Резюме. The paper considers three solutions of an interesting problem with four circles.

Ключови думи: solution, circles, geometry

The problem under consideration is the following: Given are three tangent to each other circles with radii \(a, b\) a and \(a+b\) respectively, where \(a, b \in \mathbb{R}\) are positive real numbers. Find the radius of a fourth circle tangent to each of these three circles.

Solution 1. Let the given circles be denoted as follows: \(k_{1}(A, a), k_{2}(C, b), k_{3}(B, a+b)\) and \(k_{4}(O, \rho)\); while \(A C=a+b, B C=a+2 b, A B=2 a+b\).

Let \(h, k, l\) be the altitudes of \(\triangle A B C\) and let \(x, y, z\) be the distances from the point \(O\) to the sides \(B C, A C, A B\) of \(\triangle A B C\).

We will use formula (1) from (Adamar, 1957, p. 564):

\[ \tfrac{x}{2 \rho}-\tfrac{h}{2(s-B C)}=1 ; \] (where \(s\) is the semi-perimeter of \(\triangle A B C\), i.e. \(s=2 a+2 b\) ). Analogously,

\[ \begin{gathered} \tfrac{y}{2 \rho}-\tfrac{k}{2(s-A C)}=1 \text { and } \tfrac{z}{2 \rho}-\tfrac{l}{2(s-A B)}=1, \text { i.e. } \\ \tfrac{x}{2 \rho}-\tfrac{h}{2 a}=1 ; \tfrac{y}{2 \rho}-\tfrac{k}{2(a+b)}=1 ; \tfrac{z}{2 \rho}-\tfrac{l}{2 b}=1 \\ \tfrac{x}{2 \rho}-\tfrac{h}{2 a}=1 ; \tfrac{y}{2 \rho}-\tfrac{k}{2(a+b)}=1 ; \tfrac{z}{2 \rho}-\tfrac{l}{2 b}=1 \end{gathered} \] From here:

\[ \tfrac{x}{h}-\tfrac{\rho}{a}=\tfrac{2 \rho}{h} ; \tfrac{y}{k}-\tfrac{\rho}{a+b}=\tfrac{2 \rho}{k} ; \tfrac{z}{l}-\tfrac{\rho}{b}=\tfrac{2 \rho}{l}, \]

\[ \tfrac{x}{h}+\tfrac{y}{k}+\tfrac{z}{l}=\tfrac{\tfrac{x \cdot B C}{2}}{\tfrac{h \cdot B C}{2}}+\tfrac{y \cdot A C}{\tfrac{k \cdot A C}{2}}+\tfrac{\tfrac{z \cdot A B}{2}}{\tfrac{l \cdot A B}{2}}=\tfrac{F_{\triangle O B C}+F_{\triangle O A C}+F_{\triangle O A B}}{F_{\triangle A B C}}=\tfrac{F_{\triangle A B C}}{F_{\triangle A B C}}=1:\]

(1) \[ 1=\rho\left(\tfrac{1}{a}+\tfrac{1}{a+b}+\tfrac{1}{b}+\tfrac{2}{h}+\tfrac{2}{k}+\tfrac{2}{l}\right) \]

We have:

\[ \begin{aligned} & F\left(=F_{\triangle A B C}\right)=\tfrac{(a+2 b) h}{2} \Rightarrow \tfrac{2}{h}=\tfrac{a+2 b}{F} \\ & F\left(=F_{\triangle A B C}\right)=\tfrac{(a+b) \cdot k}{2} \Rightarrow \tfrac{2}{k}=\tfrac{a+b}{F} \\ & F\left(=F_{\triangle A B C}\right)=\tfrac{(2 a+b) \cdot l}{2} \Rightarrow \tfrac{2}{l}=\tfrac{2 a+b}{F} \end{aligned} \] \[ \text { where } F=\sqrt{(2 a+2 b) \cdot a \cdot b \cdot(a+b)}=(a+b) \sqrt{2 a b} \text {. } \] If follows now from (1), that:

\[ \begin{aligned} & \rho=\tfrac{1}{\tfrac{1}{a}+\tfrac{1}{a+b}+\tfrac{1}{b}+\tfrac{a+2 b}{(a+b) \sqrt{2 a b}}+\tfrac{a+b}{(a+b) \sqrt{2 a b}}+\tfrac{2 a+b}{(a+b) \sqrt{2 a b}}}, \\ & \rho=\tfrac{1}{\tfrac{b(a+b)+a b+a(a+b)}{a b(a+b)}+\tfrac{a+2 b+a+b+2 a+b}{\sqrt{2 a b}(a+b)}}, \\ & \rho=\tfrac{1}{\tfrac{a^{2}+b^{2}+3 a b}{a b(a+b)}+\tfrac{2 \sqrt{2 a b}}{a b}}=\tfrac{1}{\tfrac{a^{2}+b^{2}+3 a b+2 \sqrt{2 a b}(a+b)}{a b(a+b)}}, \\ & \rho=\tfrac{a b(a+b)}{a^{2}+b^{2}+3 a b+2 \sqrt{2 a b}(a+b)} . \end{aligned} \]

