## October 2016 solution

No, it is not possible to tile a convex polygon in this way.

To show this, consider the following facts.

First, every quadrilateral in the tiling has exactly one angle that is over 180 degrees. (If it didn't have one, it wouldn't be concave. It can't have more than one, because the sum of the angles in a quadrilateral is 360 degrees.)

Second, if A is the set of points that serve as the location in the tiling of an over-180-degree vertex for some quadrilateral, then each point in A can only serve as the location of an over-180-degree vertex for exactly one quadrilateral. (Around each point there are at most only 360 degrees that are part of the tiled polygon.)

From the first two points we conclude that there is a one-to-one mapping between Q, the set of quadrilaterals, and A. This means, among other things, that there is the same number of elements in each set. Let us call this number n.

Next, consider that every vertex in A is located at an internal point of the polygon. (If it was at its boundary, the boundary would have included an over-180-degree vertex, contradicting the assumption that the polygon is convex.)

Let us now sum all the angles used in all the vertices in our partition. There are the vertices of the original polygon, and the angles on them sum up to at least 180 degrees. Then there are potentially some internal vertices which are not part of A, each of which contributes 360 degrees. And then there are the vertices in A, each of which contributes 360 degrees. In total, this sum cannot be any less than 360n+180.

However, we can also sum all the angles by summing them on a per-quadrilateral basis. Every quadrilateral has a total of 360 degrees in its angles. Sum this up across all n quadrilaterals and you get exactly 360n, so at least 180 degrees less than are needed.

We conclude that no such tiling is possible.

Q.E.D.

For completion:

Several solvers have worked very hard on trying to solve this problem by brute-force methods, and only a few of them actually made it to the solver list this month. The others were repeatedly told by me that their proof-attempts do not cover the full range of possibilities that need to be accounted for in order to be able to conclusively rule out the possibility that such a tiling as described exists.

In order to make this point clearer, for any solver who sent me such a partial solution and may still believe I should have accepted their attempt, I've asked Lorenz Reichel to share his solution, and Lorenz graciously agreed.

You can find Lorenz's solution here.

This is what it takes to solve this problem "the hard way". If you've gone down the brute-force path, ask yourself if you've covered all of the options and possibilities that this solution systematically goes through and rules out one by one.

Again: thanks, Lorenz!