Last week I posted the following mathematical problem:

"If you want to subdivide an obtuse triangle (i.e. one angle is > 90 deg) into acute triangles (all angles < 90deg), how many  acute triangles do you need at least ?  There must be no blunt angle left over in the end!  (note that right angles are excluded by both definitions, and the length of the sides doesn't matter. there is a solution that works for all obtuse triangles, regardless of their other properties -- so don't try to prove that there is no solution!)

If that's too easy, start with a square (handy hint: you will want to draw 2 semi-circles to crack this one!)

If both are too difficult, start with an acute triangle. "

Here are my solutions:

1) To divide an obtuse triangle, you need at least 7 acute ones. You can work that out as follows. Let's draw a triangle such that the obtuse angle alpha is at the top (points upwards) and the sides are labeled, clockwise from the base a, b, c.

Obviously, you want to draw a line that divides the obtuse angle into two acute ones. However, if you draw this line all the way down to meet the base (a), it would create either 2 right angles (forbidden) or one acute and one obtuse angle. In the latter case, we would have wasted one acute triangle without getting any closer to the solution, as the other triangle would still be an obtuse one. (In fact, this procedure could be continued ad infinitum, prompting some people to suspect that there is no solution.)

So we want to end our first line before it runs into the base (a), somewhere near the middle of the triangle. This endpoint must be a shared corner between a number of acute triangles, but how many? If we had 4 or less triangles meeting in the middle, there would be at least one obtuse angle (or 4 right angles, still forbidden). So let's make it 5 triangles meeting in the middle, requiring 5 lines to separate them. We have one line already dangling from the top. Draw 4 more in the shape ><.  There will be obtuse quadrilaterals on both sides which you can easily subdivide into acute triangles by drawing vaguely vertical lines between the points where the > meets sides a and b on the left, and also between the points where < meets c and a, creating a bow tie shape suspended from the top by the first line we drew.

The end result has 5 acute triangles grouped around the middle, plus two sitting in the acute corners of our original triangle.

2) To subdivide a square, you want to draw a pair of circles around the midpoints of the vertical sides of the square, b and d, with their diameters matching the side length of the square. Why? Well, if you now drew a triangle with the basis b or d and its tip on the corresponding circle, Thales tells us it would be rectangular. Hence, if you make sure that the tip of the triangle is just outside the circle, it is acute. Thus we create the first two acute triangles by connecting the bases b and d to a symmetrical pair of tip points just outside the circles, and less than a quarter side length above the point where the circles touch.

Link these two points to each other, and each to the mid points of sides a and c, and you will have a set of 8 acute triangles nicely dividing the square. As in the first task, there are 5 triangles meeting at vertex points inside the construction. Just this time we need two vertices, not one.  (If you want to make sure you don't  place the vertices too far up and obtain obtuse triangles dangling from the ceiling, you could use two circles half the size of the first set, centered in 1/4 and 3/4 of a, to check the vertices are outside these circles as well. But the "quarter sidelength from the middle" criterion above covers this concern nicely.)

While this isn't a strict proof that 8 is the smallest number of triangles for the square, I believe that it is, because of the nice analogy: one vertex of 5 triangles for the triangle, two such vertices for the square.

3) To subdivide an acute triangle into smaller acute triangles, you need at least 4. Find the midpoints of all three sides, and draw an upside-down smaller version of the original triangle by connecting them to each other.