If a polygon can be cut into $m$ as well as into $n$ triangular pieces of equal area, can it also be cut into $m+n$ triangles of equal area?

(I'm editing after realizing that my conjecture that a convex equidissectable polygon must have an equidissection with all triangles meeting in a common vertex is very wrong.)

Some background: An equidissection of a polygon is a dissection into triangles of equal area. The equidissection spectrum of a polygon is the set of all $n$ such that there exists an equidissection into $n$ pieces. A famous theorem of Paul Monsky states that the spectrum of the square is the set of all even positive numbers (the nontrivial part being that an equidissection into an odd number of pieces is impossible). Most quadrilaterals do not admit any equidissection, and there are quadrilaterals for which the existence of an equidissection is an open question. The wikipedia article on Equidissection describes more of the background.

If a polygon can be equidissected into $n$ pieces, then it can obviously be equidissected into any multiple of $n$ by subdividing the pieces. So the spectrum is closed under multiplication by a positive integer. On the wikipedia page mentioned above, an example is given of a quadrilateral with a "non-principal" spectrum. It is equivalent to a "kite" with corners at $(-1,0), (0,-1), (2,0)$ and $(0,1)$. The two diagonals will divide the area in ratios 1 to 1 and 1 to 2 respectively, and therefore it allows equidissections into 2 and 3 pieces.

My question is motivated by the observation (which can hardly be new, but is not mentioned in the wikipedia article or in this manuscript) that this kite also admits an equidissection into 5 triangles (or into any number $\geq 2$).

To see this, draw straight lines from the point $(4/5, 3/5)$ to the vertices $(-1,0)$ and $(0,-1)$. This will divide the kite into triangles of relative areas $1/5$, $2/5$ and $2/5$.

More generally, suppose that $P$ and $Q$ are points (of the interior or the boundary) such that the straight lines from $P$ to the corners of the quadrilateral yield triangles whose areas are multiples of $1/m$ of the total area, and similarly the lines from $Q$ give areas that are multiples of $1/n$ of the total area. Then from the point $\frac {m}{m+n}\cdot P + \frac{n}{m+n}\cdot Q$ we will get areas that are of the form $$\frac{a}{m}\cdot \frac{m}{m+n} + \frac{b}{n}\cdot \frac{n}{m+n} = \frac{a+b}{m+n}$$ and therefore multiples of $1/(m+n)$ of the total area. Starting from $P=(0,1)$ and $Q=(2,0)$, we can therefore obtain equidissections of the kite into 5, 7 etc. parts.

Similarly, the (non-convex) quadrilateral with vertices in $(-1,0)$, $(0,-1)$, $-2/3,0)$ and $(0,1)$ can be equidissected into 2 (obvious) or 3 (by the vertical line from $(-2/3,-1/3)$ to $(-2/3,1/3)$) pieces. And by continuing the boundary line between $(0,1)$ and $(-2/3,0)$ to the point where it crosses the line from $(0,-1)$ to $(-1,0)$, we divide it into triangles of relative areas $3/5$ and $2/5$, showing that an equidissection into 5 pieces is also possible.

Let us say that a dissection is star-like if all triangles have a common vertex. The argument above shows that

If a polygon has a star-like equidissection into $m$ pieces, and another into $n$ pieces, then it also has one into $m+n$ pieces.

Of course a highly non-convex polygon may have equidissections none of which are star-like, and it is easy to see that this might be the case also for convex polygons: Take a regular hexagon of unit area and label the vertices A, B, C, D, E, F. If we replace the vertex at A by a vertex at a point A' on the line through A parallel to the line BF, then the new hexagon still admits an equidissection into six pieces by cutting along the triangle BFD (and dividing the middle piece into three). But there are only countably many choices of A' for which there is some point P such that the six triangles with vertices in P and two adjacent vertices of the hexagon all have rational area.

So perhaps the answer to my question is the obvious "No, why should it?", and the reason I haven't found a counterexample is just that dissections into a small number of pieces tend to be star-like.

There is a short comment on the second page of this article that shows that the spectrum of a trapezoid is closed under addition, but I don't see how that would generalize.

  • 2
    $\begingroup$ I do not have an answer to this, but thanks to your question I learned about dissections, and Monsky's theorem. Thank you! $\endgroup$ – Boris Bukh Jun 15 '13 at 0:41
  • $\begingroup$ Consider the class C={v-shaped nonconvex quadrilaterals which are both 2-equidissectable and 3-equidissectable}. It seems that generic elements of C are not 5-equidissectable. Am I wrong by some general construction? (To prove I am right, one would have to do a finite-length but tedious case discussion of all possible 5-triangulations of C-quadrilaterals, which I didn't finish, but IMHO it looks plausible.) Note that it fits with your second statement in grey and to your forelast comment, as 3-equidissections of (generic) C-quadrilaterals are not starlike, according to your definition. $\endgroup$ – Mircea Apr 11 '16 at 11:29

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