Any luck? I’d be interested in seeing some of the computer solutions even if their scores didn’t beat mine.

By the way I can now improve my score to 14sqrt(3)-24 = 0.249… . My covering shape is a ^{1}⁄_{4} by ^{1}⁄_{7} right-angled triangle. This clearly tiles the square perfectly and you can also fit 24 of them into the equilateral triangle. To see this first divide the equilateral triangle exactly into 24 right-angled triangles of sides ^{1}⁄_{4} and 1/(4sqrt(3)), and then note that ^{1}⁄_{7} < 1/(4sqrt(3)). There’s no point in drawing a picture since you can barely see the gaps.

The cheaty solution at the end depends on what seems to me an unintended interpretation of the question (though, given that the same person wrote the question and the program that found the solution, maybe my idea of what’s intended is wrong). I took “tile both polygons” to mean “tile polygon 1 AND tile polygon 2″, not “tile the union of polygons 1 and 2”.

It is a solution similar to the one with the big shape which doesn’t cover anything, but the remainder is arbitrarily miniscule relative to the shape.

We called it “a trivial solution”, as I call this solution trivial, but maybe less trivial, since it actually covers all and it wasn’t explicitly forbidden, not to people, not to the computer.

Now, I told my chitin’ comp, don’t do that, each instance of the shape should cover either the triangle, either the square!

Interesting. We will see where this is going to go.

Any luck? I’d be interested in seeing some of the computer solutions even if their scores didn’t beat mine.

By the way I can now improve my score to 14sqrt(3)-24 = 0.249… . My covering shape is a

^{1}⁄_{4}by^{1}⁄_{7}right-angled triangle. This clearly tiles the square perfectly and you can also fit 24 of them into the equilateral triangle. To see this first divide the equilateral triangle exactly into 24 right-angled triangles of sides^{1}⁄_{4}and 1/(4sqrt(3)), and then note that^{1}⁄_{7}< 1/(4sqrt(3)). There’s no point in drawing a picture since you can barely see the gaps.https://protokol2020.wordpress.com/2017/03/30/intermezzo-problem-solutions/

The cheaty solution at the end depends on what seems to me an unintended interpretation of the question (though, given that the same person wrote the question and the program that found the solution, maybe my idea of what’s intended is wrong). I took “tile both polygons” to mean “tile polygon 1 AND tile polygon 2″, not “tile the union of polygons 1 and 2”.

It is a solution similar to the one with the big shape which doesn’t cover anything, but the remainder is arbitrarily miniscule relative to the shape.

We called it “a trivial solution”, as I call this solution trivial, but maybe less trivial, since it actually covers all and it wasn’t explicitly forbidden, not to people, not to the computer.

Now, I told my chitin’ comp, don’t do that, each instance of the shape should cover either the triangle, either the square!

We will see.

I must say, that this solutions of yours is quite impressive. Quite impressive indeed.

I promise you scores and images of solutions, whatever they will be. Calculations are under way right now and they should be available soon.