The Plemelj Construction of a Triangle: 7

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This Demonstration constructs a triangle given the length of its base, the length of the altitude from to and the difference between the angles at and . This is not Plemelj's construction, but a new one that unifies Plemelj's first construction and an alternative one.



Step 1: Draw a straight line of length and a perpendicular line segment with midpoint .

Step 2: Draw a circle with center such that is viewed at an angle from points on below the chord . Let be the midpoint of . The angle equals .

Step 3: Find a point on the circle at distance from and a point at distance from .

Step 4: Draw the isosceles trapezoid .

Step 5: The point is the intersection of the straight line through parallel to and the right bisector of and .

Step 6: The triangle meets the stated conditions.


This is similar to Plemelj's first construction, but instead of triangle , start with triangle , which is also congruent to . In the isosceles triangle , , so . The obtuse angle ; .

On the other hand, also equals . Thus .


Contributed by: Izidor Hafner, Nada Razpet and Marko Razpet (August 2017)
Open content licensed under CC BY-NC-SA



For the history of this problem, references and a photograph of Plemelj's first solution, see The Plemelj Construction of a Triangle: 1.

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