(It's perhaps easier to understand the dimensions if you actually find some Legos to play with, but the only extra piece of information you need is this: When rotated, the corner of the top Lego piece will barely brush the side of the second connection button on the bottom Lego piece. With this, can you determine the dimensions of the connection buttons in terms of the side-length of the pieces?)

Source: Original.

The angle is equal to arcsin((3 - sqrt2)/2), or about 52.4569 degrees. Solutions were received from Joseph DeVincentis, Christopher Thomson, and Colin Bown. Joseph's solution:

If the corner of a rotating brick just brushes against the second button on the lower piece, then we know the distance from the center of rotation to the edge of the second button equals half the diagonal of a 1x1 lego square, or sqrt(2)/2 when the length of the lego square is taken to be 1. This distance consists of half the length of a lego square plus the distance from a button to the edge of a lego square. Half the length is just 1/2, so the distance between the button and the edge of the lego square is (sqrt(2)-1)/2. Now the radius of a connection button is just half the length of a lego square minus the above distance, which is 1/2 - (sqrt(2)-1)/2 or 1 - sqrt(2)/2. Now, in the figure with two hinged 1x4 pieces closed as tightly as possible, the edge of the top piece is tangent to the second button on the lower piece. Draw two perpendiculars to this tangent line: the first from the center of rotation, the second from the center of the second button on the lower piece. Also connect these two centers by a line. The length of the first perpendicular is 1/2. The length of the second perpendicular is the radius of a button. The length of the line of centers is 1. Two similar triangles are formed by these three lines and the tangent edge of the top piece. A third similar triangle can be formed by constructing a line parallel to the tangent edge, from the center of the second button to the extension of the first perpendicular. This is a right triangle with hypotenuse 1 and one leg of length 1/2 + r = (3-sqrt(2))/2. The angle at the second button is the one we want -- between lines parallel to the long edges of the two pieces. The sine of this angle equals (1/2 + r)/1, or (3-sqrt(2))/2, and thus the angle is about 52.5 degrees.

Colin Bown's generic solution:

In general, two circles O1 and O2 w/ radii r1 and r2 w/ centers separated by d difine 4 lines tangent to both. we are interested in the angle between either of the lines that cross the axis between the circles and that axis. Call the points of tangency, T1 and T2, and the point on intersection of the tangent and the axis, A. let x be the length of segment O1, A1. sin theta = r1 / x now we find x by considering the similar triangles: r1/x = r2/(d-x) ==> x= r1 d / (r1 + r2) so theta = arcsin( (r1+r2)/d ) For our lego: d = 1, r2 = .5, and 1 - r1 = sqrt(2)/2 from the corner constraint. result: theta ~= 52.4569 degrees So make all the equilateral triangle based honeycombs you want but demonstrate the beauty of a 3 4 5 right triangle to your kid? Forget about it! You'll have to get out the tinker toys to even show them a right triangle. Don't get me started on Lincoln Logs!

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