geometry
What is the diameter of a circle in which are inscribed three smaller identical circles, two of which are on one side of a chord, the third on the other side? This problem came up when cutting a log into billets for turning table legs. I tried including a diagram but the reputation Nazis won't let me.
$2$ circles touch externally $\iff$ sum of their radii $=$ distance between their centers. Taking unit length to be $1$cm in Cartesian coordinate system we can express the given circles in the following way $$x^2+y^2=11^2\cdots(1)\\x^2+(y-20)^2=9^2\cdots(2)\\(x-a)^2+(y-b)^2=5^2\cdots(3)$$ As $(3)$ touches $(1),(2)$ externally,
$$a^2+b^2=(5+11)^2,a^2+(b-20)^2=(5+9)^2\Rightarrow a={3\sqrt{55}\over 2},b={23\over 2}$$ The largest circle, let's call it $(4)$, inside the region bounded by $(1),(2),(3)$, it touches $(1),(2),(3)$. Suppose $(4)$ is given by $$(x-c)^2+(y-d)^2=r^2\\\therefore c^2+d^2=(r+11)^2\cdots(5)\\c^2+(d-20)^2=(r+9)^2\cdots(6)\\(c-a)^2+(d-b)^2=(r+5)^2\cdots(7)\\
(6)-(5)\equiv d={r\over 10}+11\cdots(8)\\(5)-(8)^2\equiv c^2=99{r\over 10}\left({r\over 10}+2\right)\cdots(9)$$ Let ${r\over 10}=q$.
$$(7)-(5)\equiv ac+bd=60q+176\Rightarrow c={97q+99\over 2a}\cdots(10)$$
$(10)^2-(9)$ gives us a quadratic equation in $q$ solving which we get $$r={495(-199+30\sqrt{55})\over 9899}\approx1.17$$
"Inscribed" is the key word. You necessarily have despite $d_1$ and $d_2$ can vary a lot $$D=d_1+d_2$$ where $D,d_1,d_2$ are the diameters. Thus the equivalent equality $$D\pi=d_1\pi+d_2\pi$$ Consequently the sums are equal.
Best Answer
Let $r$ be the radius of the small circle, $s$ the distance from the center of the large circle to the chord, and $t$ the distance from the center of the large circle to the center of one of the two small tangent circles:
Clearly $t = r + s$, and also $t^2 = r^2 + \left(r-s\right)^2$, and we have
$$\left( r + s \right)^2 = r^2 + \left( r - s \right)^2$$
so that, after a tiny bit of algebra,
$$4 s = r$$
The radius of the large circle is therefore $2r+s = \frac{9}{4}r$.