Cutting A Scaled Cone

Equation of regular cone, with angle $\theta_c$ between its surface and its axis, is:

$r \cdot \hat{k} = \cos \theta_c | r |$

Let r' be the corresponding point on the stretched cone, then

$r' = S r$

where

$S = \begin{bmatrix} S_x && 0 && 0 \\ 0 && S_y && 0 \\ 0 && 0 && S_z \end{bmatrix}$

Thus $r = S^{-1} r'$

Thus for a stretched cone, its equation is obtained by replacing the vector $r$ with $S^{-1} r'$ ,

So, we have,

$((S^{-1} r') . k)^2 = \cos^2 \theta_c (S^{-1} r') \cdot (S^{-1} r' )$

hence the equation becomes

$r^T Q r = 0$

where

$Q = S^{-1} (k k^T - \cos^2 \theta_c I) S^{-1}$

at the final step, I replaced $r'$ with $r$ , for notation simplicity.

Now, the equation of the cutting plane in vector form is $r = r_0 + V u$ , where $V = [ v_1, v_2 ]$ , where $v_1$ , $v_2$ are two unit vectors that are orthogonal to each other and to the normal to the plane. $u$ is the coordinate vector with respect to $v_1$ and $v_2$ .

Substituting this into the equation of the stretched cone,

$(r_0 + V u)^T Q (r_0 + V u) = 0$

Expanding

$u^T V^T Q V u + 2 r_0^T Q V u = - r_0^T Q r_0$

the left hand side can be rewritten as

$(u - u_0)^T V^T Q V (u - u0) - u_0^T V^T Q V u_0$

where $u_0 = -(V^T Q V)^{-1} V^T Q r_0$

Hence,

$(u - u_0)^T V^T Q V (u - u_0) = u0^T V^T Q V u_0 - r_0^T Q r_0$

Define $A = V^T Q V$ and $c = u_0^T V^T Q V u_0 - r_0^T Q r_0$ , then we have

$(u - u_0)^T A (u - u_0) = c$

Dividing both sides by c,

$(u - u_0)^T B (u - u_0) = 1$

where $B = A / c$

Since B is symmetric, we can find a rotation matrix (an orthogonal matrix) R and a diagonal matrix D, such that

$R^T B R = D$

Note that

$R = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$

This suggests the change of variable, $u = u_0 + R w$ , that leads to the simpified expression

$w^T D w = 1$

Thus with respect to the w-coordinate frame, this is an ellipse with semi-major and semi-minor axes equal to the reciprocal of the square roots of the diagonal elements of D.

Note that since u and w are related by a translation and a rotation then these dimensions remain unchanged when moving from the w-coordinates to the u-coordinates.

In particular, since $u = u_0 + Rw$ , and since R is a rotation matrix by an angle $\theta$ , then with respect to the u-coordinate frame, this is an ellipse centered at u_0, with its axes rotated by the angle $\theta$ .

And finally, $r = r_0 + V u = r_0 + V (u_0 +R w) = r_0 + V u_0 + V R w$

Therefore, $r$ describes an ellipse centered at $r_0 + V u_0$ , having its major and minor axes in the plane spanned by $\{ v_1, v_2 \}$ , rotated by angle $\theta$ from the direction of $v_1$ and $v_2$ .

Making the necessary calculations, we obtain a semi-minor axis of $m = 20.70276$ and a semi-major axis of $M = 44.029848$ , making the answer $M + m = \boxed{64.733}$