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?
[Math] Divide an equilateral triangle into three similar parts
geometrytriangles
Related Solutions
The answer is "yes", it is possible to divide equilateral triangle into $5$ equal parts, see the picture below which comes from here: https://ru-math.livejournal.com/831851.html
See the picture below:
Note that: $$\triangle AB'F \sim \triangle ABJ_1 \quad (1)$$ $$\triangle AFG \sim \triangle AJ_1K_1 \quad (2)$$ $$\triangle AGC' \sim \triangle AK_1C \quad (3)$$ As triangles $\triangle ABJ_1$, $\triangle AJ_1K_1$, $\triangle AK_1C$ share common sides and triangles $\triangle AB'F$, $\triangle AFG$ and $\triangle AGC'$ too, the similarity ratio $k$ is the same for $(1)$, $(2)$ and $(3)$. In this way we have: $$BJ_1=k\,B'F = k \, r$$ $$J_1K_1=k\,FG = k \, r$$ $$K_1C=k\,GC = k \, r$$
Therefore
$$BJ_1=J_1K_1=K_1C.$$
Best Answer
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.