# Derousseau's generalization of the Malfatti circles

## Definition

Malfatti's problem is to find three circles lying inside the reference triangle such that they are tangent to one another and each of them is tangent to two of the triangle sides. For any triangle, there exists a unique solution of Malfatti's problem. The three circles in the solution is called the Malfatti circles.

Derousseau's generalization is obtained from Malfatti's problem by dropping the condition that the circles lie inside the reference triangle. More precisely, a solution of the generalized problem for a triangle $$ABC$$ is a triplet of circles $$\mathscr{C}_1$$, $$\mathscr{C}_2$$, $$\mathscr{C}_3$$ such that the following hold:

• The circle $$\mathscr{C}_1$$ is tangent to the line $$CA$$ and to the line $$AB$$.
• The circle $$\mathscr{C}_2$$ is tangent to the line $$AB$$ and to the line $$BC$$.
• The circle $$\mathscr{C}_3$$ is tangent to the line $$BC$$ and to the line $$CA$$.
• The circles $$\mathscr{C}_1$$, $$\mathscr{C}_2$$ and $$\mathscr{C}_3$$ are tangent to one another.

It is known that, for any triangle, there are exactly 32 solutions.

## Formulae

### Centers of the Malfatti circles

Let a triplet of circle $$\mathscr{C}_1$$, $$\mathscr{C}_2$$, $$\mathscr{C}_3$$ be a solution for a triangle $$ABC$$ and $$A^\prime_\triangle$$, $$B^\prime_\triangle$$, $$C^\prime_\triangle$$ their respective centers. The reference triangle $$ABC$$ and the triangle $$A^\prime_\triangle B^\prime_\triangle C^\prime_\triangle$$ are perspective. The perspector $$I_\triangle$$ is the incenter or one of the excenters.

The centers $$A^\prime_\triangle$$, $$B^\prime_\triangle$$, $$C^\prime_\triangle$$ of $$\mathscr{C}_1$$, $$\mathscr{C}_2$$, $$\mathscr{C}_3$$, respectively, can be determined by the following equations with $$\alpha$$, $$\beta$$, $$\gamma$$ and $$I_\triangle$$ determinded as below. Click 🔎 for detail.

 🔎 $$\overrightarrow{AA^\prime_\triangle}=\dfrac{\left(1+\tan\dfrac{\beta}{4}\right)\left(1+\tan\dfrac{\gamma}{4}\right)}{2\left(1+\tan\dfrac{\alpha}{4}\right)}\overrightarrow{AI_\triangle}$$, $$\overrightarrow{BB^\prime_\triangle}=\ldots$$, $$\overrightarrow{CC^\prime_\triangle}=\ldots$$. 🔎 $$\overrightarrow{AA^\prime_\triangle}=\dfrac{\cos\dfrac{\alpha}{4}\cos\dfrac{\pi-{\beta}}{4}\cos\dfrac{\pi-{\gamma}}{4}}{\sqrt{2}\cos\dfrac{\pi-{\alpha}}{4}\cos\dfrac{\beta}{4}\cos\dfrac{\gamma}{4}}\overrightarrow{AI_\triangle}$$, $$\overrightarrow{BB^\prime_\triangle}=\ldots$$, $$\overrightarrow{CC^\prime_\triangle}=\ldots$$. 🔎 $$\overrightarrow{AA^\prime_\triangle}=\dfrac{1-{\sin\dfrac{\alpha}{2}}+{\sin\dfrac{\beta}{2}}+{\sin\dfrac{\gamma}{2}}+{\cos\dfrac{\alpha}{2}}+{\cos\dfrac{\beta}{2}}+{\cos\dfrac{\gamma}{2}}}{2\left(1+{\sin\dfrac{\alpha}{2}}+{\cos\dfrac{\beta}{2}}+{\cos\dfrac{\gamma}{2}}\right)}\overrightarrow{AI_\triangle}$$, $$\overrightarrow{BB^\prime_\triangle}=\ldots$$, $$\overrightarrow{CC^\prime_\triangle}=\ldots$$.

Using $$I_\triangle = \sin\alpha:\sin\beta:\sin\gamma$$, we obtain that the barycentric coodinates are expressed as follows. Click 🔎 for detail.

