@rgx You know what's interesting and relevant about this puzzle is that the lengths of the triangles are Fibonacci numbers.

@rgx If you move on along to right triangles with longer lengths that are still consecutive Fibonacci numbers, you can repeat this puzzle with a single square missing among larger right triangles, e.g. make the blue right triangles of leg lengths 3, 8, and 5, 13 respectively, then you would have a rectangular areas of size 40 and 39, because F₂F₄ - F₁F₃ = 1 for all consecutive Fibonacci numbers F₁, F₂, F₃, and F₄.

@rgx And the illusion, of course, works because the slope approaches the square of the golden ratio.

@JordiGH @rgx Of course! Very nice. Thank you for explaining the general principle.

@clacke @rgx Now I'm kind of curious who was the one who put together these observations about the Fibonacci numbers into a clever visual puzzle. I bet it was Martin Gardner, this is his kind of thing. Let me see what I can find...

I was thinking "yeah, neat, but so what?", not realizing it was a cliffhanger. 😀
@rgx An 8x3 triangle and a 5x2 triangle do not have the same tangent (slope), so the two composite triangles are not true triangles: The hypotenuse (diagonal) is slightly concave (caving in) on the top figure and slightly convex (bulging out) on the lower figure, which makes the lower figure exactly one grid-square larger.

You can see it if you look at the top right of the dark blue triangle in the lower figure. It covers the nearest crossing in the white grid, whereas the corresponding place in the upper figure leaves the crossing visible.
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