Little explorations with HP calculators (no Prime)

[Thomas Okken]

Consider a half-unit circle jammed into the corner of the first quadrant (so its center is at (0.5, 0.5)). Now consider a radius of that circle sweeping the angles from 0 to pi/2. The tangent on the circle where it meets that radius will intersect the X axis at 1 + tan(phi), and the Y axis at 1 + cot(phi) or 1 + 1 / tan(phi). The triangle formed by the X axis, the Y axis, and this tangent, is like the four triangles in the puzzle, and the challenge is to find phi such that X = Y + 1 (or X = Y - 1). The answer to the puzzle is then obtained by scaling everything down so that the hypotenuse of the triangle OXY becomes 1, and then the diameter of the circle is 1 / sqrt(X^2 + Y^2).

EDIT: No, I screwed up. The intersections at the axes are not at 1 + tan(phi), etc., that relationship is not quite that simple. Back to the drawing board!

Second attempt:

Consider a unit circle jammed into the corner of the first quadrant (so its center is at (1, 1)). Now consider a radius of that circle sweeping the angles from 0 to pi/2. The point P on the circle where that radius intersects it is at (1 + cos(phi), 1 + sin(phi)). The tangent on the circle at that point will have a slope of -1 / tan(phi), and so it will intersect the X axis at Px + Py * tan(phi), or (1 + sin(phi)) * tan(phi) + 1 + cos(phi), and it will intersect the Y axis at (1 + cos(phi)) / tan(phi) + 1 + sin(phi). The triangle formed by the X axis, the Y axis, and this tangent, is like the four triangles in the puzzle, and the challenge is to find phi such that X = Y + 2 (or X = Y - 2). The answer to the puzzle is then obtained by scaling everything down so that the hypotenuse of the triangle OXY becomes 1, and then the radius of the circle is 1 / sqrt(X^2 + Y^2).

Because of symmetry, sweeping the angles from 0 to pi/2 is actually not necessary; you can restrict yourself to 0 through pi/4 and the case that X = Y - 2.