At some point over the last few days Danny Brown (@dannytybrown) tweeted this puzzle:
The picture shows a unit square with two congruent rectangles and a number of triangle. I looked,at the puzzle, saw no obvious answer, realised it would need a bit of thought so saved the photo until I found some time.
Yesterday I had some time, so I started by sketching the puzzle and filling in things I knew:
I quickly realised that the angles for all the triangle were the same, and thus they were all similar. (For some reason I missed one of the triangles on this page.) I couldn’t work out how I could find a missing length. I worked out the length if the diagonal of the rectangle using Pythagoras’s Theorem but it didn’t really help.
Then I drew some perpendiculars from each end of the diagonal and had a breakthrough.
If two rectangles share a diagonal, and a side length, then they must be congruent. This meant that I now knew the length of the bottom was 2x + 2y, as it’s a unit square then 2x + 2y = 1 so x+y=1/2.
I filled in these lengths:
Redrew my similar triangles and solved:
As no one likes an irrational denominator I rationalised as ended up with the satisfying solution 2-rt3.
I love this puzzle, and think it could be used as enrichment for high ability learners of many ages. After I’d solved it I looked at some solutions others had posted, and was thankful to see that other people also unnecessarily use trigonometry, and I was amazed I’d managed to avoid it, given my track record!