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The Plemelj Construction of a Triangle: 1

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 .
Construction
Step 1: Draw a straight line of length . Draw a line of length perpendicular to . Let be the midpoint of .
Step 2: Construct a circle with center such that the chord subtends an angle from points on below the chord. The inscribed angle above the chord is and the corresponding central angle is .
Step 3: Construct the point on at a distance from .
Step 4: The point is the intersection of the right bisector of and the line through parallel to .
Step 5: The triangle meets the stated conditions.
Verification
Triangle is congruent to . In the isosceles triangle , , so . Therefore the obtuse angle . On the other hand, , so and .
This Demonstration shows Plemelj's somewhat complicated construction. Fascinated, his teacher showed him the solution from a textbook unknown to the author. This is shown in The Plemelj Construction of a Triangle: 2. Plemelj then made a construction that is shown in The Plemelj Construction of a Triangle: 3. Plemelj admitted that he found the first construction using trigonometry. Three solutions of the triangle construction problem are in [2].
Here is the trigonometric proof.
The altitude from divides into two parts of length and . So , or , which can be rewritten as .
Let be the angle of at . Since , the equation can be read as
From this equation, we must determine ; it can be transformed to a quadratic equation in the unknown .
Introduce the angle as or , where .
Then , . The equation for is now .
This equation can be thought of as the law of sines of the triangle with sides and and opposite angles and .

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This problem was posed to Josip Plemelj (1873–1967) in 1891 when he was in secondary school by his mathematics teacher in Ljubljana, then in the Austro-Hungarian Empire, now in Slovenia.
Plemelj noted that he had made nine original solutions of the problem and that he knew two textbook solutions (one from his teacher's textbook and the other from [5]).
Plemelj stated that the equations
,
produced six different solutions [4].
The first publication of the construction problem by Plemelj was published in Proteus, the natural science journal for students. It included his first three solutions as well as two solutions by readers of the journal, which the editor wrote were already in Plemelj's collection.
Plemelj mentioned [2] that he had a large collection of solutions of the problem, with the last entry on December 31, 1939. The author of this Demonstration visited the Archives of the Republic of Slovenia [8] in May 2017, but found only a page of a calendar for November 1939, with three constructions on the other side (see photograph).
The Plemelj construction of Triangle 4 describes the first of these three constructions, shown at the top-left in the photograph. At the bottom-left is a construction based on a problem in Bland and Wiegand's book [5, pp. 147]. On the bottom-right is probably Plemelj's last construction.
When Professor Plemelj was in Chernivtsi (1908–1914, then in the Austrian-Hungarian Empire, now Ukraine), he talked with two students about the problem. They brought him a copy of the textbook, but Plemelj forgot for a second time to write down the title, so we still don't know its title. He looked for the book in Ljubljana after the First World War; he could not find it, but he found a similar problem in [5].
Using the construction from [5], Plemelj found a nice solution, which is given in The Plemelj Construction of a Triangle: 5. In the photograph, it is the second on the left.
The last construction on the right in the photograph is shown in The Plemelj Construction of a Triangle: 6. (The circle was drawn on the opposite side of the line segment .) This seems to be Plemelj's last construction and was found in [8]. This is the only construction that begins with the point . Our construction is adapted from [6, pp. 93]. A version of the construction by a reader of Proteus was published in [4].
So far we have mentioned eight original solutions. It seems that the ninth solution is a simple modification of the solution in Plemelj's teacher's textbook. It is published in [6, pp. 93].
It is evident that Plemelj did not mention various constructions on the basis of using geometric methods for solving quadratic equations.
Plemelj's most original contribution in mathematics is the elementary solution he provided for the Riemann–Hilbert problem about the existence of a differential equation with given monodromy group [1, 3].
References
[1] Wikipedia. "Josip Plemelj." (Aug 9, 2017) en.wikipedia.org/wiki/Josip_Plemelj.
[2] J. Plemelj, Iz mojega življenja in dela (From My Life and Work), Obzornik za matematiko in fiziko, 39, 1992 pp. 188–192.
[3] J. J. O'Connor and E. F. Robertson. "Josip Plemelj." MacTutor. www-history.mcs.st-andrews.ac.uk/Biographies/Plemelj.html.
[4] J. Plemelj, Proteus (year 12, 1949–1950), 4–5, p.166; 7, pp. 243–245; 8, pp. 285.
[5] M. Bland and A. Wiegand, Geometrische Aufgaben für Hohëre Lehranstalten, Braunschweig: Schwetschke und Sohn, 1865.
[6] D. S. Modic, Trikotniki, Konstrukcije, Algebrske Rešitve, Ljubljana: Math d.o.o., 2009.
[7] I. Pucelj, "Plemelj's Triangle and Fixed Points of Transformations," (in Slovenian), Obzornik za matematiko in fiziko, 62(1), 2015 pp.12–14.
[8] Archives of the Republic of Slovenia, Plemelj Fond (SI AS 2012), PE19, Box 3 (manuscripts).
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