In a triangle $ABC$ the base $AB=6 \ cm$. The vertex $C$ varies such that the area is always equal to $12 \ cm^2$. Find the minimum value of the sum of $CA+CB$.
My guess was to assume $ABC $ to be an isosceles triangle. After working out the numbers I get $AC+BC=10$ for this triangle and as it turns out this is the answer.
But how is this done? I am guessing I need to use the rule of cosines?. But this confuses me. Need help!
The heights always need be $C_y = 4$. You want to minimize the sum of the distances. This is $d = \sqrt{4^2 + x^2} + \sqrt{4^2 + (x+6)^2}$.
$$ d'(x) = \frac{2x}{2 \sqrt{4^2 + x^2}} + \frac{2(x+6)}{2\sqrt{4^2 + (x+6)^2}} $$
We look for $d'(0)$:
$$ \frac{2x}{2 \sqrt{4^2 + x^2}} + \frac{2(x+6)}{2\sqrt{4^2 + (x+6)^2}} = 0 $$
You could do the algebra or some analysis, but the answer is clearly $x=-3$. Check by
$$ \frac{-3}{\sqrt{4^2 + (-3)^2}} + \frac{(-3+6)}{\sqrt{4^2 + (-3+6)^2}} = 0. $$
So the center point is the minimum sum of distances. This is using Calculus. Maybe you could find a geometric argument.
The area of $ABC$ is $12$ iff the distance of $C$ from the $AB$ line is $4$, i.e. iff $C$ belongs to a fixed line $\ell\parallel AB$. On the other hand,
$CA+CB$ attains its minimum value iff the ellipse with foci at $A$ and $B$ through $C$ it tangent to $\ell$. By symmetry (with respect to the perpendicular bisector of $AB$) it follows that the only minimum is achieved at $CA=CB$. By the Pythagorean theorem, in such a case $CA=CB=5$.
