Is it possible to divide an equilateral triangle into three similar parts, in which two are identical but the third one is of different size?
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$\begingroup$ What do you mean by "parts"? Do you mean any shape at all, or a specific shape? If the shape doesn't matter, you can chop off two corners in the same way and get two identical pieces with a different-sized third piece. $\endgroup$– Jānis LazovskisCommented Dec 4, 2017 at 5:18
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2$\begingroup$ @JānisLazovskis You can do that, but the third piece won't be similar to the other two, will it? $\endgroup$– Sophie SwettCommented Dec 4, 2017 at 5:21
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$\begingroup$ @TannerSwett right, I took "similar" in the non-technical sense. Seems more difficult than I thought at first. $\endgroup$– Jānis LazovskisCommented Dec 4, 2017 at 5:28
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1 Answer
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Yes.
Start with a triangle with unit sides ABC.
Divide each side into thirds, so the sides are ADEB, BFGC, CHIA.
Colour triangle AEH blue; BEF red; CGH yellow. The blue triangle has sides twice as long as the red and yellow ones.
That leaves trapezium HEFG that has not been coloured yet. Divide it into four similar trapeziums. Three of the smaller trapeziums have their long side along one of the short sides of the largest trapezium. Colour the middle two trapeziums blue.
That leaves two trapeziums. Repeat the division, but assign the middle ones to red and yellow respectively.
In the third stage, colour them blue; in the fourth stage, red and yellow.
At each stage, half the remaining area is assigned, leaving twice as many trapeziums as before.
EDIT:
I think this can be tweaked, so the width of the blue shape is any multiple of the other two, greater than $\phi=(1+\sqrt{5})/2$. When the trapezium is divided, only the two remaining trapeziums have to be the same shape as the original. The bounding value is when the two remaining trapeziums start to overlap.
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$\begingroup$ So, if I understand correctly, this is a fractal with infinite iterations but bounded size, and three copies of the fractal, one of which has four times the area of the other two, fit together to form an equilateral triangle. That's really rather nice. $\endgroup$– nickgardCommented Dec 6, 2017 at 15:03
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$\begingroup$ Yes, that's the idea. I wonder what the dimension of the boundary line is? $\endgroup$– Empy2Commented Dec 7, 2017 at 6:53