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Let $p_i=(r_1\cos\phi_i,r_1\sin\phi_i)$ and $q_i=(r_2\cos\theta_i,r_2\sin\theta_i)$. Then rotating by $\psi$ leads to \begin{align} \sum_{i=1}^nd(p_i,q_i')^2 &= \sum_{i=1}^n\left((r_1\cos(\phi_i+\psi)-r_2\cos\theta_i)^2+(r_1\sin(\phi_i+\psi)-r_2\sin\theta_i)^2\right) \\ &= n(r_1^2+r_2^2)-2r_1r_2\sum_{i=1}^n\cos(\phi_i-\theta_i+\psi) \\ &= ... 4 For brevity, I will not go into details about how to construct perpendiculars and 30-degree angles using straightedge and compass. Same for doubling lengths. Translating those operations to compass-and-straightedge primitives is left to the reader. Given the three parallel lines and a vertex C arbitrarily fixed on one of those lines, I will find the two ... 3 Well, I'm late to this, but I've been obsessing over this far too much to back out now. So: Label your parallel lines a, b, and c from bottom to top. Construct a line d with a "positive slope" that crosses b such that the top right angle of their intersection is 60^\circ. Extend the line so that is crosses c. Call the intersection of d ... 3 You don't need Euclidean geometry to prove that lines intersect once, no times, or infinitely many times. This can be proven in affine geometry, the point being that affine geometry has a notion of lines and parallel lines but not a notion of length. Affine geometry in turn is more or less linear algebra, which is thoroughly embedded into modern mathematics. ... 3 Take C centered at the origin, and let p_i = (R_i \cos \phi_i, R_i \sin \phi_i) and q_i = (\cos \psi_i, \sin \psi_i). Say we rotated the circle by an angle \theta. Then you want to minimize the sum \sum_{i=1}^n (R_i \cos \phi_i - \cos( \psi_i + \theta))^2 + (R_i \sin \phi_i - \sin (\psi_i + \theta))^2 = \sum_{i=1}^N R_i^2 + 1 - 2R_i(\cos \phi_i ... 2 You must be knowing that the director circle subtends right angle tangents to the the circle. Here the director circle for [x^2 + y^2 = a^2] is [x^2 + y^2 = 2a^2] (you may prove it by simple geometry) 1.From an arbitary point on DC, P(a*sqrt2*cos(w),a*sqrt2*sin(w)) make a Chord Of Contact on original circle as T=0 for P. 2.Take (h,k) as the mid point of ... 2 Great question! There actually is a general theory for this, or more accurately, many general theories. The different theories depend on what feature you want to generalize. Here are some examples, though I am sure the list is far from complete: Riemannian manifolds: This is a huge area with much work in it! Here, you say that the key feature of these ... 1 Different geometries denote different sets of axioms, which in turn result in different sets of conclusions. I'll concentrate on the planar cases. Projective geometry is pure incidence geometry. The basic relation expresses whether or not a point lies on a line or not. One of its axioms requires that two different lines will always have a point of ... 1 1) One method you can use to find line L_1 is to make equations in x, y, z for each plane and solve them, instead of finding the cross product:x=\lambda_1+2\mu_1, y=\lambda_1-\mu_1, z=-\lambda_1+\mu_1x=\lambda_2+3\mu_2, y=\lambda_2+\mu_2, z=\lambda_2-\mu_2 What we'll do is find the line of intersection; this line is clearly parallel to both ...