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:

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

Examples

Click each triangle for details.

mqs4331i Martin's solution
fukyu14 Ajima's example
pyt0 The Smallest Pythagorean Triangle
pyt1 Pythagorean Triangle U
eis0 The Smallest Eisenstein Triangle
halfeq Half of an Equilateral Triangle
equil Equilateral Triangle
iso4 Isosceles Triangle with 90° Top Angle
iso5 Isosceles Triangle with 108° Top Angle
iso6 Isosceles Triangle with 120° Top Angle
iso8 Isosceles Triangle with 135° Top Angle
etc Test Case in ETC

Formulae

Centers of the Malfatti circles

There are many formulae known to solve Derousseau's generalization of Malfatti's problem. Let a triplet of circle \(\mathscr{C}_1\), \(\mathscr{C}_2\), \(\mathscr{C}_3\) be a solution for a triangle \(ABC\). 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.

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

For the solution (\(l\)\(m\)\(n\)),   \(\alpha\), \(\beta\), \(\gamma\) and \(I_\triangle\) are determined as follows.

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.