# Partition of polygons into 'strongly acute' and 'strongly obtuse' triangles

Definition: Let us refer to obtuse triangles with the largest angle strictly above a given cutoff value as 'strongly obtuse' - the definition is parametrized by the cutoff value. Likewise, strongly acute triangles are acute triangles with the smallest angle strictly less than a specified cutoff.

General Question: Given an n-gon, how does one triangulate it into a finite number of strongly obtuse/ strongly acute pieces with specified cutoff - resulting in least number of pieces? And before that, how to decide whether such a triangulation exists?

Example: it seems impossible to cut a square into finitely many strongly obtuse triangles with cutoff 120.

Guess: Consider partitioning a given n-gon into strongly obtuse triangles with some cutoff greater than or equal to 120 degrees. It appears that to check if the n-gon can be so triangulated, we need to see only its triangulations using only segments connecting the vertices of the input n-gon - adding extra vertices for the triangulation does not help. And for any cutoff less than 120 degrees, any triangle and hence, any n-gon can be easily cut into strongly obtuse triangles with extra vertices.

Note: For partition into strongly acute triangles, the 'first' case seems to be a cutoff of 36 degrees.

• Oh, I see, I was a sloppy reader. Sorry. Oct 29, 2022 at 11:48
• Just a comment on the definitions: it seems more natural to me to describe strongly acute triangles as those whose maximum angle is less than some $\theta>60^\circ$. The condition of max angle greater than some threshold automatically implies obtuseness (for angles over $90^\circ$), but the condition of min angle under some threshold doesn't imply acuteness, so it's a little weird to label such a condition "strongly acute" when the property they satisfy is not actually stronger than acuteness. Oct 29, 2022 at 17:10
• if we follow your definition - strongly acute to mean maximum angle of an acute triangle is less than some limit greater than 60 - then, it appears to indicate a certain fatness of the triangle. by strongly acute, i was trying to think of triangles that are both acute and thin and for that one felt a need to minimize the smallest angle. note that strongly obtuse triangles too are necessarily thin. Oct 29, 2022 at 19:36
• Earlier attempts to achieve partition into triangles that are both acute and isosceles ( for example, Hoggatt, V. E. Jr. and Denman, R. "Acute Isosceles Dissection of an Obtuse Triangle." Amer. Math. Monthly 68, 912-913, 1961) appear to yield many acute triangles that are 'fat' and that motivated the present query on partitioning into thin acute and obtuse triangles. Oct 30, 2022 at 3:03
• Since every triangle can be partitioned into at most three triangles with the maximal angle $120^\circ$ or higher, the "obtuse" possibility for lower cutoffs is always there. Acute triangulation is trivially possible no matter what (just split an angle into as many parts as you wish. The count minimization problem may be hard though unless you don't mind a factor of 3. Oct 30, 2022 at 21:33

It is, indeed, true that if we have a polygon that can be partitioned into $$120^\circ$$-obtuse triangles, then at least one angle of the original polygon exceeds $$120^\circ$$. However we may need extra vertices in the partition into $$\alpha$$-obtuse triangles even for a symmetric convex trapezoid. On the picture below, the angles $$A,B,C$$ satisfy $$A,B>C,$$ so if we choose $$\alpha\in(C,\min(A,B))$$ we will have an admissible partition (given by red lines) but not an admissible partition without extra vertices. Thus the question of existence of an $$\alpha$$-obtuse partition with $$\alpha\ge 120^\circ$$ looks quite non-trivial, indeed. The proof is fairly simple. The first step is to upgrade the given partition to a regular one where any two triangles can share either a vertex or a full edge. That is done by triangulating the triangles having some vertices of other triangles on the boundary further. The key pictures here are (they allow take care of one bad point a time; the top angle is the large one; in the second case the maximal angle in one of the new triangles is a bit smaller than that of the original one, but the drop is arbitrarily small, so the strict inequality survives).

Now it becomes the usual Euler's formula mumbo-jumbo. Let $$V_0$$ be the number of the vertices of the starting polygon (we prohibit to use them as large angle vertices), $$V_i$$ the number of inner vertices and $$V_b$$ the number of extra boundary vertices. Let $$F$$ be the number of faces (without the infinite one) and $$E$$ be the number of edges. By Euler, $$V_0+V_i+V_b+F-E=1.$$ On the other hand, $$V_b+2V_i\ge F$$ (each face has a large angle vertex and the inner vertex can serve at most two faces while the boundary one at most one). Also we have $$2E=V_0+V_b+3F$$. Thus $$2=2V_0+2V_i+2V_b+2F-2E= \\ 2V_0+2V_i+2V_b-F-V_0-V_b=V_0+(V_b+2V_i-F)\ge V_0\,$$ which is nonsense because the minimal number of the vertices of the original polygon is $$3$$.

The pentagon example of RavenClawPrefect is possible because he used $$V_0$$ for the large angles 3 times (one vertex once and one twice), which is exactly what finally compensated the undercount $$V_0-2=3$$ for the pentagon.

• thanks. i understand the guess given in question is not valid even when convex polygons are being triangulated! and that would make the question of deciding whether a given polygon admits a triangulation into obtuse triangles with max angles above a certain cutoff more difficult! Oct 31, 2022 at 16:04

It is not the case that all strongly obtuse subdivisions with a threshold over $$120^\circ$$ are possible through vertex-to-vertex connections.

Consider the following dissection of a pentagon into triangles that are all very close to $$10-10-160$$ angles (they can't all be exactly congruent, but they're close enough for our purposes and get closer as the angles get more extreme): The reflex angle here must be connected to at least one other vertex of the pentagon if we're to have a vertex-only dissection. But it's easy to see that both available choices yield triangles below the threshold.

• Thanks! Interesting example. I see that a threshold of 160 clearly cannot be achieved if we do not add extra vertices while triangulating. Can't decide though what would happen to the guess if we consider only convex n-gons being partitioned. Oct 29, 2022 at 19:31