The Plemelj Triangle via the Fixed Point of a Transformation

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This problem was posed to Plemelj by his math teacher Borstner when Plemelj was about 16 years old. The problem asks for the construction of a triangle using ruler and compass given the length of the base , the length of the altitude from to , and , where and are the angles at and . In this Demonstration, and .


Draw two parallel lines separated by a distance . Mark off the line segment of length on the lower parallel line. The line drawn through with angle intersects the upper parallel line at , and the line through with angle intersects the upper parallel line at . What is the relationship between and ?

Introduce a coordinate system with the origin at and the axis along so that . The line has the equation , and has the equation . Let and . Then, using the addition formula for the tangent gives the following formulas for the abscissas of and :





So the construction problem is to find the fixed point of the transformation to give . For this to happen, . If , this equation has the solution . In this Demonstration, we take the solution with the minus sign. Geometrically, the fixed point is determined by the values of and such that points and are coincident.

Since quadratic equations can be solved by construction with a straight edge and a compass, Plemelj's problem has a purely geometric solution.


Contributed by: Izidor Hafner (June 2017)
Open content licensed under CC BY-NC-SA



Plemelj (1873–1967) was Vidav's (1918–2015) dissertation adviser and later a professor at the University of Ljubljana.

This Demonstration shows a way to solve the problem that Vidav mentioned to students of mathematics in 1952. The explanation here is based on [1].


[1] I. Pucelj, "Plemelj's Triangle and Fixed Points of Transformations," (in Slovenian), Obzornik za matematiko in fiziko, 62(1), 2015 pp. 12–14.

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