 🔎 $$A^\prime_\triangle=\left(\dfrac{4\left(1+\tan^2{\dfrac{\alpha}{4}}\right)}{\left(1+\tan^2{\dfrac{\beta}{4}}\right)\left(1+\tan^2{\dfrac{\gamma}{4}}\right)}-1\right){\sin{\alpha}}:{\sin{\beta}}:{\sin{\gamma}}$$, $$B^\prime_\triangle=\ldots:\ldots:\ldots$$, $$C^\prime_\triangle=\ldots:\ldots:\ldots$$. 🔎 $$A^\prime_\triangle=\left(4\sec^2\dfrac{\alpha}{4}\cos^2\dfrac{\beta}{4}\cos^2\dfrac{\gamma}{4}-1\right){\sin{\alpha}}:{\sin{\beta}}:{\sin{\gamma}}$$, $$B^\prime_\triangle=\ldots:\ldots:\ldots$$, $$C^\prime_\triangle=\ldots:\ldots:\ldots$$. 🔎 $$A^\prime_\triangle=\left(\dfrac{2\left(1+\cos\dfrac{\beta}{2}\right)\left(1+\cos\dfrac{\gamma}{2}\right)}{1+\cos\dfrac{\alpha}{2}}-1\right){\sin{\alpha}}:{\sin{\beta}}:{\sin{\gamma}}$$, $$B^\prime_\triangle=\ldots:\ldots:\ldots$$, $$C^\prime_\triangle=\ldots:\ldots:\ldots$$.

A solution of Malfatti's problem can be encoded as $$(lmn)$$ where $$l$$, $$m$$, $$n$$ are integers with $$l+m+n\equiv0\pmod2$$. For the solution ($$l$$$$m$$$$n$$),   $$\alpha$$, $$\beta$$, $$\gamma$$ and $$I_\triangle$$ are determined as follows.

• If $$l+m+n\equiv0\pmod4$$, then $$\alpha=l\pi+A$$, $$\beta =m\pi+B$$, $$\gamma=n\pi+C$$.
• If $$l+m+n\equiv2\pmod4$$, then $$\alpha=l\pi-A$$, $$\beta =m\pi-B$$, $$\gamma=n\pi-C$$.
• If $$(l,m,n)\equiv(0,0,0)\pmod2$$, then $$I_\triangle$$ is the incenter $$I$$.
• If $$(l,m,n)\equiv(0,1,1)\pmod2$$, then $$I_\triangle$$ is the $$A$$-excenter $$I_{\mathsf{a}}$$.
• If $$(l,m,n)\equiv(1,0,1)\pmod2$$, then $$I_\triangle$$ is the $$B$$-excenter $$I_{\mathsf{b}}$$.
• If $$(l,m,n)\equiv(0,1,1)\pmod2$$, then $$I_\triangle$$ is the $$C$$-excenter $$I_{\mathsf{c}}$$.

## Brief History

Sometime before 1773 (precise date unknown), Naonobu Ajima (1732?–1798), who was a samurai, or a member of the military class in old Japan, solved Malfatti's problem by giving a procedure to compute the diameters of the Malfatti circles from the side lengths of the reference triangle. 

In 1803, Gianfrancesco Malfatti (1731–1807) posed the Malfatti problem and solved it. 

In 1895, Derousseau generalized Malfatti's problem and found 32 generalized solutions. He considered all of the 32 cases and obtained formulae for each ab initio. 

In 1904, Pampuch obtained another proof. It gives all solutions concurrently by solving a single system of equations. 

In 1930, Lob and Richmond gave all solutions systematically by replacing the sizes of the angles in a set of formulae which solves the original problem. 

In 2003, Stevanović gave the coordinates of some triangle centers associated with the Malfatti circles. 

In 2007, Guy pointed out that the existence of the 32 solutions is an application of the lighthouse theorem. 

## References

1. 安島直円 (Ajima, Naonobu), 不朽算法 (Fukyū Sanpō) in classical Chinese, unpublished, prepared for publication in 1799.
2. Jullian Lowell Coolidge, A Treatise on the Circle and the Sphere, Oxford University Press, 1916. Reprinted by AMS Chelsea Publishing, 2004.
3. J. Derousseau, “Historique et résolution analytique complète du problème de Malfatti”, Mémoires de la Société royale des sciences de Liège, 2-18:1–52, 1895.
4. Richard K. Guy, “The lighthouse theorem, Morley & Malfatti: A budget of paradoxes”. The American Mathematical Monthly, 144(2):97–141, Feb. 2007.
5. H. Lob and H. W. Richmond, “On the solution of Malfatti's problem for a triangle”. Proc. London Math. Soc., 2:287–304, 1930.
6. Gianfrancesco Malfatti, “Memoria sopra un problema sterotomico”. Memorie di Matematica e di Fisicà della Societa Italiana delle Scienze, 10-1:235–244, 1803.
7. A. Pampuch, “Die 32 Lösungen des Malfattischen Problems”. Archiv der Mathematik und Physik, 8:36–49, 1904.
8. Milorad R. Stevanović, Triangle Centers Associated with the Malfatti Circles. Forum Geometricorum 3:83–93, 2003.