Determine the radius of circle $(C)$ which is less than those of circles $(C_1), (C_2), (C_3)$ that are tangent to $(C)$ and to one another.

Question

Consider four circles $(C_1), (C_2), (C_3)$ and $(C_4)$ which all lie on a sphere of radius $2$ and are tangent to one another. The radius of circle $(C)$ is less than those of circles $(C_1), (C_2), (C_3)$ which are equal to $1$. Determine the radius of circle $(C)$.

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)

By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]

"Someone call Apollonius or Descartes. How? I don't know, galvanism?"

"Mom, pick me up, I'm scared."

A clarification before we start - This problem is set in the Euclidean space/three-dimensional Euclidean geometry, not spherical geometry. The centres of all aforementioned circles don't lie on the surface of the sphere, but rather within the sphere.

Sorry for not being able to provide an illustration for this problem. You just have to use your ~imagination~.

For sake of convenience, the equation of sphere $(S)$ will be $x^2 + y^2 + z^2 = 4$. Furthermore, let $A(m - n; m; m), B(m; m - n; m), C(m; m; m - n)$ and $M(m; m; m)$ be the centres of circles $(C_1), (C_2), (C_3)$ and $(C_4)$ in that order $(m > 0, n > 0)$. This is based on the observation that $\triangle ABC$ is equilateral and $M$ lies on the half-line starting from $O$ and going through the centre of $\triangle ABC$.

I'm going to attempt to solve this problem in two-dimensional slices. (Imagine a spherical cake. Actually, you just need two bake two batches of cake in dome pans.) First of all, let's consider plane $(OCA)$.

As $OA = OB = OC = \sqrt{2^2 - 1^2} = \sqrt{3}$, $A, B, C$ all lie on sphere $(S')\colon \; x^2 + y^2 + z^2 = 3$. From which it can be inferred that $$2m^2 + (m - n)^2 = 3 \iff 3m^2 - 2mn + n^2 = 3 \implies n = m \pm \sqrt{3 - 2m^2}$$

Additionally, it can be calculated that $CA = 2 \times \sqrt{\dfrac{1}{\left(\dfrac{1}{\sqrt 3}\right)^2 + \left(\dfrac{2}{2}\right)^2}} = \sqrt{3}$.

And since $$CA = \sqrt{[m - (m - n)]^2 + (m - m)^2 + [m - (m - n)]^2} = n\sqrt 2 = \sqrt 2 \times \left(m \pm \sqrt{3 - 2m^2}\right)$$, it can be inferred that $n\sqrt 2 = \sqrt 3 \implies n = \dfrac{\sqrt 6}{2}$ and $\sqrt 2 \times \left(m \pm \sqrt{3 - 2m^2}\right) = \sqrt 3 \implies m = \dfrac{\sqrt 6}{2}$.

The coordinates of point $M$ are $\left(\dfrac{\sqrt 6}{2}; \dfrac{\sqrt 6}{2}; \dfrac{\sqrt 6}{2}\right)$ and $OM = \sqrt{3} \times \dfrac{\sqrt 6}{2} = \dfrac{3\sqrt 2}{2}$, which is larger than $2$, hmmm~ this is wrong.

I forgot to put in the condition, $3m^2 < 4 \iff -\dfrac{2\sqrt 3}{3} < m < \dfrac{2\sqrt 3}{3}$.

Well then, that's all for now. As always, thanks for reading, (and even more so if you could help), and have a great tomorrow~

By the way, the options were $\dfrac{3 - \sqrt{6}}{3}, \dfrac{3 + \sqrt{6}}{3}, \dfrac{3 - \sqrt{6}}{6}$ and $\dfrac{3 + \sqrt{6}}{6}$.

Answer 1

I'm just posting an image that might be helpful.

Answer 2

Since the radius of $C_1$ is $1$ then the distance between the center of the sphere and the plane which contains $C_1$ is $\sqrt{3}$(orange line segment). Assume that $C_1, C_2, C_3$ are placed symmetrically about the $z$-axis. Further assume that the center of $C_1$ lies in the $xz$ plane. Let the angle between the $z$ axis (the axis of symmetry, and the vector extending from the origin (the center of the sphere) to the center of $C_1$ be $\theta$, then the equation of $C_1$ is

$p(t) = \sqrt{3} (\sin(\theta), 0, \cos(\theta)) + \cos(t) ( \cos(\theta), 0, -\sin(\theta) ) + \sin(t) (0, 1, 0 ) $

And we want the intersection with the plane $y = \sqrt{3} x $ (so that the circle is contained in one third of the $2 \pi$ when projected onto the $xy$ plane) to be at a single point, therefore,

$ \sin(t) = \sqrt{3} (\sqrt{3} \sin(\theta) + \cos(t) \cos(\theta) ) $

has only one solution in $t$. Re-arrange,

$ \sin(t) - \sqrt{3} \cos(t) \cos(\theta) = 3 \sin(\theta) $

This will have a single solution if and only if

$ 1 + 3 \cos^2(\theta) = 9 \sin^2(\theta) $

so, $ \sin^2(\theta) = \dfrac{1}{3} $ , and $ \cos^2(\theta) = \dfrac{2}{3} $

Hence, it follow that the $x$ coordinate of the top of the circle is

$ \sqrt{3} \sin(\theta) - \cos(\theta) = 1 - \sqrt{ \dfrac{2}{3} } = 1 - \dfrac{\sqrt{6}}{3}= \boxed{\dfrac{ 3 - \sqrt{6}}{3}} $