![]() Since the right angled triangle was handled similarly to the middle case above, we get at most nine pieces in every case. ![]() 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. 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. 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.įinally, if A+C is not less than 90, then we have an acute or right angle triangle. ![]() 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. If all three quantities are less than 90, we are done with a seven triangle partition. 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. Suppose we have a non isosceles triangle with two smallest angle measures A less than C. 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. I found a path to decompose any triangle into at most nine parts. Perhaps this will inspire someone to characterize those nonisosceles triangles which admit a decomposition into seven or fewer parts. This may lead to an eight part decomposition of the triangle. 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 results in 5N - 10 triangles and N-2 quadrilaterals which have a regular form. 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. One can settle for a partial decomposition. 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. One potential analysis which I will not do here is to find which triangles lend themselves to a seven part decomposition this way. 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. We now get nine isosceles triangles, with the potential to get all of them acute. The problem is that at least three are obtuse (or right). It touches the triangle at three points of tangency, and these along with the circle center help divide the triangle into six isosceles triangles. Look at the inscribed circle of an arbitrary triangle. Combining it with a greedy approach of choosing the largest acute isosceles triangle in the remaining polygon may get an even better upper bound. I suspect it can lead to a bound of 8N or better. The following is a worthy line of approach. Thus we have a total of $2N + 7 \times 2N \sim 16 N$ acute isosceles triangles. By the symmetry of the input obtuse isosceles triangle, it appears that each of the $7$ acute triangles is also isosceles. Now, partition each of the $2N$ obtuse isosceles triangles into $7$ acute pieces as described in the above linked page. So, at the end of this step, we have $2N$ acute isosceles triangles and $2N$ obtuse isosceles triangles. Triangulate the $N$-gon into $\sim N$ triangles (actually $N-2$ triangles).ĭivide 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. It is to be cut into the least number of acute isosceles triangles.īased on this MathSE discussion, one can think of a method to get $\sim 16N$ acute isosceles triangles for any given $N$-gon (given below).Įven if the method is valid, it the bound of $16N$ acute isosceles triangles a tight one? ![]() Question: Given an $N$-vertex polygon (not necessarily convex).
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