Find the minimum perimeter of a triangle given one of the sides 1250/h^2 and its altitude 2h.

Question

Find the minimum perimeter of a triangle given one of the sides 1250/h^2 and its altitude 2h.

I already tried an attempt by using Heron's formula. Then I looked at some triangle constructions similar problems. But to construct a triangle given its base and altitude you need the angle in front of the base or the sum of the two remaining sides .

Answer 1

But to construct a triangle given its base and altitude you need the angle in front of the base or the sum of the two remaining sides.

True, but you need to set a parameter $\theta$ the angle in front of the base or $S$ the sum of the two remaining sides. Then you can study the function, and hopefully you will find a minimum for $\frac{\partial P}{\partial \theta} = 0$ or $\frac{\partial P}{\partial S} = 0$ with $P$ the perimeter.

Answer 2

There are 2 cases possible with the lengths of the sides that you have given.

Case 1: Altitude and given side share a vertex $A$ and the altitude is not for the given side.

For this case, the minimum perimeter is that of a right triangle $T$ with the hypotenuse equal to the side with length $1250/h^2$ and the altitude with length $2h$ equal to one of the sides of the triangle. This is because if you consider, instead, a triangle in which the vertex of the altitude other than $A$ does not coincide with one of the vertices of the side for which it is the altitude you will get a triangle with a larger perimeter than that of triangle $T$. This perimeter comes out to, using Pythagoras theorem, $\frac{1250}{h^2} + 2h + \frac{\sqrt{1562500-4h^6}}{h^2}$.

Case 2: Altitude and side do not share a vertex

In this case, the given altitude is perpendicular to the given side. It turns out that the minimum perimeter in this case is that of an isosceles triangle with base the given side, altitude the given altitude, and where the remaining two sides are equal. Each of the two equal sides has length $\frac{\sqrt{4h^6+390625}}{h^2}$. So the perimeter is $2\frac{\sqrt{4h^6+390625}}{h^2}+\frac{1250}{h^2}$

Case 1 imposes a constraint on $h$ viz. $1250/h^2>2h \implies h<8.55$. If you check the values of the expressions for perimeters for small values of $h$ case 2 gives you a lower number, and for large values of $h$ case 1 gives you a lower number. At $h=6.5383$ both numbers are roughly the same. So for $h<6.5383$ the answer is the expression for perimeter in case 2, and for $6.5383<h<8.55$ the answer is the expression for perimeter in case 1.