From the people behind the Alan Turing Cryptography Competition.
You are reading the website of the 2020 edition of the MathsBombe Competition, which ended on Sunday 26 April at 11:59 pm

# Puzzle 7

Wacław sat at the marble tables of the Scottish Cafe in Lwów wondering how best to draw the triangle he had been thinking about.

The construction of the triangle was quite easy. Start with a filled equilateral triangle of side length 1cm, including the edges, call this triangle $$E_0$$

Inside $$E_0$$ there exists an upside down equilateral triangle of side length .5cm and with the same centre as $$E_0$$. By throwing away this upside down triangle from $$E_0$$ we are left with three equilateral triangles of side length .5cm, each of which include the edges. Let $$E_1$$ be the union of these three triangles.

Proceed iteratively, if $$E_n$$ consists of $$3^n$$ triangles of side length $$\frac{1}{2^n}$$cm, obtain $$E_{n+1}$$ by discarding an upside down triangle of side length $$\frac{1}{2^{n+1}}$$cm from the centre of each triangle in $$E_n$$.

The triangle Wacław was thinking about was the set $$E$$ of points in $$E_0$$ that never get thrown away under the iterative process described.

Drawing the thing wasn't so easy though. However finely he sharpened his pencil, the tip would have positive area, and so any picture drawn of $$E$$ would have positive area, while the actual area of $$E$$ is zero. If someone drew a really accurate picture of $$E$$ it would have zero area, so we probably wouldn't be able to see it. Maybe the best picture is just to leave the paper blank?

Eventually Wacław decided to let $$F$$ be the set of all points $$x$$ such that there exists a point $$y$$ in $$E$$ with the distance between $$x$$ and $$y$$ being less than 0.001cm, and to draw $$F$$ instead of $$E$$. This seemed at least to convey the basic geometry of $$E$$. What is the area in $$cm^2$$ of the shape $$F$$? Enter a decimal accurate to 3dp, so if you think the answer is 2.564432cm$$^2$$ enter 2.564.