# Minimum quadrilateral inscribed in a square

A problem that I’ve seen lately reduces to the following problem:

We have a square, and we put a point on each side of the square. Then we connect the four points to create a quadrilateral. How can we make this quadrilateral have the smallest possible perimeter?

Intuitively, you may believe that this natural, obvious configuration should produce the least perimeter:

### Attempt with Calculus

How can we prove that this indeed gives us the smallest possible perimeter?

A first attempt might be to give variables to the side lengths, and somehow find the minimum perimeter using algebra and calculus tools. So there are four independent points — let’s parameterize them with four variables, and assume the side length of the square is 1:

Then we want to minimize this expression:

$\sqrt{a^2+(1-d)^2} + \sqrt{b^2+(1-a)^2}+ \sqrt{c^2+(1-b)^2}+ \sqrt{d^2+(1-c)^2}$

At this point, it isn’t clear how to proceed — there doesn’t seem to be any way to minimize this expression of four variables.

### Proof by Net

We’ll have to try something different. It’s hard to make sense of anything when there are four independent variables. Instead, if we expand things out a bit, things start to become more manageable:

What we did was reflect the square three times, and each time the square is reflected, the inscribed quadrilateral goes with it. By taking only the relevant parts of the quadrilateral, we get the green path.

Now we might have a solution. If we had a different green path, can we reverse the steps and get the original quadrilateral back? Basically, the following requirements have to be met:

• The path has to cross all three of the internal lines BC, BA, and DA.
• The path’s position on the bottom-most line, DC must be the same when reflected onto the top-most line DC.

With these requirements in mind, the shortest green path that satisfies these requirements is a straight line connecting a point on the bottom left to its reflected point on the top right:

Our intuition at the start was well-founded.

Now notice that this isn’t the only possible shortest path. If we move the entire green line to the left or right, we get a different path of the same length!

For instance, the degenerate ‘quadrilateral’ formed by connecting two opposite corners has the same perimeter as the one we get by connecting the midpoints. Neat, huh?

# Varignon’s theorem proved in one line with vectors

Today I was reading some math book when the author mentions Varignon’s theorem, and gives a proof. The proof was not very long, but it was somewhat confusing. On Wikipedia, several more short proofs were given, but they were all more confusing than need be.

I remembered seeing the theorem proven using vector geometry before, but I couldn’t find the text (nor any other page / book that proves it this way) —

[image shamelessly taken from Wikipedia]

Varignon’s theorem states that in any quadrilateral, if we join the midpoints of the sides, then we get a parallelogram.

In the diagram, it suffices to prove that vector HG is equal to vector EF — vectors must have both the same orientation and length to be equal. This works since any method that proves HG = EF can also prove HE = GF. The proof goes as follows —

$\vec {HG} = \vec{HD} + \vec{DG} = \frac{1}{2} (\vec{AD} + \vec{DC}) = \frac{1}{2} \vec{AC} = \vec{EF}$

And we’re done. (the last step is due to symmetry of HG and EF)