The Plemelj Construction of a Triangle: 15

This Demonstration shows the construction of a triangle , given the length of its base , the length of the altitude from to and the difference between the angles and at and .
1. Choose a point on a line . Draw a vertical segment of length . Draw a point so that .
2. Draw a point so that is the midpoint of . On the orthogonal bisector of draw a point so that the angle equals . Draw a circle with center and radius .
3. The point is an intersection of and .
4. Draw the point on line so that .
5. The triangle meets the stated conditions.
Draw the point so that is the midpoint of . The quadrilateral is a parallelogram with diagonals and . Since the central angle equals , the angle equals . So the angle . But then the angle and . Then the triangle is isosceles and . Then .


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As far we know, this problem appeared first in [1]. Also, see [2].
For the history of Plemelj's solutions of this problem see The Plemelj Construction of a Triangle 1.
This solution was found in [3, p. 19, problem B21, solution p. 121]. It is also possible to construct the point before the point .
A variation to construct the point before is also possible along the same lines.
[1] L. H. von Holleben and P. Gerwien, Aufgaben-Systeme und Sammlungen aus der Ebenen Geometrie: Aufgaben, Berlin: G. Reimer, 1832.
[2] The Ohio Journal of Education, 4, 1855 pp. 278 and 369; 5, 1856 p. 112; 6, 1857 pp. 56–57, 145 and 184.
[3] E. Specht, 300 Aufgaben zur Geometrie und zu Ungleichungen insbesondere zur Vorbereitung auf Mathematik-Olympiaden, Version 2.5 (Dezember 2000), Otto-von-Guericke-Universität Magdeburg, Fakultät für Naturwissenschaften.
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