On one side of a given segment, we drew two externally tangent semicircles with radii $r$ and $R$, and a semicircle that is tangent to both of these, as shown in the figure. What is the radius of the circle that is tangent to these three semicircles?

![](https://cdn.mathpix.com/cropped/2024_05_02_6cffc49493b86b44560cg-1.jpg?height=258&width=439&top_left_y=237&top_left_x=824)
I. solution. Let the radius of the tangent circle be $x$. Using the additional notations of Figure 1. Applying the cosine theorem to the triangles $O_{2} O_{1} O$ and $O_{1} O_{3} O$:

$$
(R+x)^{2}=(R+r-x)^{2}+r^{2}+2(R+r-x) \cdot r \cdot \cos \gamma, \quad (r+x)^{2}=(R+r-x)^{2}+R^{2}-2(R+r-x) \cdot R \cdot \cos \gamma
$$

Multiply the first equation by $R$ and the second by $r$, then add them together:

$$
(R+x)^{2} R+(r+x)^{2} r=(R+r-x)^{2}(R+r)+r^{2} R+r R^{2}
$$

from which, after some calculation:

$$
x=\frac{R r(R+r)}{R^{2}+R r+r^{2}}
$$

Ferenc Lobozár (Szeged, Csonka J. Műsz. Szki., III. o.t.)

II. solution. The following theorem is attributed to R. Descartes: if any of four circles touches the other three, then twice the sum of the squares of their curvatures is equal to the square of the sum of their curvatures.

The curvature of a circle with radius $r$ is $\frac{1}{r}$, and the curvature of a line is zero. The theorem is also valid when a line appears instead of one of the circles. From the perspective of its application, the curvatures of circles that touch each other externally have the same sign, and those that touch internally have opposite signs. Its proof can be found on pages 31-32 of H. S. M. Coxeter's book "Introduction to Geometry". According to the theorem:

$$
2\left(\frac{1}{r^{2}}+\frac{1}{R^{2}}+\frac{1}{(r+R)^{2}}+\frac{1}{x^{2}}\right)=\left(\frac{1}{r}+\frac{1}{R}-\frac{1}{r+R}+\frac{1}{x}\right)^{2}
$$

Solving this quadratic equation for $\frac{1}{x}$, $x$ can be determined:

$$
x=\frac{R r(R+r)}{R^{2}+R r+r^{2}}
$$

Zsolt Terék (Fazekas M. Főv. Gyak. Gimn., III. o.t.)

Remarks. 1. Descartes' theorem mentioned above fell into obscurity over time and was rediscovered by several others, for example, in 1936 by Sir Frederick Soddy, a Nobel laureate in physics, who also showed that an analogous theorem holds for five spheres in space. Soddy wrote a poem titled "The Kiss Precise" about the theorems, one of whose verses expresses the theorem used in the second solution as follows:

Four circles to the kissing come,  
The smaller are the bolder some;  
The bend is just the inverse of  
The distance from the center, of course.  
Euclid could not have seen it,  
Nor could the compass rule it;  
For the straight line does not curve,  
And the concave bend is minus,  
The sum of the squares of the four bends  
Is half the square of their sum.

2. Gabriella Zsombori (Csíkszereda, Márton Á. Gimn., IV. o.t.) based her solution on the Stewart theorem. The theorem, with the notations of Figure 2, can be stated as follows: $b^{2} p+c^{2} q-a r^{2}=a p q$. Its proof can be found on pages 21-22 of H. S. M. Coxeter and S. L. Greitzer's book "New Horizons in Geometry". Applying the theorem to the triangle $O O_{2} O_{3}$ in Figure 1, we obtain equation (1), from which $x$ can be determined.
3. In Ferenc Lobozár's second solution, he chose the point $B$ in Figure 1 as the pole of an inversion. He defined the radius of the inversion circle by the relation $d^{2}=B C \cdot B A=2 r(2 r+2 R)=4 r(r+R)$. Thus, the images of the two circles passing through point $B$ became two lines, and he calculated the desired radius based on the significantly simpler figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_81590bba0a50f1329fcfg-1.jpg?height=279&width=499&top_left_y=2365&top_left_x=802)

![](https://cdn.mathpix.com/cropped/2024_05_02_81590bba0a50f1329fcfg-2.jpg?height=244&width=271&top_left_y=125&top_left_x=908)