### Evolution of a problem and an answer

In grades 1 through 5, my little boy went to a private Hebrew school. Every day I drove him there and then picked him up at the end of the day - half an hour drive each way. We spent the combined hour mostly talking of mathematics. For the next three years he took the bus to a public school - our math hours fell a victim to his now hectic schedule. Things changed this summer. His daily swim practice starts at 7 am, six days a week. We again have to share about an hour talking of math. (By the way, I wonder why all athletes at the London Olympiad thank their moms. In our family it's the pop who does the driving.)

Today we talked of a problem I picked from M. Gardner's *The Colossal Book of Short Puzzles and Problems* (9.20):

A cake has been baked in the form of a rectangular parallelopiped with a square base. Assume that the square cake is frosted on the top and four sides and that the frosting's thickness is negligible (zero). We want to cut the cake into N pieces so that each piece has the same volue and the same area of frosting. The slicing is conventional. Seen from above, the cuts are like spokes radiating from the square's center, and each cutting plane is perpendicular to the cake's base.

If N is 2, 4, 8, the problem is easily solved by slicing the cake into two, four, or eight congruent solids. Suppose, however, that N = 7. How can we locate the required seven points on the perimeter of the cake's top. If you solve it for 7, you'll be able to generalize to any N.

A clue to the intended solution comes from another problem (5.5) sent to M. Gardner by Piet Hein:

From the corners of a square extend two lines that exactly trisect the square's area. Into what ratios do the trisecting lines cut the two sides of the square?

Since, clearly, the problem is intended to be generalized, I decided to start with a simple modification

Cut an equilateral triangle into four parts of equal area.

The boy came up with this solution:

Not what I wanted to lead him to, but a legitimate solution. I used the opportunity and we agreed that the same solution would work for any triangle.

Next I asked about cutting the triangle into five pieces. He volunteered to cut it into 6. This is how:

We agreed that for a general triangle the incenter should be taken as a starting points for the cuts, while the midpoints must still be used on the sides:

That was close to my intent but the number was 6 not 5. What about five? Here's his answer:

Trying to swerve him in the "right" direction I asked if there is a construction where all the cuts emanate from an internal point. His answer was, "Yes, just cut each of the six triangles into five parts."

The time was ripe to talk of another shape. A square into 3 pieces? That was too easy; just make two parallel cuts. A regular hexagon into five pieces? "Same as with the triangle: first cut into six triangles and then cut them into five pieces each." We agreed that there was no need to cut all six triangles: cutting just one should suffice.

On the way back from swim practice, we also agreed that this approach appears to work for all tangential polygons: divide a K-gon into K triangles from the incircle to the vertices and then all of so obtained triangles (or as many as may be needed) into N triangles, with cuts from one vertex.

We made progress solving the problem but were still missing the solution implied by the Piet Hein problem. We managed to discuss cutting the square from a corner into parts of equal area just before returning home. We now have all the requisite pieces for solving the cake-and-frosting problem. But this will have to wait till tomorrow.

I did to find the triangle that divides into 6ths and I can not find it is hard to find and I need help to find it

November 3rd, 2017 at 10:50 amThank you so much have a great day.

There's a picture below the sentence "We agreed that for a general triangle the incenter should be taken as a starting points for the cuts, while the midpoints must still be used on the sides"

November 7th, 2017 at 7:28 am