Some questions on the intersection of three cones.

Question

I have three cones in $\mathbb{R}^3$, explicitly defined by the equations: $$ (x-\alpha_x)^2+(y-\alpha_y)^2=(z-r_1)^2 \,, \\ (x-\beta_x)^2+(y-\beta_y)^2=(z-r_2)^2 \,, \\ (x-\gamma_x)^2+(y-\gamma_y)^2=(z-r_3)^2 \,. \tag{1} $$ The parameters obey the conditions: $$ \alpha_x^2+\alpha_y^2 = \beta_x^2+\beta_y^2 = \gamma_x^2+\gamma_y^2 \leq 1 \,, \\ r_1+r_2+r_3 = 1 \,, \quad 0<r_1,r_2,r_3<1 \,. $$

Side question 1: By demanding $z>c$ with $0<c<1$, what are the conditions such that the cones have a unique point of intersection? What is the solution? What if I add more cones into the game?

Assume now that a unique solution exists and let's denote it with $x_0, y_0, z_0$. Let $\kappa \in [0,1]$ denote a parameter and consider the closely related problem $$ (x-\kappa\alpha_x)^2+(y-\kappa\alpha_y)^2=(z-r_1)^2 \,, \\ (x-\kappa\beta_x)^2+(y-\kappa\beta_y)^2=(z-r_2)^2 \,, \\ (x-\kappa\gamma_x)^2+(y-\kappa\gamma_y)^2=(z-r_3)^2 \,. \tag{2} $$ That is, the projection on the $x-y$ plane of the vectors to the origins of the cones have been rescaled; or equivalently the origins of the cones are moved towards the $z$ axis by the same amount (the distance of the origins to the $z$ axis is the same for all three cones, which follows from the conditions on the parameters). This implies that in fact the origins of the cones have come closer to each other.

Main Question: Assuming that solutions exist for all $\kappa$ (or at least for some subset of [0,1]), is there a way to relate the solutions of the second problem to the solutions of the first. That is, can I obtain $$ x_\kappa = f_1(\kappa, x_0, y_0,z_0) \,, \\ y_\kappa = f_2(\kappa, x_0, y_0,z_0) \,, \\ z_\kappa = f_3(\kappa, x_0, y_0,z_0) \,. $$ Similarly, can one obtain the curve traced by the points of unique intersection with varying $\kappa$?

Side Question 2: Are there any complications when I go to $\mathbb{R}^4$ or is it a direct generalisation?

Regarding the main question, for some specific parameters I have numerically solved the problem and found a curve that looks like in the following graphic, which confirms my suspicion that at least in certain cases a curve like that exists.

Answer 1

Hint:

If I am right, your problem is not so terrible as it seems. Subtract one equation from the two others and you get the equations of two planes. Hence you in fact discuss the intersection of a straight line and a cone.

Find a point on the line (such as the intersection with a perpendicular plane through the origin) and the direction vector (cross-product of the normals). This gives you a parametric equation of the line. Then, plugging in the implicit equation of the first cone, you'll get a quadratic polynomial.

Tedious computation, but giving an analytical solution.

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WLOG, $\alpha_x=\alpha_y=r_1=0$ (if not, translate the apexes).

By subtracting the second and third equations from the first,

$$\beta_xx+\beta_yy-r_2z=\frac{\beta_x^2+\beta_y^2-r_2^2}2, \\\gamma_xx+\gamma_yy-r_3z=\frac{\gamma_x^2+\gamma_y^2-r_3^2}2.$$

Now we solve $x,y$ for $z$ and plug in the equation of the first cone. The relation is linear

$$x=u_xz+v_x,\\y=u_yz+v_y,$$ giving

$$(u_xz+v_x)^2+(u_yz+v_y)^2=z^2$$

or

$$(u_x^2+u_y^2-1)z^2+2(u_xv_x+u_yv_y)z+v_x^2+v_y^2=0.$$

The discriminant vanishes when

$$(u_xv_x+u_yv_y)^2-(u_x^2+u_y^2-1)(v_x^2+v_y^2)=v_y^2+v_x^2-u_y^2v_x^2+2u_xu_yv_xv_y-u_x^2v_y^2=0.$$

Remains to draw the expressions of the solution of the $2\times2$ system and expand.

Answer 2

WLOG, $\alpha_x=\alpha_y=r_1=0$.

