Cutting regular polygon

Obviously, the triangle (into four triangles) and the square (into four squares) can be dissected as required. Let us restrict our attention to $n$ -gons with $n > 4$ , so that the interior angles are obtuse and the exterior angles acute.

If, in a dissection, one of the angles of the $n$ -gon is not "cut", then an $n$ -gon must be used to fill that angle. Since more than one subpolygon is required, this means that the two sides adjacent to that corner must be "cut", and hence that the exterior angle of the $n$ -gon must be fillable by the interior angles of a set of regular polygons. Since this exterior angle is acute (and the only acute interior angle of a regular polygon is $60^\circ$ ),. we deduce that the exterior angle of the polygon must be $60^\circ$ , and hence $n=6$ , so that we have the hexagon, which can be dissected into hexagons and triangles each of a third the original side length, as shown on the right.

Thus, otherwise, every angle of the regular $n$ -gon must be "cut", and so the (obtuse) interior angle of the regular $n$ -gon must be splittable into a sum of two or more interior angles of regular polygons. There are only three options for expressing an obtuse angle as the sum of two or more regular polygon interior angles, namely $120^\circ = 60^\circ + 60^\circ$ , $150^\circ = 60^\circ + 90^\circ$ and $168^\circ = 108^\circ + 60^\circ$ .

• The first of these gives the hexagon, dissected into six triangles.
• The second of these gives the dodecagon, dissected into six squares, six triangles and one hexagon.
• The third forces us to consider the regular $30$ -gon, with each angle split into a pentagon and a triangle. The exterior angle of a pentagon cannot be split into a sum of interior angles of regular polygons, and so the pentagon cannot "cut" any side of the $n$ -gon it touches. Thus the side-length of each pentagon must be the side-length of the original $n$ -gon. The only way this is possible is for alternate faces of the $30$ -gon to be sides of pentagons, and for the other faces to be sides of triangles. These triangles could be further subdissected, but that is not important. But, "one layer in", this creates angles of $360^\circ - 60^\circ - 2\times108^\circ = 84^\circ$ , which cannot be filled by the interior angles of regular polygons. This option is not possible.

Thus the answer is $3+4+6+12 = \boxed{25}$ .

I love this problem... Thanks for posting Khang!

Here is my rough solution:

The following regular polygons can be further divided into regular polygons:

• Triangle: four triangles
• Square: four squares
• Hexagon: six triangles
• 12-gon: six triangles, six squares, and a hexagon

Here is my first cut argument as to why none of the remaining polygons don't work...

Cuts have to go through either the vertices or the sides or both. I persuaded myself that only the triangle and square can have cuts going through the sides. (Will try to come up with a more concrete proof)

For the remainder of the solids (cuts going through the vertices) the angle would have to add up to angles of other regular polygons.

The interior angles of the first few regular polygons are:

• Triangle: 60 degrees
• Square: 90 degrees
• Pentagon: 108 degrees
• Hexagon: 120 degrees

(No need to go beyond the hexagon since the limit on the interior angle of any n-gon is 180 degrees, so the remaining angle after a cut would be less than 60 degrees, the angle of the smallest n-gon)

So, only regular polygons whose interior angles add up to these numbers can work. That includes:

• Hexagon: 120 degrees (60 + 60)
• 12-gon: 150 degrees (90 + 60)

That leaves the four regular polygons I mentioned at the beginning.

So, the answer is $3+4+6+12=\boxed{25}$

Make sense?