# Proving that a square inscribed in a circle has a strictly smaller area

I'm trying to (for self-study) follow a proof that $$\pi \neq 4$$. The exercise allows me to use basic geometric facts and constructions, but I can't use any specific facts about $$\pi$$.

The first step of the instructor's solution considers a circle of radius $$1$$ inscribed in a square of side length $$2$$. He says that "we will admit that the area of the circle is less than or equal to the area of the square," but I don't know how to prove this without appealing to the properties of $$\pi$$. The standard way I would convince myself of this is as follows:

The circle has radius $$1$$ and therefore area $$\pi \cdot 1^2 = \pi$$. The square has area $$2^2 = 4$$. The ratio of the area of the circle to the area of the square is then $$\frac{\pi}{4}$$. As $$\pi < 4$$, $$\frac{\pi}{4} < 1$$, hence the area of the circle is strictly smaller than the area of the square.

I can't use this argument in this specific proof because I'm using a specific fact about $$\pi$$. My question, therefore, is: is there a way to notice that the area of the inscribed circle is less than or equal to the square without knowing anything specific about $$\pi$$? The other argument is to just look at the picture and say, "well, clearly it's smaller because the square takes up every bit of area of the circle, plus some more," but that seems very un-rigorous to me.

Your "look at the picture" argument is in fact essentially correct.

To make it a little more formal, find a small number $$s$$ such that a square of side $$s$$ fits in the corner of a square of side $$2$$ without overlapping the inscribed circle. Then clearly $$\text{area of circle} + s^2 \le 4.$$

• Frankly, I would drop the "essentially" and just call it correct. It's only un-rigorous if you're going to go to the (considerable) trouble of introducing measure theory, and even then, this is just the monotonicity axiom ($A \subseteq B \implies \mu(A) \le \mu(B)$). You don't really need to introduce the $s^2$ argument, although I suppose it does no harm. Commented Mar 1, 2023 at 3:45

We can suppose that both the circle and the square are centered at the origin.

If the square has sides of length two, then the distance between the center and any of the vertices is $$\sqrt{2}$$ (this is by Pythagoras' theorem). On the other hand, the radius of the circle is $$1$$. As $$1 < \sqrt{2}$$ we have that $$l=\sqrt{2}-1 > 0$$.

Consider the triangle with vertices $$(1,1), (1-l,1)$$ and $$(1,1-l)$$. The triangle is between the circle and the square, and it's area is $$\frac{(1-l)^2}{2}>0$$. Let's call this triangle $$T$$, the circle $$C$$ and the square $$S$$, then $$area(C) < area(T)+area(S) < area(S).$$

PS: A drawing will help a lot here.