A circle of radius R inscribes three circles of radius T
Find T in terms of R
I have found that the three circles with radius T form an equilateral triangle
The answer is supposed to be T=0.464R
If we connect the vertices of that equilateral triangle into the center we have the radius point of the larger circle. Then, we create 3 isosceles triangles.
Each of these triangles we can use the law of cosines.
$\displaystyle (2T)^{2}=a^{2}+b^{2}-2abcos(\frac{2\pi}{3})$
But a=b, so:
$\displaystyle 4T^{2}=2a^{2}(1-cos(\frac{2\pi}{3}))$
$\displaystyle a=\frac{\sqrt{2}T}{\sqrt{1-cos(\frac{2\pi}{3})}}$
Then radius of the large circle is R, so we have:
$\displaystyle R=\frac{\sqrt{2}T}{\sqrt{1-cos(\frac{2\pi}{3})}}+T$
Solve for T and we have:
$\displaystyle T=(2\sqrt{3}-3)R\approx .464R$