How to find minimal perimeter of a triangle.

Question

I have area A and length of the base s.
I need to describe a triangle minimal perimeter using this information.
from given equation I can find height.
and final equation will look like this
Since my perimeter is equal to
$ s + \sqrt{x^2 + h^2} + \sqrt{(x - s)^2 + h^2} $
problem is to change orthocenter of h (x) in order to minimize equation above?
How do I do that?

Answer 1

If I understand the problem correctly, both $s$ and $h$ are fixed, and you want to choose $x$ so that $p(x)=s + \sqrt{x^2 + h^2} + \sqrt{(s-x)^2 + h^2}$ is minimized. Note that $x$ can be negative, in which case one of the base angles is obtuse.

Computing $p'(x)$ and simplifying gives a fraction whose numerator is $$ (x-s)\sqrt{h^2+x^2}+x \sqrt{h^2+(s-x)^2}.$$ Set equal to zero and solve for $x$, giving $x=\frac{s}{2}$. Thus the minimal-perimeter triangle is isosceles on its base $s$.

Answer 2

Do you know any other way to approach this problem?

For an alternative approach, note that the given area and base determine the altitude $h=2A/s\,$, so the third point must lie on a line parallel to the base and at distance $h$ from it. Also, the base is fixed, so minimizing the perimeter of the triangle is equivalent to minimizing the sum of distances to the two endpoints of the base.

Once reformulated this way, the question reduces to a special case of a more general "shortest path" problem which has a simple and purely geometric proof due to Heron, see for example this or this.

In the case at hand, let $AB$ be the fixed side, the dashed line $L$ the parallel to $AB$ at distance $h$ on which the third vertex $C$ must lie, and $B'$ the symmetric of $B$ across $L$. Let $C = AB' \cap L\,$, then:

• $\triangle ABC$ is isosceles in $C$;

• for $\,\forall \,C' \in L$ the sum of distances $C'A+C'B = C'A+C'B'\ge AB' = CA+CB$.