Prove that any triangle can be cut into $2019$ quadrilaterals such that each quadrilateral is both inscribed and circumscribed. (Nairi Sedrakyan)
Problem
Source: Tournament of Towns, Senior O-Level , Spring 2019 p3
Tags: bicentric quadrilateral, Tiling, combinatorics, combinatorial geometry, cut
XbenX
11.05.2020 14:00
First, we show that any triangle can be cut into $3$ quadrilaterals being both cyclic and inscribed. Let the triangle be $ABC$, $I$ its incenter, and $D,E,F$ be the feet of the perpendicular from $I$ to the sides of the triangle, it is clear that the $3$ quadrilaterals formed fulfill the conditions. Now, divide the triangle into $\frac{2019}{3}$ smaller triangles and do what we did above to each of these triangles and we're done.
P1kachu
03.06.2021 06:33
It can also be shown that for any integers $n\ge3$, any triangle can be cut into $n$ bicentric quadrilaterals.
We prove this by induction. The base case, $n=3$ has been shown above by @XbenX.
For the inductive step, it suffices to show that any triangle can be cut into a bicentric quadrilateral and a triangle.
Let the triangle be $ABC$ and its incircle be $\omega$. Let $P$ and $Q$ be variable points on sides $AB$ and $AC$ respectively such that $PQCB$ is concyclic.
If $P$ and $Q$ are close to $A$, then $PQ$ and $\omega$ do not intersect at any point, while if $P$ is close to $B$ and $Q$ is close to $C$, then $PQ$ and $\omega$ intersects at two points.
This means, by IVT, there exists points $P$ and $Q$ on sides $AB$ and $AC$ such that $PQ$ is tangent to $\omega$.
This proves the existence of a bicentric quadrilateral $PQCB$, which can be used to cut triangle $ABC$ to $PQCB$ and $APQ$, and this completes the induction.