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There are a few steps in a constructive proof that $\pi \neq 4$ that I cannot fully justify. Here is a sketch of the proof, with the step I don't fully understand bolded.

Inscribe a circle of radius $1$ inside a square of side length $2$. The area of the circle is $\pi$ and the area of the outer square is $4$. Call the center of the circle $O$ and the upper left corner of the square $A$, and draw a line segment $OA$ between $O$ and $A$. Along with the radial length from $O$ to the left side of the square (of unit length), this forms a right triangle with both legs having length $1$. So by the Pythagorean theorem, $|OA| = \sqrt{2}$. Let $C$ be the pint at which $OA$ intersects the circle. Then $|OC| = 1$, the radius of the circle, so $|AC| = \sqrt{2} - 1$. Now, draw a vertical segment from $C$ to the boundary to the square, and call the point where it touches the square $B$; this forms a right angle. Extending a horizontal segment from $C$ to the left edge of the square forms a quadrilateral $ABCD$. This quadrilateral is a square with diagonal length $\sqrt{2} - 1 > 0$, so its area is positive but does not contribute to the area of the circle. This means that the combined area of the circle with center $O$ and the square $ABCD$ is less than or equal to the area of the square which is $4$. Let $c$ denote the area of $ABCD$. Then $\pi + c \leq 4$, so $\pi \leq 4 - c < 4$.

I'm not fully sure if my language is sufficiently precise. I'm trying to justify that I'm actually forming a square and that these angles are in fact right angles. In order to show that, I believe I can show one of a few possible facts: that all the angles are right angles or that the diagonals are perpendicular. The figure itself is a square, so $\langle A$ is $90$ degrees. I tried to justify drawing segments from $C$ upward to the square and to the left of $C$ gives a right angle, so that would give $\langle C = 90$ degrees. I don't know how to justify, other than by looking at the picture, that $\langle B$ and $\langle D$ are $90$ degrees. Despite all of this, I think I don't need the figure to be a square (even though it surely is): the fact that it has a diagonal of positive length is enough to say that it contributes positive area.

Mathematical Endeavors
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    A picture would help a lot – lhf Dec 18 '23 at 20:52
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    In that setting, isn’t it much simpler to consider perimeters? The perimeter of the circle is $2\pi$ and is smaller than the perimeter of the square, which is $8$. – lhf Dec 18 '23 at 20:54
  • @lhf: looks the same to me. If anything, the area one seems more obvious to me. But, for a formal proof, you need to somehow argue that there are parts of the interior of the square that are not in the circle. – Martin Argerami Dec 18 '23 at 21:05
  • @lhf I thought a lot about posting a picture. I have one drawn by hand, but I'm not sure if it's common practice to include hand-drawn pictures over pictures drawn in Tikz. I'm not very skilled in Tikz unfortunately. – Mathematical Endeavors Dec 18 '23 at 21:20
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    Angles $B$ and $D$ are constructed right angles. Vertical and horizontal are perpendicular/parallel to the sides of the square by definition. – Daniel Mathias Dec 18 '23 at 21:32
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    They are right angles because you are consciously constructing them to be right angles. If you have a point not on a line you can always drop a perpendicular from the point to the line and that perpendicular will always intersect the line at a specific point. The instructions are asking you to do just that. The one thing it seems to be assuming is that the point of perpendicular intersection will be within the side of the square rather than on an extension outside the square. (It seems like it is trying to prove something intuitively but making intuitive assumptions it doing so.) – fleablood Dec 18 '23 at 22:11
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    @lhf: you can fit a closed curve with arbitrary large perimeter inside any square. I agree with you about the picture: do post one if you can, please Mathematical Endeavors: hand-drawn pictures are fine. – Rob Arthan Dec 18 '23 at 22:16
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    "But, for a formal proof, you need to somehow argue that there are parts of the interior of the square that are not in the circle." The very first line is "Inscribe a circle of radius 1 inside a square of side length 2". But why can we assume that is even possible? It seems to me if we can assume that can be done and we assume that $\pi$ is defined as the area of a unit circle we should just stop then and there. Because it is inscribed in a square the area is less than the square's and there's no need to do anything more. – fleablood Dec 18 '23 at 22:22
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    @RobArthan " you can fit a closed curve with arbitrary large perimeter inside any square" but wouldn't the area of any such region be less than the square's? The assumption that the area of a unit circle is $\pi$ seems almost to be a definition. In this entire excercise I'm not sure what our basic axioms and definitions are. Surely by the time we've shown that a constant such as $\pi$ even exists and that the area of a circle is $\pi r^2$ I'd think $\pi < 4$ would have fallen out by the side of the road miles ago. – fleablood Dec 18 '23 at 22:27
  • @fleabllood: I was replying to lhf's comment which suggests looking at perimeters rather than areas. (Aside: I agree with you that the proof is heavily dependent on what axiomatic approach you take.) – Rob Arthan Dec 18 '23 at 22:33

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If we define $\pi$ as the ratio of a circle’s circumference to its diameter, then for a circle of radius $1$, $\pi$ is the length of the semi-circumference, and proving that $\pi<4$ would require proving that$$\overset{\huge\frown}{DEC}=\pi<DA+AB+BC=1+2+1=4$$ pi<4 However, if this is not as easy as it first appears, it seems better to compare areas instead. As suggested by @fleablood, assuming the area of the unit circle is $pi$, then since a square of side $2$ and area $4$ has portions of its area outside the inscribed circle of diameter $2$ (tangents lie outside their circle), then evidently$$\pi<4$$The proof as posted seems sound, but perhaps goes beyond what is necessary.

Edward Porcella
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    See also https://math.stackexchange.com/questions/2639679/prove-sinx-x-tanx-when-0-x-cfrac-pi2-using-only-geometry-trigo – lhf Dec 19 '23 at 23:46