This is from another question which I started answering but which has been closed before I could finish my answer. Right now there is still a gap in my answer, so it is now my turn to ask about this problem in an attempt to fill that gap.
We have a ring of six circles with radii $a,b,c,a,b,c$, each of them touching both the previus and the next one from the outside. They also all touch the inside of a circle of radius $r$. Show that $a+b+c=r$.
The original question described this as a sangaku, so I'll keep that tag. But I don't know a source where this came from. (Update: See comment from brainjam for a source.) Below I'll show my approach for how to prove this fact, but there is one gap that I don't have a good handle for. Also, my approach used a lot of computer algebra, so there is a high chance someone has a more intuitive approach which would be better suited to a paper and pencil computation, or perhaps even avoid computations altogether.
The whole configuration has to be point symmetric with respect to the center of the circumcircle. Let's use $O$ to denote that center, $A,B,C$ for the centers of the first three circles, and $A',B',C'$ for those of their point reflections in $O$. Define $\alpha:=\measuredangle AOB, \beta:=\measuredangle BOC, \gamma:=\measuredangle COA'$. Due to the point symmetry, $\alpha+\beta+\gamma=\pi$. As we want to use the cosine rule eventually, let's first use angle sum formulas to rewrite this in terms of trigonometric functions.
\begin{align*}
\alpha+\beta+\gamma&=\pi \\
\cos(\alpha+\beta+\gamma) &= -1 \\
\cos(\alpha+\beta)\cos\gamma-\sin(\alpha+\beta)\sin\gamma &= -1 \\
\cos\alpha\cos\beta\cos\gamma-\sin\alpha\sin\beta\cos\gamma
-\sin\alpha\cos\beta\sin\gamma-\cos\alpha\sin\beta\sin\gamma &= -1
\end{align*}
Now the sines are in the way here. So let me combine $\sin^2\alpha+\cos^2\alpha=1$ with the equation above to eliminate $\sin\alpha$, then do the same with $\beta$ and $\gamma$. I used Sage and resultants for this elimination. The result I got was some polynomial of total degree $12$ in the cosines of the angles, but factoring it returned just a single factor with exponent $4$ which represents the following equivalent equation:
$$2\cos\alpha\cos\beta\cos\gamma + \cos^2\alpha+\cos^2\beta + \cos^2\gamma = 1$$
Now we can apply the cosine rule to find a relation between these cosines of the angles and the radii of the circles. That is because of the triangles seen in the figure above, the edges of which are sums or differences of radii.
\begin{align*}
(a+b)^2 &= (r-a)^2 + (r-b)^2 – 2(r-a)(r-b)\cos\alpha \\
(b+c)^2 &= (r-b)^2 + (r-c)^2 – 2(r-b)(r-c)\cos\beta \\
(c+a)^2 &= (r-c)^2 + (r-a)^2 – 2(r-c)(r-a)\cos\gamma
\end{align*}
I can use the first of these to eliminate $\cos\alpha$, the second to eliminate $\cos\beta$ and the third to eliminate $\cos\gamma$, again using resultants. When I factor the resulting degree $12$ polynomial I get:
$$
(a + b + c – r)\cdot r^2\cdot (a – r)^2 \cdot (b – r)^2 \cdot (c – r)^2 \cdot (4abc – ar^2 – br^2 – cr^2 + r^3) = 0
$$
The first of these factors is the condition $a+b+c=r$ we are trying to prove. So we need to show that all the other factors are non-zero to conclude that the first one has to be zero. This will hopefully boil down to some reasonable non-degeneracy assumptions.
$r\neq 0$ is reasonable to assume since otherwise we don't have a proper circumcircle anyway. $r\neq a, r\neq b, r\neq c$ follow from the assumption that the inner circles are actually smaller, so their radius can't be equal to that of the circumcircle.
But how do we know $4abc – ar^2 – br^2 – cr^2 + r^3\neq 0$? Why can we rule that out? Does that term have any intuitive geometric interpretation? Also, do you have any alternative approaches to suggest for this whole problem?
Update: Rewriting that last factor as $4abc – (a+b+c-r)r^2$ (as some of the proposed answers did) we can see that if the assumption holds, that parenthesis becomes zero and the whole term becomes positive. So we might want to prove that $4abc > (a+b+c-r)r^2$ holds without building that proof on the assumption $a+b+c=r$. Perhaps some clever chain of well-argued inequalities can achieve that. I'm convinced that $\frac r6\le a,b,c\le\frac r2$ but my reasoning for the lower bound currently involves the assumption we try to prove, and anyway doesn't lead to the desired inequality in an obvious way.
Update 2: After reading the answers it becomes clear that $a+b+c=r$ can be concluded if we assume that the six inner circles don't overlap, don't have any inner points common to two of them. That condition obviously holds in the example figure, but hasn't been explicitly stated in my original wording of the problem statement.
Best Answer
By symmetry, $A'C$ is parallel and congruent to $AC'$, $BC$ is parallel and congruent to $B'C'$ and $AB$ is parallel and congruent to $A'B'$. So $A'CBAC'B'A'$ is a hexagon with parallel and equal opposite sides. Such hexagon is centrally symmetric, the center of symmetry being the point of intersection of diagonals $BB'$, $CC'$ and $AA'$. Also by symmetry $BC$ is parallel to the diagonal $AA'$ etc. As a consequence, $A'CBO$ is a parallelogram. Its opposite sides are congruent: $a+b=r-c$, which is the claim ($a$ is the radius of the blue circle, $b$ is the radius of the green one and $c$ is the radius of the red one.