Remark 1. For the radius of the circle \(k^{\prime}\) with center \(O^{\prime}\) we get analogously:

\[ \rho^{\prime}=\left|\tfrac{F}{2 s-\left(r_{a}+r_{b}+r_{c}\right)}\right| \] where \(F\left(=F_{\triangle A B C}\right)=(a+b) \sqrt{2 a b}, 2 s=4(a+b), r_{a}=\tfrac{F}{s-(a+2 b)}=\tfrac{(a+b) \sqrt{2 a b}}{a}\), \[ r_{b}=\tfrac{F}{s-(a+b)}=\tfrac{(a+b) \sqrt{2 a b}}{a+b} ; r_{c}=\tfrac{F}{s-(2 a+b)}=\tfrac{(a+b) \sqrt{2 a b}}{b} \]

Solution 2. The semi-perimeter of \(\triangle A O C\) is \(\tfrac{1}{2}(a+b+a+\rho+b+\rho)=a+b+\rho\) and we have

\[ \cos ^{2}\left(\tfrac{1}{2} \measuredangle A O C\right)=\tfrac{\rho(a+b+\rho)}{(a+\rho)(b+\rho)} ; \quad \sin ^{2}\left(\tfrac{1}{2} \measuredangle A O C\right)=\tfrac{a b}{(a+\rho)(b+\rho)} \]

We will use the following well known theorem (Grozdev, 2007):

If \(\alpha^{\prime}+\beta^{\prime}+\gamma^{\prime}=180^{\circ}\), then

\(\sin ^{2} \alpha^{\prime}-\sin ^{2} \beta^{\prime}-\sin ^{2} \gamma^{\prime}+2 \sin \beta^{\prime} \sin \gamma^{\prime} \cos a^{\prime}=0\)

and from here by \(\alpha^{\prime}=\tfrac{1}{2} \measuredangle A O C, \beta^{\prime}=\tfrac{1}{2} \measuredangle B O C\) and \(\gamma^{\prime}=\tfrac{1}{2} \measuredangle A O B\), we get: \(\tfrac{a b}{(a+\rho)(b+\rho)}-\tfrac{b(a+b)}{(b+\rho)(a+b+\rho)}-\tfrac{a(a+b)}{(a+\rho)(a+b+\rho)}+2 \cdot \tfrac{(a+b) \sqrt{a b \rho(a+b+\rho)}}{(a+b+\rho)(a+\rho)(b+\rho)}=0\), i.e. \(\tfrac{a+b+\rho}{a+b}-\tfrac{a+\rho}{a}-\tfrac{b+\rho}{b}+2 \sqrt{\tfrac{\rho(a+b+\rho)}{a+b}}=0\).

Further, we divide by \(\rho\) and use the substitutions \(\alpha=\tfrac{1}{a}, \beta=\tfrac{1}{b}, \gamma=\tfrac{1}{a+b}\) and \(\delta=\tfrac{1}{\rho}\) :

(∗) \[ \alpha-\beta-\gamma-\delta+2 \sqrt{\beta \gamma+\gamma \delta+\delta \beta}=0 \]

Since

\[ \begin{aligned} (\alpha+\beta+\gamma+\delta)^{2} & =(\alpha-\beta-\gamma-\delta)^{2}+4(\alpha \beta+\alpha \gamma+\alpha \delta) \\ & \stackrel{(*)}{=} 4(\beta \gamma+\gamma \delta+\delta \beta)+4(\alpha \beta+\alpha \gamma+\alpha \delta) \\ & =2(\alpha+\beta+\gamma+\delta)^{2}-2\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right) \\ & \Rightarrow 2\left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)=(\alpha+\beta+\gamma+\delta)^{2} \end{aligned} \]

