Partitioning convex polygons into quadrilaterals of equal area and perimeter

Question

This post records a little bit more on this question: Partitioning convex polygons into triangles of equal area and perimeter.

The basic question of the above linked post was about this claim: ""For any convex polygon, there is some finite value of a positive integer n such that the polygon allows partition into n triangles all of which are of same area." The claim may not be true (please see above post).

Question: What happens if we replace 'triangles' in above claim with 'convex quadrilaterals'? The new claim thus generated appears more likely to be true.

Note: In the new claim, one can replace "area" with "perimeter" or diameter or combinations such as "area and perimeter", " area and diameter",... and generate further questions.

Answer 1

Convex quadrilateral decompositions of equal area are always possible. Given a convex polygon $P$ with at least $5$ sides, draw a line from one of the vertices of $P$ to a point $Q$ on one of the edges furthest from $P$ (in the graph sense). As we vary the location of $Q$ continuously, we'll find some decomposition into two convex pieces with strictly fewer sides whose areas are in a rational ratio.

Repeating as necessary, we obtain a collection of quadrilaterals whose areas are rational multiples of one another; from there, we just need to subdivide the quadrilaterals into smaller pieces the size of their least common divisor $d$, which we can do by cutting off quadrilateral-shaped chunks of area $d$ from each piece by connecting two points on opposite sides. (As we slide the two points freely, we can obtain any fraction of the area between $0$ and $1$.)

This only leaves the case where we start with a triangle, which is easy: since decompositions are affinely equivalent, we can assume our triangle is equilateral and draw three line segments from the center to the midpoints of the side to obtain our decomposition.