, 2019

Solution:

Lemma. For any polygon including its boundary, there exists a largest circle contained inside it.

Proof. It's easy to see that for any circle inside the polygon, it can be increased in size until it is tangent to at least three sides of the polygon. Then for any three sides of the polygon, there is only one circle tangent to all three, so there are only finitely many possibilities. Therefore there exists a largest one.

Alternatively, one can show that the space of valid $(x, y, r)$ such that the circle with center $(x, y)$ and radius $r$ is compact, e.g. by showing the complement is open and that the complement is open. Then the map $(x, y, r) \rightarrow r$ is continuous and therefore has a maximum.

Now, take the largest circle. It clearly must be tangent to three sides. If the circle lies inside the triangle made by the three lines, we can expand the polygon to that triangle and solve it for the triangle instead. Otherwise, we have the following diagram:

Here the circle is an excircle of the triangle $ABC$ made by the lines $AB$, $AD$, and $BE$. (Note that $AD$ and $AB$ don't have to be consecutive sides of the polygon, but the ones in between don't really matter.)

Then since the circle is an excircle, we can consider a homothety at $C$ with power $1+\epsilon$, which sends the circle to a slightly larger circle which does not touch line $AB$. If this homothety causes the circle to leave the polygon for small enough $\epsilon$, it must be because the circle was initially tangent to another line $\ell$, for which it would be an incircle of the triangle made by $\ell$ and lines $AD$, $BE$, bringing us back to the first case.

Thus we can reduce to a case where we have a triangle with each height at least $1$, and we want to show the inradius is at least $1/3$. Let $K$ be the area of the triangle, so the heights $\frac{2K}{a}$, $\frac{2K}{b}$, $\frac{2K}{c}$ are all at least $1$. Then the inradius $r$ satisfies

$$r=\frac{K}{s}=\frac{2K}{a+b+c} \geq \frac{2K}{2K+2K+2K}=\frac{1}{3}$$
as desired.

Solution:

Consider the center of mass $G$. We will use the notion of support lines for convex shapes. (Support lines are the lines that touch the shape but do not cut through it.) If a circle centered at $G$ with radius $1/3$ cannot be contained inside the polygon, then there exists a point $P$ on the boundary such that $GP < 1/3$. Let $\ell_1$ be the support line passing through $P$, $\ell_2$ be the line parallel to $\ell_1$ and passing through $G$, and $\ell_3$ be the other support line that is parallel to $\ell_1$, touching the polygon at $P'$. Suppose $\ell_2$ intersects the polygon at $A$ and $B$. Extend $P'A$ and $P'B$, intersecting $\ell_1$ at $A'$ and $B'$. Then, if we consider the two parts of the polygon that $\ell_2$ divides the polygon into, we have:

- the part of the polygon that contains $P'$ contains the triangle $P'AB$;
- the part of the polygon that contains $P$ is contained in the quadrilateral $AA'B'B$.

Then we conclude that the center of mass $G'$ of the triangle $P'A'B'$ lies between $\ell_2$ and $\ell_1$, which by assumption is less than $\frac{1}{3}$ away from $\ell_1$. However, because the height from $P'$ to $\ell_1$ is at least $1$, the distance from $G'$ to $\ell_1$ is at least $\frac{1}{3}$, so we have a contradiction. Therefore no such $P$ exists and the circle can be placed inside the polygon.