Another Symmetry Puzzle
I recently posted a symmetry puzzle from Donald Bell. He just sent me another one.
Puzzle. Start with a 30-60-90 triangle (half of an equilateral triangle). Divide it into two 30-60-90 triangles of different sizes by dropping a perpendicular from the right-angled corner to the opposite side. Put the resulting two pieces together to form a symmetrical shape. There are two solutions.
It took me some time to find the second solution. I love this puzzle.
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Lazar Ilic:
One could compose a programme fairly easily. I immediately observed a solution where one moves the smaller triangle over so that there is a spiral symmetry formed at the 30 degree vertices taking the smaller hypotenuse on to the larger long leg. At this point it became quite clear that partitioning the larger triangle in to 3 copies of the smaller triangle naturally leads to the other solution. Both of the 2 internal lines in that partitioning can be used as lines of symmetry. So in particular the other solution looks like 2 30 30 120 triangles stuck together to also form 2/3rds of an equilateral triangle from the centre via the other thirds splitting.
12 July 2024, 2:37 am