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Picture is attached. I want to use similarity to find the ratio but am not sure where to start. Hints are appreciated.

enter image description here

ACB
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sato
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    hint: Let $u_n,v_n$ be the length of left/right sides of $n^{th}$ gray triangle (counting from left to right). Both $u_n$ and $v_n$ are arithmetic progressions. What are $u_1 = v_{\color{red}{0}}$ and $u_{\color{red}{5}} = v_{\color{red}{5}}$? – achille hui Sep 30 '21 at 03:49

1 Answers1

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Let $a$ denote side length of an equilateral triangle. Then, the area of one equilateral triangle is $A=\frac12 (a)(a)\sin60^\circ$.

By similarity of triangles, we can find marked lengths in the figure below.

enter image description here

Therefore the area of shaded triangle on left is $\frac12(a)(\frac45a)\sin 60^\circ=\frac45A$

Using the property, "similar triangles are to one another in the squared ratio of (their) corresponding sides" [1], we can find other shaded areas as well.

Can you take it from here?

ACB
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  • See this image if the above is not clear. – ACB Sep 30 '21 at 09:18
  • Which triangles have you taken similar for example for the first triangle from left, that is, for finding it to be $4a/5$? – sato Sep 30 '21 at 12:15
  • @sato Do you see the line parallel to $\frac45a$ with length $a$ on the left?(which I constructed) Those are the corresponding sides of those similar triangles with a common vertex on the upper right corner of apperant parallelogram. (I couldn't label points because it will make the picture a mess!) – ACB Sep 30 '21 at 12:35
  • yes i got it now – sato Sep 30 '21 at 12:52