# Morley's Miracle

B. Bollobas' proof

This proof appears in *The Art of Mathematics* by B. Bollobás (Cambridge University Press, 2006, p. 127-128) accompanied by a remark that the author was used to offer Morley's theorem in the early 1970s as a problem to his better freshmen in Cambridge, who never found it easy. They were able to find the solution (and also the trigonometric one) after getting some hints.

Let us work backwards. To prove the theorem, it suffices to show that if PQR is an equilateral triangle and we erect on its sides triangles RQA, PQB and QPC as in the diagram then we get a triangle ABC with angles 3α, 3β, 3γ and appropriate trisectors.

Let us reflect P in BR to get R_{1}, and Q in AR to get R_{2}; construct the points P_{1}, P_{2}, Q_{1} and Q_{2} similarly as in the diagram. All we need then is that the points R_{1} and R_{2} are on AB (and so P_{1}, P_{2}, Q_{1} and Q_{2} are also on the appropriate sides). That this so is easily seen by computing some angles. For example, _{2} = β^{+},_{1} = β^{+} - (π - 2γ^{+}), where "^{+}" is a shorthand for

Let us assume that _{1} and R_{2} are in the order shown in the diagram. Then

∠R_{1}RR_{2} | = β^{+} + α^{+} - γ^{++} | |

= β + α - γ | ||

= π/3 - 2γ, |

so ∠RR_{2}R_{1} = (π - (π/3 - 2γ))/2 = γ^{+}. This implies that point R_{1} is on the segment AR_{2}. Similarly, R_{2} is on the segment BR_{1}, and we are done.

If α + β < γ then R_{1} and R_{2} are interchanged on AB; to see that, we note that _{1}RR_{2} = 2γ - π/3_{2}R_{1} = π - γ^{+},_{1} is on BR_{2}. Similarly, R_{2} is on AR_{1}, completing the solution.

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