Lemoine's problem
In geometry, Lemoine's problem is a straightedge and compass construction problem posed by French mathematician Émile Lemoine in 1868:[1][2]
- Given one vertex of each of the equilateral triangles placed on the sides of a triangle, construct the original triangle.
The problem was published as Question 864 in Nouvelles Annales de Mathématiques (Series 2, Volume 7 (1868), p 191). The chief interest in the problem is that a discussion of the solution of the problem by Ludwig Kiepert published in Nouvelles Annales de Mathématiques (series 2, Volume 8 (1869), pp 40–42) contained a description of a hyperbola which is now known as the Kiepert hyperbola.[3]
Ludwig Kiepert's solution
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Kiepert establishes the validity of his construction by proving a few lemmas.[3][4]
- Problem
- Let A1, B1, C1 be the vertices of the equilateral triangles placed on the sides of a triangle Given A1, B1, C1 construct A, B, C.
- Lemma 1
- If on the three sides of an arbitrary triangle one describes equilateral triangles then the line segments are equal, they concur in a point P, and the angles they form one another are equal to 60°.
- Lemma 2
- If on one makes the same construction as that on there will have three equilateral triangles three equal line segments which will also concur at the point P.
- Lemma 3
- A, B, C are respectively the midpoints of
- Solution
- Describe on the segments the equilateral triangles respectively.
- The midpoints of are, respectively, the vertices A, B, C of the required triangle.
Other solutions
[edit | edit source]Several other people in addition to Kiepert submitted their solutions during 1868–9, including Messrs Williere (at Arlon), Brocard, Claverie (Lycee de Clermont), Joffre (Lycee Charlemagne), Racine (Lycee de Poitiers), Augier (Lycee de Caen), V. Niebylowski, and L. Henri Lorrez. Kiepert's solution was more complete than the others.[3]
References
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- ^ a b c The details of the construction as given by Kiepert in French can be read here. [1]
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