We obtain:

\[ \begin{aligned} & 2\left(\tfrac{1}{a^{2}}+\tfrac{1}{b^{2}}+\tfrac{1}{(a+b)^{2}}+\tfrac{1}{\rho^{2}}\right)=\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}+\tfrac{1}{\rho}\right)^{2} \\ & \Leftrightarrow \tfrac{2}{\rho^{2}}+2\left(\tfrac{1}{a^{2}}+\tfrac{1}{b^{2}}+\tfrac{1}{(a+b)^{2}}\right)=\tfrac{1}{\rho^{2}}+2\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}\right) \cdot \tfrac{1}{\rho}+\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}\right)^{2} \\ & \Leftrightarrow \tfrac{1}{\rho^{2}}-2\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}\right) \cdot \tfrac{1}{\rho}+2\left(\tfrac{1}{(a+b)^{2}}+\tfrac{1}{a^{2}}+\tfrac{1}{b^{2}}\right)-\left(\tfrac{1}{a+b}+\tfrac{1}{a}+\tfrac{1}{b}\right)^{2}=0 \end{aligned} \]

If \(\tfrac{1}{\rho}=t\), then:

\[ \begin{aligned} & \Rightarrow t=\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b} \pm \sqrt{2\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}\right)^{2}-2\left(\tfrac{1}{(a+b)^{2}}+\tfrac{1}{a^{2}}+\tfrac{1}{b^{2}}\right)} \\ & \Rightarrow t=\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b} \pm \sqrt{4\left(\tfrac{1}{a b}+\tfrac{1}{a(a+b)}+\tfrac{1}{b(a+b)}\right)} \\ & \Rightarrow t=\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b} \pm 2 \sqrt{\tfrac{a+b+b+a}{a b(a+b)}} \\ & \Rightarrow t=\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b} \pm 2 \sqrt{\tfrac{2}{a b}} . \end{aligned} \]

We have \[ \tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}=\tfrac{a+b}{a b}+\tfrac{1}{a+b} \stackrel{A M-G M}{\geq} 2 \sqrt{\tfrac{1}{a b}} \] Since \(2 \sqrt{\tfrac{1}{a b}} \lt 2 \sqrt{\tfrac{2}{a b}}\), then \(t=\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}-2 \sqrt{\tfrac{2}{a b}}\). It follows that there is only one solution:

\[ \begin{gathered} t=\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{a+b}+2 \sqrt{\tfrac{2}{a b}}, \text { i.e. } \\ t=\tfrac{b(a+b)+a(a+b)+a b}{a b(a+b)}+\tfrac{2 \sqrt{2}}{\sqrt{a b}} \\ t=\tfrac{a^{2}+b^{2}+3 a b}{a b(a+b)}+\tfrac{2 \sqrt{2 a b}}{a b} \\ t=\tfrac{a^{2}+b^{2}+3 a b+2 \sqrt{2 a b}(a+b)}{a b(a+b)}, \text { i.e. } \\ \rho=\tfrac{a b(a+b)}{a^{2}+b^{2}+3 a b+2 \sqrt{2 a b}(a+b)} \end{gathered} \]

Solution 3. Let \(B\) and \(A\) be the tangent points of the circles \(k_{2}\) and \(k_{3}, k_{1}\) and \(k_{3}\), respectively. By the law of cosines for \(\Delta S_{1} A B\) we have:

\[ (A B)^{2}=\left(S_{3} B\right)^{2}+\left(S_{3} A\right)^{2}-2 \cdot S_{3} B \cdot S_{3} A \cos \alpha \text {, i.e. } \]

(2) \[ (A B)^{2}=2 r_{3}^{2}(1-\cos \alpha) \]

Analogously, by the law of cosines for \(\Delta S_{1} S_{2} S_{3}\) :

\[ \begin{aligned} & \cos \alpha=\tfrac{\left(S_{1} S_{3}\right)^{2}+\left(S_{2} S_{3}\right)^{2}-\left(S_{1} S_{2}\right)^{2}}{2 S_{1} S_{3} \cdot S_{2} S_{3}}, \text { i.e. } \\ & \cos \alpha=\tfrac{\left(r_{1}+r_{3}\right)^{2}+\left(r_{2}+r_{3}\right)^{2}-\left(r_{1}+r_{2}\right)^{2}}{2\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)} \end{aligned} \] or

(3) \[ \cos \alpha=\tfrac{r_{3}^{2}+r_{1} r_{3}+r_{2} r_{3}-r_{1} r_{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)} \]

Now, it follows from (2) and (3):

(4) \[ \begin{gathered} (A B)^{2}=2 r_{3}^{2}\left(1-\tfrac{r_{3}^{2}+r_{1} r_{3}+r_{2} r_{3}-r_{1} r_{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)}\right) \\ \Rightarrow(A B)^{2}=\tfrac{4 r_{1} r_{2} r_{3}^{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)} \end{gathered} \]

