Here is my solution to M/A2. It’s straightforward in theory, but messy in its numbers. To simplify them a little, I divided all of them by 3; at the end I multiply them by 3. The locus of possible medallion positions is the lower portion of an ellipse lying below its foci (x’,y’) = (0,0) and (5, -4). The parameters of the ellipse as shown on the left figure, using the Pythagorean relation, are semi foci separation f = (42+52)1/ 2/2 = 411/2/2, major axis a = 5 (half the string’s length), and semi-minor axis b = (52- 41/4)1/2 = 591/2 /2. The equilibrium position of the medallion will be the point of tangency of a horizontal tangent to the ellipse. Referring to the right figure of the usual handbook drawing of an ellipse centered at the (x,y) = (0,0) origin, the equation of its tangent line, with (x1,y1) its point of tangency, is xx1/a2 + yy1/b2 = 1. From the equation recast in the form of slope-intercept, the slope m = -x1b2/a2y1. It’s key now is to make m = 4/5 so that the tangent line will be horizontal after the usual figure’s orientation of the ellipse is rotated to the actual ellipse on the left figure. To complete the solution for the position (x2’,y2’) where the medallion rests, solve simultaneously for x1 and y1 using the equation for m and the relation x12/a2 + y12/b2 = 1. The result is x1 = a/(1 + b2/m2a2)1/2 = 40/1231/2 = 3.607 and y1 = y1’ = -x1b2/ma2 = -29.5/1231/2 = -2.660. Completing the conversion to primed coordinates, x1’ = x1 + f = 6.809. Finally, rotating these primed coordinates clockwise by -tan-1 (4/5) = -38.66 (T) degrees using conventional formulas: x2’ = x1’ cos T – y1’ sin T = 3,655 and y2’ = x1’ sin T + y1’ cos T = -6.333. Multiplying by 3 gives the result for the medallion coordinates (11.0,-19.0).