Given a circle, construct three circles tangent to it and tangent to each other.
Surprisingly, I can find no reference to this (although there are many similar sounding problems, and Apollonius' Problem, none of which solve this.)
My solution is below. I request verification, improvements to the exposition, or alternate solutions. In particular, the construction has many moving parts, and I struggled to simplify it. Is there a simpler construction, or a simpler way of writing this one?
Construction: Circumscribe a [added: equilateral] triangle $\triangle ABC$ around given circle $\bigcirc O$. Label the midpoint of $AB$ as $P$, of $BC$ as $Q$, and of $CA$ as $R$. By symmetry, these midpoints are the three points of tangency between $\triangle ABC$ and $\bigcirc O$.
Construct the bisector of $\angle PAO$ as ray $A_p$, and the bisector of $\angle PBO$ as ray $B_p$. Likewise construct bisector rays $B_q, C_q, C_r, A_r$. Let $A_p$ and $B_p$ intersect at point $p$, rays $B_q$ and $C_q$ at point $q$, and $C_r$ and $A_r$ at $r$. (See figure. For simplicity, only some objects are shown, with the others following the same scheme.)
Now draw circles $\omega_p, \omega_q, \omega_q$ with centers $p, q, r$ respectively, each tangent to $\bigcirc O$ at $P,Q,R$ respectively. We will show that $\omega_p, \omega_q, \omega_r$ are tangent to each other.
The reflection of ray $AB$ over ray $A_p$ is ray $AO$. Therefore, since $AB$ is tangent to $\omega_p$, so is $AO$. By a similar argument, $AO$ is tangent to $\omega_r$ as well. By symmetry, these two points of tangency must be equidistant from $O$ and therefore coincide, showing that $\omega_p$ and $\omega_r$ are tangent to each other along $AO$. A similar argument works for each pair of circles, completing the proof.
Update
Based on @aschepler's suggestion, I rewrote the solution, which I think is clearer, more succinct, and more insightful. Further feedback is appreciated.
Circumscribe an equilateral triangle $\triangle ABC$ around the given circle $\bigcirc O$. Construct $\omega_{abo}, \omega_{bco}, \omega_{cao}$ as the incircles of $\triangle ABO, \triangle BCO,$ and $\triangle CAO$ respectively. These circles are the required circles, tangent to each other and to $\bigcirc O$, as we now show.
Recall that the incircle of an isosceles triangle is tangent to the midpoint of its base. Therefore, $\omega_{abo}, \omega_{bco}, \omega_{cao}$ are tangent to the midpoints of $AB, BC, CA$ respectively. For the same reason, $\bigcirc O$ is tangent to each of these same points.
Likewise, both $\omega_{abo}$ and $\omega_{cao}$ are tangent to $AO$. By symmetry, these two points of tangency must be equidistant from $O$ and therefore coincide, showing that $\omega_{abo}$ and $\omega_{cao}$ are tangent to each other. A similar argument works for each pair of circles, completing the proof.
Best Answer
I would mention why the circles actually are tangent to $\bigcirc O$. This is simple, but worth justifying. If nothing else, it checks the requirement off the list, and makes sure it's not read as an assumption or something to be proved later.
I would replace the second paragraph with just "Construct $\omega_p$ as the incircle of $\triangle ABO$, $\omega_q$ as the incircle of $\triangle BCO$, and $\omega_r$ as the incircle of $\triangle CAO$."
Constructing a triangle's incircle is a standard construction, and readers familiar with compass-and-straightedge construction will know that it's normally done using angle bisectors, or at least will know that it can be done. If this is in the context of a book or class, you can refer back to some previous construction or exercise. If you think it's absolutely necessary, describe constructing a general triangle's incircle as a lemma here, rather than describing the steps three times.
Proving the four circles actually are mutually tangent will require naming some of the points and lines omitted there, but only the ones you need for this.