geometrytrigonometry
The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the solution.
Let's get a figure in here, eh?
We can exploit the fact that the points of mutual tangency of the three inner circles (red points in the figure) form an equilateral triangle; we also know that the red points are the midpoints of the segments formed by joining any two of the three blue points (the centers of the inner circles).
We then deduce that the triangle formed by the red points is half the scale of of the triangle formed by the blue points, and find that the equilateral triangle formed by the blue points has a side length of 6 ft., and that the inner circles have a radius of 3 ft. Using the law of cosines, we reckon that the distance from a blue point to the center of the triangle formed by the blue points is $2\sqrt{3}$ ft. Adding to that the radius of an inner circle, we find that the radius of the outer circle is $3+2\sqrt{3}$ ft.
The area of the outer circle is $\approx$ 131.27 square feet.
In general, the radius $r$ of a circle inscribed by any two touching circles of radii $a$ & $b$ and their common tangent, is given by Generalized Formula as follows $$\boxed{r=\frac{ab}{(\sqrt a+\sqrt b)^2}}$$ Now, substituting the values of radii $a=2$ of blue circle and $b=1$ of green circle, the radius $r$ of red circle is calculated as follows $$r=\frac{2\cdot1}{(\sqrt 2+\sqrt 1)^2}$$ $$=\frac{2}{3+2\sqrt2}$$ $$=\frac{2(3-2\sqrt2)}{(3+2\sqrt2)(3-2\sqrt2)}$$ $$=2(3-2\sqrt2)$$
Best Answer
Radius is exactly $\dfrac{72}{23}$, pretty neat. See the figure below, where dotted lines from the center of the circumcribing circle passes through the midpoints of the sides of the triangle.
Let : $B$ be the origin $(0,0)$, and the center of circle of unknown radius $r$ be $(x,y)$. Then we solve the following three equations to find $r$
$\dfrac{3}{2}+\sqrt{x^2+(\dfrac{3}{2}-y)^2}=r$
$2+\sqrt{(2-x)^2+y^2}=r$
$\dfrac{5}{2}+\sqrt{(2-x)^2+(\dfrac{3}{2}-y)^2}=r$
so that $(r, x, y)= \left(\dfrac{72}{23}, \dfrac{36}{23}, \dfrac{24}{23}\right)$
Note: The midpoints of the sides $(0,\dfrac{3}{2})$, $(2,0)$, and $(2,\dfrac{3}{2})$ are centers of circles of radii $\dfrac{3}{2}, 2, \dfrac{5}{2}$. Given any two tangent circles, their centres and point of tangency are colinear.
Bonus: If right triangle : $\triangle ABC$ has rational sides, then $(r,x,y)$ are also all rationals.