Polygon with Inscribed Circle – Proof It Is Convex

euclidean-geometrygeometry

In many elementary (and not-so-elementary) Euclidean geometry texts, a (simple) polygon is said to be tangential if it is convex and has an inscribed circle (i.e., a circle that intersects and is tangent to each side of the polygon). The assumption of convexity is not needed: I've come up with a rather laborious proof that every polygon with an inscribed circle is convex. But I'd like to find either a simple elementary proof or a reference to a proof in the literature. (By "elementary," I mean using only standard facts of axiomatic Euclidean geometry.)

Does anyone know of a reference for a proof of this fact (elementary or not)? Or can anyone think of a straightforward elementary proof? You can use any definition of "convex polygon" that you like, but the easiest one to work with is that for each edge, the vertices not on that edge lie on one side of the line through that edge.

(Interestingly, the corresponding fact for circumscribed circles–i.e., that every polygon with a circumscribed circle is convex–is quite easy to prove: If P has a circumscribed circle, any two nonadjacent sides of P are non-intersecting chords of the circle; it is easy to show that both endpoints of each chord lie on the same side of the line through the other, and from there it is an easy matter to prove that P is convex.)

Best Answer

Thanks to everyone who suggested approaches to this problem. In the end, none of the suggested approaches fit into the axiomatic framework that I was working in, so I had to write up my own rather laborious proof. It's a bit long to post here, but in order to close this question, I just want to post the reference and a quick summary of the approach.

You can find the complete proof in my textbook Axiomatic Geometry (Theorem 14.31). The basic idea is first to prove the following lemma:

Lemma. Let $\mathscr P$ be a polygon circumscribed about a circle $\mathscr C$. Suppose $A$ is any vertex of $\mathscr P$, and $E$ and $F$ are the points of tangency of the two edges containing $A$. Then there are no points of $\mathscr P$ in the interior of $\triangle AEF$.

To prove that a tangential polygon $\mathscr P$ must be convex, the basic idea is to show that if $\ell$ is any edge line of $\mathscr P$, then there can't be any vertices of $\mathscr P$ on the "wrong" side of $\ell$ (the side not containing the inscribed circle), because that would violate the lemma.

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[Math] A simple proof that a polygon circumscribing a circle overestimates its perimeter

You may use a general fact:

If $A,B\subset\mathbb{R}^2$ are two convex bounded shapes and $A\subset B$, the perimeter of $A$ is less than the perimeter of $B$.

Proof: if $A\neq B$, you may "cut out" a slice of $B$ without touching $A$. By convexity, the perimeter of the "reduced set" $B$ is less than the perimeter of the original set $B$. If $A$ is a polygon, by iterating this argument a finite number of times you get that $A$ is a reduced version of $B$, hence $\mu(\partial A)<\mu(\partial B)$ as wanted.

[Math] Finding the interior angles of an irregular polygon inscribed on a circle

[Improved answer, based on excellent comment from David]

We have two different cases: one with the center inside the polygon and one with the center outside of it.

WLOG, we can assume that $a_1$ is the length of the longest edge. If the length of the $i^{th}$ edge of the polygon is $a_i$ and the radius of the circumscribed circle is $R$, the corresponding central angle would be (in both cases):

$$\alpha_i=2\arcsin \frac{a_i}{2R}$$

Case on the left: Sum of all central angles is $2\pi$:

$$2\pi=\sum_{i=1}^{N}\alpha_i=2\arcsin \frac{a_1}{2R}+\sum_{i=2}^{N}2\arcsin \frac{a_i}{2R}$$

$$\pi=\arcsin \frac{a_1}{2R}+\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\pi-\arcsin \frac{a_1}{2R}=\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\frac{a_1}{2R}=\sin(\sum_{i=2}^{N}\arcsin \frac{a_i}{2R})\tag{1}$$

Case on the rigth: $$2\pi=(2\pi-2\arcsin \frac{a_1}{2R})+\sum_{i=2}^{N}2\arcsin \frac{a_i}{2R}$$

$$\arcsin \frac{a_1}{2R}=\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\frac{a_1}{2R}=\sin(\sum_{i=2}^{N}\arcsin \frac{a_i}{2R})\tag{2}$$

Actually, both equations (1),(2) are the same. The equation (1,2) has only one unknown ($R$) but I doubt that it is possible to solve it in a closed form for an arbitrary $N$. So numerical approach seems to be your only option. Once you have $R$ you can calculate pretty much anything, all internal angles included.

Example for the case on the left with $a_1=a_2=a_3=6$, $a_4=3$:

a1 = 6
a2 = 6
a3 = 6
a4 = 3
Solve[a1/(2 R) == Sin[ArcSin[a2/(2 R)] + ArcSin[a3/(2 R)] + ArcSin[a4/(2 R)]], R]

The result is $R=6\sqrt\frac25$.

Example for the case on the right with $a1=4$, $a2=3$, $a3=2$:

a1 = 4
a2 = 3
a3 = 2
Solve[a1/(2 R) == Sin[ArcSin[a2/(2 R)] + ArcSin[a3/(2 R)]], R]

This gives one solution: $R={8\over\sqrt{15}}$

Best Answer

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[Math] A simple proof that a polygon circumscribing a circle overestimates its perimeter

[Math] Finding the interior angles of an irregular polygon inscribed on a circle

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