Grégoire Nicollier, University of Applied Sciences of Western Switzerland
November 24, 2016

Yesterday I read three interesting papers [1, 2, 3] addressing the following problem: What are the sides of a triangle whose angle bisectors have any prescribed positive lengths? This brought me to a further question: How long are the angle bisectors of a degenerate triangle? Here is the answer.

We first consider a nondegenerate triangle ABC with sides a=BC, b=CA, c=AB and interior angle bisector AD, D being on BC. Let us establish the well-known formula


Proof. As BD+DC=a and BD/DC=c/b by the angle bisector theorem [4] one has BD=ac/(b+c) and DC=ab/(b+c). Formula (1) follows at once from the cosine rule in BCA and BDA:

\cos B=\frac{a^2+c^2-b^2}{2ac}\quad\text{and}\quad AD^2=BD^2+c^2-2BD\cdot c\cos B.

Let now ABC be degenerate with three nonzero sides and C on segment AB, hence c=a+b. We consider this as the limit position of a point C hardly above AB: the angle bisector of the straight angle \angle C has length 0 and the angle bisector of the zero angle \angle A has by (1) squared length

AD^2= b(a+b)\left(1-\frac{a^2}{(a+2b)^2}\right)=\left(\frac{2bc}{b+c}\right)^2.

The following result is proven:

In a degenerate triangle, the length of the bisector of a zero angle is the harmonic mean of the adjacent sides.

(Note that AD=b+CD=b+ab/(b+c)=b(a+b+c)/(b+c) is a more direct proof!) In general, the bisectors of the two zero angles have thus different lengths!

When the two nonzero angle bisectors u=AD and v=BE are any prescribed positive numbers, the sides a, b, and c=a+b are given by a=b=\frac34u if u=v and

a= v\frac{2u-v-\sqrt{u^2-u v+v^2}}{2(u-v)},\quad b=u\frac{u-2 v+\sqrt{u^2-uv+v^2}}{2(u-v)}\quad\text{if }u\ne v.

S. Osinkin pointed out in a personal communication that these formulae can be unified for all u,\,v>0 as

c=\frac{u+v+\sqrt{u^2-uv+v^2}}2,\quad a=\frac{cv}{2c-v},\quad b=\frac{cu}{2c-u}.

[1] A. Zhukov and N. Akulich, Is the triangle defined uniquely? (Odnoznachno li opredeliaetsia treugol'nik?), Kvant No. 1 (2003) 29-31 (in Russian).

[2] S.F. Osinkin, On the existence of a triangle with prescribed bisector lengths, Forum Geom. 16 (2016) 399-405.

[3] G. Heindl, How to compute a triangle with prescribed lengths of its internal angle bisectors, Forum Geom. 16 (2016) 407-414.