Let \(k(S, r)\) be the fourth circle tangent to each of the circles \(k_{1}, k_{2}\) and \(k_{3}\). Let \(A, B\) and \(C\) be the tangent points of circles \(k_{1}\) and \(k_{3}, k_{2}\) and \(k_{3}, k_{3}\) and \(k\), k2 and k3 , k3 and k , respectively. Let the point \(H\) be the foot of the normal from the point \(A\) to the line \(B C\). Using (4) for the circles \(k_{1}, k_{2}, k_{3}\) firstly, then for the circles \(k_{2}, k_{3}, k\) and finally for the circles \(k_{1}, k_{3}, k\), we get: \[ (A B)^{2}=\tfrac{4 r_{1} r_{2} r_{3}^{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)} , \]

(5) \[ (B C)^{2}=\tfrac{4 r_{2} r_{3}^{2} r}{\left(r_{2}+r_{3}\right)\left(r+r_{3}\right)} , \]

(6) \[ (A C)^{2}=\tfrac{4 r_{1} r_{3}^{2} r}{\left(r_{1}+r_{3}\right)\left(r+r_{3}\right)}. \]

The similarity of the right triangles \(\triangle A B H\) and \(\triangle C D A\) gives:

\[ \tfrac{A C}{C D}=\tfrac{A H}{A B} \] and from here by (4) and (6), wd obtain:

(7) \[ (A H)^{2}=\tfrac{(A B)^{2} \cdot(A C)^{2}}{4 r_{3}^{2}}=\tfrac{4 r_{1}^{2} r_{2} r_{3}^{2} r}{\left(r_{1}+r_{3}\right)^{2}\left(r_{2}+r_{3}\right)\left(r+r_{3}\right)} , \]

(8) \[ (C H)^{2}=(A C)^{2}-(A H)^{2}=\tfrac{4 r_{1} r_{3}^{2} r}{\left(r_{1}+r_{3}\right)\left(r+r_{3}\right)}\left(1-\tfrac{r_{1} r_{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)}\right). \]

By (4), (5), (6), (7) and (8):

\[ \begin{aligned} & (A B)^{2}=(B H)^{2}+(A H)^{2}=(B C+C H)^{2}+(A H)^{2}=(B C)^{2}+2 B C \cdot C H+(C H)^{2}+(A H)^{2}= \\ & =(A C)^{2}+(B C)^{2}+2 B C \cdot C H, \text { i.e. } \\ & \quad \tfrac{4 r_{1} r_{2} r_{3}^{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)}=\tfrac{4 r_{1} r_{3}^{2} r}{\left(r_{1}+r_{3}\right)\left(r+r_{3}\right)}+\tfrac{4 r_{2} r_{3}^{2} r}{\left(r_{2}+r_{3}\right)\left(r+r_{3}\right)}+2 \cdot \tfrac{2 r_{3} \sqrt{r_{2} r}}{\sqrt{\left(r_{2}+r_{3}\right)\left(r+r_{3}\right)}} \cdot \\ & \quad \cdot \tfrac{2 r_{3} \sqrt{r_{1} r}}{\sqrt{\left(r_{1}+r_{3}\right)\left(r+r_{3}\right)}} \sqrt{1-\tfrac{r_{1} r_{2}}{\left(r_{1}+r_{3}\right)\left(r_{2}+r_{3}\right)}} . \end{aligned} \] We get: \(r_{1} r_{2}\left(r+r_{3}\right)=r_{1} r\left(r_{2}+r_{3}\right)+r_{2} r\left(r_{1}+r_{3}\right)+2 r \sqrt{r_{1} r_{2} r_{3}\left(r_{1}+r_{2}+r_{3}\right)}\) and finally \(r=\tfrac{r_{1} r_{2} r_{3}}{r_{1} r_{2}+r_{2} r_{3}+r_{1} r_{3}+2 \sqrt{r_{1} r_{2} r_{3}\left(r_{1}+r_{2}+r_{3}\right)}}\).

If \(r_{1}=a, r_{2}=b, r_{3}=a+b\), , then:

\[ \begin{aligned} & r=\tfrac{a b(a+b)}{a b+b(a+b)+a(a+b)+2 \sqrt{a b(a+b)(2 a+2 b)}}, \text { i.e. } \\ & r=\tfrac{a b(a+b)}{a^{2}+b^{2}+3 a b+2(a+b) \sqrt{2 a b}} \end{aligned} \]

REFERENCES

1. Adamar, Ž. (1957). Elementarnaja geometrija, Knjiga 1., Planimetrija. Moskva.

2. Grozdev, S. (2007). For High Achievements in Mathematics. The Bulgarian\

3. Experience (Theory and Practice) . Sofia: ADE. (ISBN 978-954-92139-1-1), 295 pages.