4c-20=0$. Now we can calculate the two values of $c$. They even happen to be rational.

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Wouldn't they only have one coordinate in common? If the sphere is centered at the origin and an inclined plane intersects it, it seems that the points of intersection would both be in the xz plane, and they would only have y in common. – Ovi May 6 '13 at 0:44
It is just a question of tilting one's head to see the geometry. If one is more algebraically minded, use Lagrange multipliers. We get $\lambda +2\mu a=0$, $\lambda+2\mu b=0$, $1+\lambda+2\mu c=0$, giving $a=b$. – André Nicolas May 6 '13 at 1:43
Haha I have no idea about Lagrange multipliers, and I've never learned 3D geometry, but from what I know the situation would look like this:… If you draw the standard x and y axis, it looks like they will have totally different x, y coordinates. It looks like there is only one dimension in which they can have a common coordinate. I know your answer is right, but I am just trying to understand it. – Ovi May 6 '13 at 2:25

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