[alogo] 1. Pappus triangle construction

Construct triangle ABC from its side a=BC, opposite angle A and bisector-length bA of the bisector from that angle.

[0_0] [0_1]
[1_0] [1_1]

Quadrangle AGFD is cyclic having two opposite angles at A and F equal to a right angle.
(AD+DE)*(DE) = EF*EG = EB2 = e2, leads to the quadratic equation  x2 + bA*x - e2 = 0.
Solve this to find x = DE and from this the solution to the problem.
The value of e is easily determined from the given data and can be used to find a necessary condition which must be satisfied by these data in order to have a solution.

[alogo] 2. Related problems

1) The same reasoning can be applied for the construction of the triangle from its elements: {A, R, bA}, where R is the circumradius.
2) The corresponding problem {A, r, bA}, where r is either the inradius or one of the exradii  (radii of escribed circles) is trivial.
Following figure illustrates such a construction from {A, rB, bA}.


Points {A,G,H,D} can be constructed from the given data and point E is found drawing the tangent from D to the known circle c.

[alogo] 3. Non constructible cases

Related to bisectors are many construction problems that can be not carried out by ruler and compasses as required in euclidean geometry.
The following list of non-constructible cases is taken from Fursenko.
1) { a, A, bB}.
2) { a, bA, bB}.
3) { a, bA, r}.
4) { a, bA, rA}.
5) { a, bA, rB}.
6) { a, bB, rB}.
7) { a, bB, bC}.
8) { a, bB, R}.
9) { a, bB, rA}.
10) { a, bB, rB}.
11) { a, bB, rC}.

See Also



Fursenko F. B. Lexicographical account of constructional problems of triangle geometry problems Mathematics in school, 1937, no. 5 pp. 4-30, no. 6 pp. 21-45, Moscow
Problem E2499, American Mathematical Monthly 82(1975) p. 1015.

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