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:
It is known that, for any triangle, there are exactly 32 solutions.
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.
| 🔎 | \(\overrightarrow{AA^\prime_\triangle}=\dfrac{\left(1+\tan\dfrac{\beta}{4}\right)\left(1+\tan\dfrac{\gamma}{4}\right)}{2\left(1+\tan\dfrac{\alpha}{4}\right)}\overrightarrow{AI_\triangle}\), | \(\overrightarrow{BB^\prime_\triangle}=\ldots\), | \(\overrightarrow{CC^\prime_\triangle}=\ldots\). |
| 🔎 | \(\overrightarrow{AA^\prime_\triangle}=\dfrac{\cos\dfrac{\alpha}{4}\cos\dfrac{\pi-{\beta}}{4}\cos\dfrac{\pi-{\gamma}}{4}}{\sqrt{2}\cos\dfrac{\pi-{\alpha}}{4}\cos\dfrac{\beta}{4}\cos\dfrac{\gamma}{4}}\overrightarrow{AI_\triangle}\), | \(\overrightarrow{BB^\prime_\triangle}=\ldots\), | \(\overrightarrow{CC^\prime_\triangle}=\ldots\). |
| 🔎 | \(\overrightarrow{AA^\prime_\triangle}=\dfrac{1-{\sin\dfrac{\alpha}{2}}+{\sin\dfrac{\beta}{2}}+{\sin\dfrac{\gamma}{2}}+{\cos\dfrac{\alpha}{2}}+{\cos\dfrac{\beta}{2}}+{\cos\dfrac{\gamma}{2}}}{2\left(1+{\sin\dfrac{\alpha}{2}}+{\cos\dfrac{\beta}{2}}+{\cos\dfrac{\gamma}{2}}\right)}\overrightarrow{AI_\triangle}\), | \(\overrightarrow{BB^\prime_\triangle}=\ldots\), | \(\overrightarrow{CC^\prime_\triangle}=\ldots\). |
Using \(I_\triangle = \sin\alpha:\sin\beta:\sin\gamma \), we obtain that the barycentric coodinates are expressed as follows.
| 🔎 | \(A^\prime_\triangle=\left(\dfrac{4\left(1+\tan^2{\dfrac{\alpha}{4}}\right)}{\left(1+\tan^2{\dfrac{\beta}{4}}\right)\left(1+\tan^2{\dfrac{\gamma}{4}}\right)}-1\right){\sin{\alpha}}:{\sin{\beta}}:{\sin{\gamma}}\), | \(B^\prime_\triangle=\ldots:\ldots:\ldots\), | \(C^\prime_\triangle=\ldots:\ldots:\ldots\). |
| 🔎 | \(A^\prime_\triangle=\left(4\sec^2\dfrac{\alpha}{4}\cos^2\dfrac{\beta}{4}\cos^2\dfrac{\gamma}{4}-1\right){\sin{\alpha}}:{\sin{\beta}}:{\sin{\gamma}}\), | \(B^\prime_\triangle=\ldots:\ldots:\ldots\), | \(C^\prime_\triangle=\ldots:\ldots:\ldots\). |
| 🔎 | \(A^\prime_\triangle=\left(\dfrac{2\left(1+\cos\dfrac{\beta}{2}\right)\left(1+\cos\dfrac{\gamma}{2}\right)}{1+\cos\dfrac{\alpha}{2}}-1\right){\sin{\alpha}}:{\sin{\beta}}:{\sin{\gamma}}\), | \(B^\prime_\triangle=\ldots:\ldots:\ldots\), | \(C^\prime_\triangle=\ldots:\ldots:\ldots\). |
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.
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.