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⟵ Steiner − LehmusRevisionism

Let the bisector \overrightarrow{AD} of angle ^{}B̸AC′ and the bisector \overrightarrow{BE} of angle ^{}A̸BC′′ be such that E ∈ \overrightarrow{AC′} and D ∈ \overrightarrow{BC′′}. If m^{}B̸AC′ + m^{}A̸BC′′ < π we may take C′ = C′′ := C, and we are in the setting of a question posed in 1840 by Daniel Christian Ludolf Lehmus of Berlin to Jakob Steiner: if the bisectors of the base angles of a triangle are congruent, must the triangle be isosceles? In the above notation, does AD = BE imply AC = BC? The answer, quickly supplied by Steiner, is yes, and Lehmus himself supplied a different proof in short order.

Proving the theorem anew became a perennial pastime of the “mathematical public” of the 19th Century. Mathematicians of the time tended to focus on the search for a “direct” proof, some (most notably Sylvester) suspecting that, in the strictest sense, this would be impossible. The sometimes duplicated claims and counterclaims over the next 150 years have essentially validated the skeptics, along the way revealing that a much smaller part of Euclidean geometry follows by purely direct proof from first principles than had been assumed in the pre-Hilbert era. There is an enormous literature on the synthetic solutions of this problem. In this note we aim to show how an analytic method leads to geometric generalizations. Our basic question must be phrased, “Does AD = BE imply m^{}B̸AC′ = m^{}A̸BC′′?”

Lehmus’ statement has a history in the Monthly Problems Section - sometimes an embarrassing one. The problem was first proposed in March, 1895; the May issue gave the proposer’s solution, and the June issue contained five solutions (numbered II through VI).

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It was proposed again in November, 1897, and three more solutions were published in April, 1898. A problem requesting the generalization to trisectors, etc. appeared in November, 1899. The solution, in October, 1900, began with criticisms of the nine previously published solutions by B.F.Yanney, a regular contributor to the Monthly. He concluded that only two of the nine were complete, with more than one containing glaring errors. Unfortunately, the solution printed after Yanney’s critique would not have met his standards, either!

In this note, we attempt to resuscitate Solution II (June, 1895). It is credited to three individual solvers, but it is unclear whether it was submitted in concert or is an editor’s composite. The method is from the theory of equations, but the argument is not easy to follow. Yanney, who often used similar methods in his notes on the Pythagorean theorem, declared himself unconvinced.

If we set m^{}B̸AC′ = 2α and m^{}A̸BC′′ = 2β, then the Law of Sines in △ADB and △BEA quickly produces the necessary condition

sin 2α sin(α + 2β) = sin 2β sin(β + 2α)

for the equality AD = BE. Applying double-angle formulas and setting x = sin α, y = sin β, we get

\sqrt{1 − x^{2}}(2xy − 2xy^{3} − x^{2} + 2x^{2}y^{2}) = \sqrt{1 − y^{2}}(2xy − 2x^{3}y − y^{2} + 2x^{2}y^{2}),

an equation we refer to as f(x, y) = g(x, y). Because of the solution points (1, 0), (0, 1), (−1, 0), (0, −1), we cannot conclude categorically, as seems to have been done in the 1895 solution, that this implies x^{2} = y^{2}, but we will prove that all solutions are of one of these types. In the geometric setting we began with, both α and β are less than π/2, so we need only consider the first quadrant to generalize the Lehmus statement as above. In fact, we mayrelax the requirement that E and D are on the same side of \overline{AB} as C′ and C′′. (For instance, if α and β are both greater than π/3 we must extend AC′ and

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BC′′ into the other halfplane to intersect the bisectors.) The equation f(x, y) = g(x, y) is still the correct condition.

Now to show that the only solutions in Q1 = {(x, y)|0 < x < 1, 0 < y < 1} are on x = y, we first note that, by symmetry, we need only check this for y > x. Squaring both sides

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