Question: Given an $N$-vertex polygon (not necessarily convex). It is to be cut into the least number of acute isosceles triangles.


Based on this MathSE discussion, one can think of a method to get $\sim 16N$ acute isosceles triangles for any given $N$-gon (given below). Even if the method is valid, it the bound of $16N$ acute isosceles triangles a tight one?

The method:

  1. triangulate the $N$-gon into $\sim N$ triangles (actually $N-2$ triangles).

  2. Divide each triangle into $2$ right triangles (yielding a total of $2N$ right triangles) and then partition each right triangle into $2$ isosceles triangles by joining the midpoint of its hypotenuse to the opposite vertex - note that except when the right triangle is itself isosceles, one of these $2$ isosceles triangles is acute and the other is obtuse. So, at the end of this step, we have $2N$ acute isosceles triangles and $2N$ obtuse isosceles triangles.

  3. Now, partition each of the $2N$ obtuse isosceles triangles into $7$ acute pieces as described in the above linked page. By the symmetry of the input obtuse isosceles triangle, it appears that each of the $7$ acute triangles is also isosceles. Thus we have a total of $2N + 7 \times 2N \sim 16 N$ acute isosceles triangles.

  • 1
    $\begingroup$ Try this approach. For each edge (or interior diagonal, if you have the processing power) do a test: find the nicest acute isosceles triangle with base lying on the edge that is contained in the polygon. A really nice triangle will use the edge exactly, not divide any other edges, and split the problem into two smaller equalsized problems at the cost of adding one edge to the total. Sometimes you have to add three edges to the total for less nice triangles. Use this to reduce to solving the problem on pentagons cheaply. Gerhard "As If Pentagon Meets Budgets" Paseman, 2019.09.13. $\endgroup$ Sep 13 '19 at 15:00

The following is a worthy line of approach. I suspect it can lead to a bound of 8N or better. Combining it with a greedy approach of choosing the largest acute isosceles triangle in the remaining polygon may get an even better upper bound.

Look at the inscribed circle of an arbitrary triangle. It touches the triangle at three points of tangency, and these along with the circle center help divide the triangle into six isosceles triangles. The problem is that at least three are obtuse (or right). To fix this, we grow the circle radius.

We now get nine isosceles triangles, with the potential to get all of them acute. Further, when the grown circle intercepts a vertex of the original triangle, one gets seven isosceles triangles, with the potential of all seven being acute. One potential analysis which I will not do here is to find which triangles lend themselves to a seven part decomposition this way. One easily observes that if the obtuse angle of the triangle is greater than 120 degrees, then one cannot get a nice decomposition into seven acute parts this way.

There are two ways to go. One can settle for a partial decomposition. I believe it is the case for any triangle that there is a decomposition into six parts, five of which are acute isosceles triangles, and the sixth a diamond figure which may break down into acute isosceles triangles upon further consideration and examination. This results in 5N - 10 triangles and N-2 quadrilaterals which have a regular form.

The other approach is to take really obtuse triangles and carve out a large acute isosceles triangle, leaving an acute or at least less obtuse triangle. This may lead to an eight part decomposition of the triangle.

Perhaps this will inspire someone to characterize those nonisosceles triangles which admit a decomposition into seven or fewer parts.

Edit 2019.09.14:

I found a path to decompose any triangle into at most nine parts. It is based on using the inscribed circle and growing it to form six or five isosceles triangles inside the circle, leaving three or two outside.

Suppose we have a non isosceles triangle with two smallest angle measures A less than C. If we take a circle concentric with the inscribed circle, and passing through the nearest vertex (not having A or C as angle measure), then we get five triangles inside the circle with three of them having angle measure A+C, one 180 - A - 2C, and one 180-C -2A, measured in degrees.

If all three quantities are less than 90 , we are done with a seven triangle partition. Otherwise, if A+C is less than 90 and one of the other quantities is not, then cut this triangle in two, given an acute isosceles piece with angle A, and the other with angles C, 90 + A/2, and 90-C-A/2. This smaller triangle is seen to satisfy the desired inequalities if A+C is less than 90, and we end up with an eight piece partition into acute isosceles triangles.

Finally, if A+C is not less than 90, then we have an acute or right angle triangle. We can use the circle construction above, but keep all vertices of the triangle outside the circle, and bring three of the angles as close to ninety as we want, but keep them acute. It is an exercise to show the other three inside the circle can be made acute, and we either have a nine piece desired decomposition, or we have a right angled triangle. Since the right angled triangle was handled similarly to the middle case above, we get at most nine pieces in every case.

End Edit 2019.09.14.

Gerhard "Not Yet Inspired To Finish" Paseman, 2019.09.13.

  • $\begingroup$ Hoggatt, V. E. Jr. and Denman, R. "Acute Isosceles Dissection of an Obtuse Triangle." Amer. Math. Monthly 68, 912-913, 1961, shows that any obtuse triangle can be dissected into eight acute isosceles triangles. $\endgroup$ Oct 9 '20 at 6:59

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