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This puzzle is an illustration of a slice group symmetry.
Q, the radius of the yellow circle, is 'outside' the circle and the square, P.
This is a numerical map of the rationals, over a section of Pi, or how quadrature of circle Y, fails when Q - P is beyond a/2 the outer 'products' diameter for Q.
And Kepler's golden ratio, since the trigonometry is, when you do use it it shows that a Kepler distance is the root of any q, along the intersection of P, for radius Q. Q is 'on' X where the tangent of Y is.
The letters have not been chosen arbitrarily.
I might be pushing my feeble grasp of symmetry groups, but I think Q0 is the centroid, and P is a helicoid over S3 the sphere, the slice group is this map of the circle U(1) subgroup with the intersection (slice) of P through Q. Q = Y, and X is all tangents on Y and roots in Q on P.
Could you 'not use' trigonometry by using a ruler to extend Q to the tangent on the outer semicircle, radius a/2? Also produce line P' from this new point, a reflection of Q, i.e. from Q' to the lower tangent on a/2? Quadrature is in the a/2 upper half plane of left-right, - +, + +, quadrants, divided at Y = 0 = a - a/2.
Then Q' is less than any X,X' values for P, along -a/2. When P = a/2, the blue square is larger in area than Y, and Q' is the (invariant/variant) radius of Y. The triangle QQ'X, where X is the P' intersection, is the eccentricity for (Q,P) the Archimedes spiroid over X,Y.
Or something. Might need a bit more gold.
Hm, note also that as it's drawn, P doesn't slice Y enough that if Q rotates Y, P will section Y completely - it isn't deep enough to be a 'deep-slice' puzzle, ala Rubik's group on the cube. So until P approaches Q and the centroid, no cube symmetry groups appear since Y is only shallow-cut.
Now consult
inertial frames of reference on plane of ecliptic for planet earth that extend to 'infinity', or to a fixed set of celestial objects called quasars and binary neutron stars, so that VLBI astronomy measures nutational changes in the earth's tilt angle along QQ' as it precesses in the total solar moment P(m) over Q, for mE the energy of the geogyroscopic helicoid, relative to at least P' the lunar moment, and S' the solar moment. Use Kepler distance-measure again to verify Archimedes spiroid exists in planetery motion, up to 0.001(pi)C the circular measure of e in the rationals.
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