By subtracting the second and third equations from the first,

$$\beta_xx+\beta_yy-r_2z=\frac{\beta_x^2+\beta_y^2-r_2^2}2=:\delta_2, \\\gamma_xx+\gamma_yy-r_3z=\frac{\gamma_x^2+\gamma_y^2-r_3^2}2=:\delta_3.$$

Now we solve $x,y$ for $z$ and plug in the equation of the first cone. The relation is linear

$$x=u_xz+v_x,\\y=u_yz+v_y,$$ giving

$$(u_xz+v_x)^2+(u_yz+v_y)^2=z^2$$

or

$$(u_x^2+u_y^2-1)z^2+2(u_xv_x+u_yv_y)z+v_x^2+v_y^2=0.$$

The discriminant vanishes when

$$(u_xv_x+u_yv_y)^2-(u_x^2+u_y^2-1)(v_x^2+v_y^2)=v_y^2+v_x^2-u_y^2v_x^2+2u_xu_yv_xv_y-u_x^2v_y^2=0.$$

Answer 3

Considering only Side Question 1 ...

I'll replace your $\alpha_x$ and $\alpha_y$ with $a\cos\alpha$ and $a\sin\alpha$ (for $0\leq a\leq 1$), and likewise for your $\beta_{-}$ and $\gamma_{-}$; also, I'll write $p$, $q$, $r$ for your $r_i$.

As noted in @Yves' answer, you can subtract the second and third equations from the first to get plane equations, and parameterize their line of intersection. Explicitly, we can write that parameterization as $$(x,y,z) = \frac{ \left( \begin{align}&\phantom{\;\;\;\;} a \left( b^2 - c^2 - q^2 + r^2 +2t(q-r)\right)\;(\sin\alpha,-\cos\alpha,0) \\ &+ b \left( c^2 - a^2 - r^2 + p^2 +2t(r-p)\right)\;(\sin\beta,-\cos\beta,0) \\ &+ c \left( a^2 - b^2 - p^2 + q^2 +2t(p-q)\right)\;(\sin\gamma,-\cos\gamma,0) \end{align} \;\;\right)}{2 \left(\;a b \sin(\alpha-\beta) + b c \sin(\beta-\gamma) + c a \sin(\gamma-\alpha) \;\right)} + (0,0,t) \tag{1}$$

Substituting into the first cone equation gives ... a mess ... but it is quadratic in $t$, and we conclude that the line meets the cone in a unique point when the discriminant of that quadratic vanishes. Surprisingly, that discriminant factors nicely, and we have:

$$\begin{align} 0 &= \phantom{\cdot}\left( a^2 + b^2 - 2 a b \cos(\alpha - \beta ) - (p-q)^2 \right)\\ &\;\phantom{=}\cdot\left( b^2 + c^2 - 2 b c \cos(\beta - \gamma) - (q-r)^2 \right)\\ &\;\phantom{=}\cdot\left( c^2 + a^2 - 2 c a \cos(\gamma - \alpha) - (r-p)^2 \right)\\ &\;\phantom{=}\cdot\left(a b \sin(\alpha-\beta) + b c \sin(\beta-\gamma) + c a \sin(\gamma-\alpha) \right)^2 \end{align}\tag{2}$$

Note that if we define $A'$, $B'$, $C'$ as the projections of the cone apices $A$, $B$, $C$ into the $xy$-plane, with $d := |B'C'|$, $e := |C'A'|$, $f := |A'B'|$, and $T := |\triangle A'B'C'|$. $$\begin{align} d^2 &= b^2 + c^2 - 2 b c \cos(\beta-\gamma)\\ e^2 &= c^2 + a^2 - 2 c a \cos(\gamma-\alpha)\\ f^2 &= a^2 + b^2 - 2 a b \cos(\alpha-\beta) \\ T &= \frac12 \left( b c \sin(\beta-\gamma) + c a \sin(\gamma-\alpha) + a b \sin(\alpha-\beta) \right) \end{align}$$

Since the denominator of $(1)$ is a multiple of $T$, it must be non-zero; that is, $\triangle A'B'C'$ must be non-degenerate. Thus, the final factor of $(2)$ is non-zero, too, and can be ignored, so that we have

$$\begin{align} \left( d^2 - ( q - r )^2 \right)\left( e^2 - ( r - p )^2 \right)\left( f^2 - (p - q )^2 \right) = 0 \end{align} \tag{3}$$

We deduce that the "vertical" distance (in the $z$ direction) between some pair of apices is equal to the "horizontal" distance (in the $xy$-plane). That is,

At least one of $\overline{AB}$, $\overline{BC}$, $\overline{CA}$ makes a $45^\circ$ angle with the $xy$-plane.

For instance, supposing $\overline{BC}$ makes the $45^\circ$ angle, this says exactly that (presumably-distinct) $B$ and $C$ lie on either other's cones, with the line joining them being a line-of-tangency of those cones and thus their intersection. Clearly, that line meets $A$'s cone exactly once, if at all, unless $A$ itself is on the line (which would make that entire line the common intersection of all three mutually-tangent cones).