The Plemelj Construction of a Triangle: 7
Requires a Wolfram Notebook System
Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.
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.[more]
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 .[less]
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.