Given any triangular region and two integers $n$ and $p$ which can be large and $p > 4$. It is needed to cut the triangle into $n$ $p$-gons (e.g., cut a triangle into 10 heptagons). Among the $p$-gons, we need the maximum possible number to be convex. No other requirements on the pieces.


To our knowledge, for any high value of $n$ and $p$, it is possible to cut any triangle into $n$ $p$-gons such that $\lfloor n/2\rfloor$ of the pieces are convex. Can this upper bound on the number of convex $p$-gons be raised?


**Update(31st January, 2021):** Thanks to Gerry Myerson and his comments (please see below) on some specific values of $p$. Summary:


For $p$ = 4, for any $n$, any triangle can be cut into $n$ quadrilaterals, *all* of which are convex.


For $p$ = 5, there is a neat recursive scheme that can cut a triangle into $n$ pentagonal pieces out of which only 1 is non-convex. For $p$ = 6 (hexagonal pieces), there is a somewhat similar recursive scheme that cuts a triangle into $n$ hexagons out of which only 1 is non-convex.


The question is open, as of now, for $p$ > 6.


**Note:** The basic question above can be modified with an additional equal area requirement on the pieces in which case, the question appears open for $p$ > 5. A different possibility is to replace the input triangle by a square (say).


Note added on March 23rd, 2021): This earlier discussion: <https://puzzling.stackexchange.com/questions/92892/is-it-possible-to-divide-a-square-into-convex-pentagons>
is on a closely related question.



To cut a triangle into $n$ $p$-sided polygonal regions
This is nothing like a complete answer to the question, but it takes care of a few cases. It summarizes the string of comments I posted.


Quadrilaterals:


Any triangle can be cut up into $n$ quadrilaterals, all of them convex, for any $n\ge3$, as follows:


Take a point in the interior of the triangle, and for each side of the triangle, draw a line segment connecting that point to that side. This cuts the triangle into three convex quadrilaterals.


Now any line segemnt connecting two opposite sides of a convex quadrilateral cuts it into two convex quadrilaterals. By iterating this construction, we cut the original triangle into $n$ convex quadrilaterals, for any $n\ge3$.


Pentagons:


Any triangle can be cut up into $n$ pentagons, all but one or two of them convex, for any $n\ge5$, except possibly $n=7$.


First, we cut a triangle up into three convex pentagons and a triangle, as follows: draw a small triangle upside down inside the given triangle, and connect each vertex of the small triangle by a line segment to the nearest edge of the given triangle.


Next, we cut up a triangle into four convex pentagons and a triangle, as follows: draw a small triangle upside down inside the given triangle, and connect two of the vertices of the small triangle by a line segment to the nearest edge of the given triangle. Draw a short line segment from the third vertex of the small triangle toward the third edge of the original triangle, and then connect the free end of this short line segment to the other two edges of the original triangle.


Now by iterating the first construction $a$ times, and the second one $b$ times, we get $3a+4b$ convex pentagons and one triangle. $3a+4b$ can be any integer three or greater, except for five. Then the remaining triangle can be cut into two pentagons by a three-segment crooked line from any vertex to the opposite side. Depending on how this three-segment line is drawn, either one pentagon will be convex, and one nonconvex, or both will be nonconvex.


Hexagons:


A given triangle can be cut up into an arbitrarily large number of convex hexagons, with only one nonconvex hexagon.


Starting with a triangle, draw a short line segment into the triangle from the midpoint of each side. Connect the free end of each of these short line segments to each of the others by a chain of two line segments, each chain bending in toward the center of the given triangle. We now have three convex hexagons, and one nonconvex, with alternating acute and reflex angles. At each of the reflex angles, draw a short line segment pointing inward, and then connect the ends of those segments as was done in the previous step. So now we have three more convex hexagons, and one nonconvex of the sort we just dissected. Now iterate.


