Triangle From Angle Bisectors

Vladimir Zajic
July 15, 2003

Apart from special cases (such as all bisectors are of equal length, i.e., equilateral triangle, and a few more cases), it is impossible to construct a triangle given the lengths of its 3 bisectors.

It is impossible to construct an isosceles triangle given the lengths of its 2 different bisectors (again, apart from the equilateral triangle). It is sufficient to prove this and it follows that the general triangle is also impossible to construct.

Proof:

Isosceles triangle ΔABC, angles at vertices A and B equal, denote α. I is the incenter, i.e., the intersection of angle bisectors. la = lb, lc are the bisector lengths. La, Lb, Lc are the intersections of sides a, b, c with the bisectors.

Consider triangle ΔABC (or ΔALcC):

c/a = 2 cos(α)

Consider triangle ΔABLa:

angle ∠BLaA = π - α - α/2 = π - 3/2 α

la/c = sin(α)/sin(3/2 α)

Consider triangle ΔALcC:

lc/a = sin(α)

Combining:

lc/la = a/c sin(3/2 α) = sin(3/2 α) / (2 cos(α))

sin(3/2 α) = sin(α) cos(α/2) + cos(α) sin(α/2)

sin2(3/2 α) = sin2(α) cos2(α/2) + cos2(α) sin2(α/2) + 2 sin(α) cos(α) sin(α/2) cos(α/2)

Using sin2(α/2) = (1 - cos(α)) / 2 and cos2(α/2) = (1 + cos(α)) / 2,

sin2(3/2 α) = 1/2 + (3/2 sin2(α) - 1/2 cos2(α)) cos(α) = 1/2 + (3/2 - 2 cos2(α)) cos(α).

Let us denote x = cos(α) and k = lc/la. The problem is reduced to determining (constructing) cos(α) from the bisector length ratio. This is equivalent to the following cubic equation

k2 = (1/2 + 3/2 x - 2 x3) / 4 x2 or

2 x3 + 4 k2 x2 - 3/2 x - 1/2 = 0 or

4 x3 + 8 k2 x2 - 3 x - 1 = 0

It is easy to see that for k = 1 (equilateral triangle), the equation has rational root x = cos(α) = 1/2, α = π/3. Suppose that the quadratic coefficient is some positive integer (of course, except 8), for example some prime number:

8 k2 = q > 0

Then the ratio k is always constructible (as a square root of rational number q/8). We can construct the bisector lengths la and lc and try to construct cos(α) to solve the problem for these particular lengths. This is generally impossible. It is sufficient to show that for some integer quadratic coefficient the above cubic equation has no rational root. Then no roots are constructible.

By the rational root theorem for polynomials with integer coefficients, the possible rational roots are r = p/q, where p is a divider of the absolute term (1 for our equation) and q is a divider of the highest power term coefficient (4 for our equation). The possible rational roots are -1, +1, -1/2, +1/2, -1/4, +1/4 and no others.

It is easy to find a positive integer for the quadratic term, for which our cubic equation has none of the above roots (almost any integer will do). Indeed, by calculating the value of 4 x3 - 3 x - 1 for all possible rational roots, which must be compensated by the quadratic term, we can quickly convince ourselves that the above cubic equation has rational root(s) for the following integer values of q only:

q = 0: r1 = +1, r2 = -1/2, r3 = -1/2
q = 2: r = -1
q = 5: r = -1/4
q = 8: r = +1/2
q = 27: r = +1/4

Therefore, cos(α) is not constructible for all possible bisector length ratios. Q.E.D.

Remark

The problem has a long history. In 1903 a Ph.D. thesis has been submitted at the University of Chicago by Richard Philip Baker titled The Problem of the Angle-Bisectors. (The thesis has been published by the UC in 1911 and can be found at the University of Michigan Historical Math Collection) Inter alia, the author mentions the previous efforts of Pascal, Euler, Brocard, among others. He also mentions E. Catalan and an unnamed Russian mathematician as being used to propose the problem as an elementary exercise or as a standard set-back for ambitious young geometers.


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