So here is the problem: you have an acute-angled triangle. You are supposed to find a triangle inscribed within the triangle that has the minimum perimeter.

Fagnano first solved this problem in the year 1775 where he used calculus to arrive at the solution. Many other mathematicians later proved the same using various other methods. But now we will look at an interesting correlation that I discovered between this problem and classical mechanics.

So let's start with a triangle ABC made of three rigid rods. Now consider three beads that are allowed to slide (without friction) on the three rods AB, AC, and BC. Now take a rubber band (whose length is much smaller than the perimeter of the triangle) that passes through the three beads. The arrangement would look something like this:

Let us deviate from reality and say the rubber-band is allowed to slide through the beads with no friction and that it obeys Hooke's law.

We now have a triangle inscribed inside the original triangle whose sides are $x_1$, $x_2$, and $x_3$. Let's say the rubber band has a spring constant $k$. Now the rubber-band is divided into three parts which are in the ratio $x_1 : x_2 : x_3$. We need to find the spring constants for each of the three segments. You can try this as an exercise, but I will tell you that they will be $ \frac{kl}{x_1}$, $ \frac{kl}{x_2}$, and $ \frac{kl}{x_3}$ where $l$ is the length of the rubber-band. The beads try to arrange themselves in such a way that they minimize the elastic potential of the spring. That is, they try to minimize:

$$ \frac{1}{2} k_1 {x_1}^2 + \frac{1}{2} k_2 {x_2}^2 + \frac{1}{2} k_3 {x_3}^2$$

Which can be simplified to:

$$ \frac{1}{2} \frac{kl}{x_1}{x_1}^2 + \frac{1}{2} \frac{kl}{x_2} {x_2}^2 + \frac{1}{2} \frac{kl}{x_3}{x_3}^2 $$

Which is nothing but:

$$ \frac{1}{2} kl (x_1 + x_2 + x_3)$$

By minimizing the spring energy, we are minimizing the perimeter.

For the rubber-band to have the least potential energy, The beads must be in stable equilibrium. That is, all the forces in the beads must be balanced.

Let us allow the bead in rod BC to move while keeping the other two fixed. The bead would experience three forces: two spring forces and a normal force.

The two spring forces would have magnitudes $k_1 x_1$ and $k_2x_2$ which is nothing but $kl$ and $kl$. So we see that the forces exerted by the springs are the same. Let's call that $F$.

Here is the free body diagram for the bead:

For the bead to stay at equilibrium, the horizontal forces must cancel out each other. Therefore

$$ F \text{cos}(\theta_1) = F \text{cos}(\theta_2) $$

Which gives us $$ \theta_{1}=\theta_{2}$$

We can do the same thing for the other beads and get the same relation for each of them.

So we have it: a triangle inscribed inside the original triangle where $$ \theta_{1}=\theta_{2}$$ for each vertex.

And an orthic triangle has the property that $\theta_{1}=\theta_{2}$. You can find the proof for it online, but since the converse is also true, we can say that the orthic triangle is the triangle inscribed within a triangle that has the minimum perimeter.

If you are wondering what an orthic triangle is, it is the triangle that is formed when you join the foot of altitudes.

We finally have it — a physics proof for a mathematical theorem. If you are still not convinced, go ahead and grab a few sticks or pencils and a rubber-band and see for yourself.