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The center of the bottom square rotates so that a line cast from the highlighted corner perpendicular to the right face will intersect with the highlighted corner of the other square. I'm trying to find how to calculate that angle of rotation.

image

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  • $\begingroup$ It's not clear how you are manipulating the two squares to get the new picture, and it's not clear how their movement is constrained. Why do you need to rotate the bottom square at all if you can just stretch one corner over to the other? $\endgroup$ Mar 3, 2014 at 1:48
  • $\begingroup$ The center of the bottom square rotates so that a line cast from the corner perpendicular to that face will intersect with the highlighted corner of the other square. I'm trying to find how to calculate that angle of rotation. $\endgroup$
    – Waffle
    Mar 3, 2014 at 1:54
  • $\begingroup$ Presumably you mean that the bottom square rotates about its center, but that' much clearer, thanks. $\endgroup$ Mar 3, 2014 at 2:34

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You'll have to provide more information about what information it is exactly that you're starting with if you want a formulaic answer, but here's a diagram that should help

enter image description here

Easiest way I can think to find the point on the circle:

Let $A$ be the distance from the center of the bottom square to the corner of the top square. Let $s$ be the length of the (bottom) square. The center, point of tangency, and radius of the inscribed circle form the green right triangle. Let $B$ be the distance from the point of tangency to the corner. We have $$ A^2 = B^2 + (s/2)^2 $$ Use this to find the length $B$. Draw a circle of radius $B$ around the corner. This intersects the inscribed circle (which has radius $s/2$) at two points. Find the two points. One of those points is the point of tangency, which is what you want.

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  • $\begingroup$ Is there an easy way to find that point on the circle? $\endgroup$
    – Waffle
    Mar 3, 2014 at 17:14
  • $\begingroup$ See my latest edit. $\endgroup$ Mar 3, 2014 at 18:11

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