If four circles are inscribed in the same angle, or in the corresponding vertical angle, and they are also tangent to a fifth circle, then their radii r1, r2, r3, r4 form a proportion. (One observes that these circles can be arranged in pairs which correspond to each other in the same inversion.)

Question

If four circles are inscribed in the same angle, or in the corre-sponding vertical angle, and they are also tangent to a fifth circle, then their radii [latex]r_{1},\;r_{2},\;r_{3},\;r_{4}[/latex] form a proportion. (One observes that these circles can be arranged in pairs which correspond to each other in the same inversion.)

Accepted Answer

Lemma. If two tangent circles are inverted around a point P, the mode of tangency (internal or external) is preserved whenever P is outside both circles, and is reversed if P is inside one of the circles.
We leave this proof for the reader. Observe that if P is on one of the circles (neither outside nor inside), then the image of that circle is a line tangent to the image of the other circle, and the mode of tangency is not well-defined. Another special case occurs if P is at the point of tangency of the two original circles.

Suppose (fig. t413bi) that the four circles are [latex]C_{1},\ C_{2},\ C_{3},\ C_{4},[/latex] and that they are all tangent to a fifth circle D as well as to two lines [latex]m_{1},\;m_{2},[/latex] which intersect at point P. Let k be the power of point P with respect to circle D. We invert around P with power k. (The red circle in figure t413bi is the circle of inversion.) Then clearly lines [latex]m_{1},\;m_{2}[/latex] are their own image (220, remark), and it is not hard to see that circle D is also its own image.

Then what is the image of circle [latex]{}C_{1}[/latex]? It must remain tangent to the two lines, and also to circle D, so it must be one of the other three given circles. In figure 413bi, it is clear from our lemma that [latex]{}C_{1}[/latex] inverts onto [latex]{}C_{4}[/latex], and [latex]{}C_{2}[/latex] inverts onto [latex]{}C_{3}[/latex]. This implies that the points of tangency are images: [latex]{T}_{1}[/latex] inverts into [latex]{T}_{4}[/latex] and [latex]{T}_{2}[/latex] into [latex]{T}_{3}[/latex]. Then, if [latex]t_{1},\ t_{2},\ t_{3},\ t_{4}[/latex] denote the segments [latex]P T_{1},\ P T_{2},\ P T_{3},\ P T_{4}[/latex] respectively, we have

(1) [latex]t_{1}t_{4}=t_{2}t_{3}=k.[/latex]

Now these circles are all homothetic with respect to point P, so we have [latex]r_{1}:r_{2}:\,[/latex] [latex]r_{3}:r_{4}=t_{1}:t_{2}:t_{3}:t_{4}.[/latex] Using this relationship, we can rewrite (1) as [latex]r_{1}r_{4}=r_{2}r_{3}\ ,[/latex] so that [latex]r_{1}:r_{2}=r_{3}:r_{4},[/latex] which is what we wanted to prove.

Note. Figure t413bii shows a situation where some of the circles are inscribed in the vertically opposite angle, so that their centers are on opposite sides of the bisector of the angle formed by [latex]m_{1},\;m_{2}.[/latex] In this case, the proof must be reworded a bit. We can again invert around P so that circle D is its own image, but we must use a negative power of inversion (i.e. inverting in the red circle in figure 413bii, then reflecting in point P).

In this case, our lemma tells us that circle [latex]{C}_{1}[/latex] inverts onto circle [latex]{C}_{2}[/latex] , and circle [latex]{C}_{3}[/latex] onto circle [latex]{C}_{4}[/latex] . The proof follows as before, but the roles of the circles are changed, and the proportion is now [latex]r_{1}:r_{4}=r_{2}:r_{3}.[/latex] Details are left to the reader.

Question P.413b: If four circles are inscribed in the same angle, or in the c......

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