# Taylor and Marr's Proof - R. A. Johnson's Version

### Problem

If the trisectors of the angles of a triangle are drawn so that those adjacent to each side intersect, the intersections are vertices of an equilateral triangle.

### Hint

Morley's theorem or, as it is often referred to, Morley's Miracle has a long history and multiple proofs, many - if not all of which - have been documented at this site. Each of the proofs (and one that follows is no exception) sheds an extra light on Morley's wonderful discovery. Anyone who has an ambition to add to the collection, should exercise his or her ingenuity. Good luck and enjoy.

### Solution

Let the trisectors adjacent to $A_{2}A_3$ be $A_2P_1$ and $A_3P_1,$ etc.; it is to be proved that $P_1P_2P_3$ is an equilateral triangle.

Extend $A_2P_1$ and $A_1P_2$ to meet at $L.$ Draw the incircle of $\Delta A_1A_2L,$ whose center is obviously $P_3.$ Let $Q$ and $R$ denote its points of contact on $LA_2$ and $LA_1,$ respectively, and let $P_3R$ meet $A_1A_3$ at $K$, and $P_3Q$ meet $A_2A_3$ at $N$; let the tangent from $K$ to the circle touch it at $P,$ and meet $A_2L$ at $F.$

So, by the construction,

$P_3R=RK,$ $\displaystyle P_3P=\frac{1}{2}P_3K,$

$\angle PP_3K=60^{\circ},$ $\angle P_3KP=30^{\circ},$

$\displaystyle\angle QP_3R=180^{\circ}-\angle QLR=120^{\circ}-\frac{2}{3}\angle A_3.$

It follows that

\begin{align} \angle FNQ &=\angle FP_3Q=\frac{1}{2}QP_3P \\ &=\frac{1}{2}(\angle QP_3R-60^{\circ})=30^{\circ}-\frac{1}{3}\angle A_3. \end{align}

Also, $\angle P_3NK=\angle P_3KN=\frac{1}{2}\angle QLR=30^{\circ}+\frac{1}{3}\angle A_3,$ implying $\angle FNK=\frac{2}{3}\angle A_3$ and $\angle FKN=\frac{1}{3}\angle A_3,$ so that $F,$ $K,$ $A_3,$ and $N$ are concyclic. Therefore, $F$ coincides with $P_1,$ and the tangent from $K$ passes through $P_1.$

By the same token, the tangent from $N$ to the same circle passes through $P_2.$

Also, by the construction, quadrilateral $P_3QLR$ (including the arc $QR$) and $\Delta P_3KN$ are symmetric with respect to the bisector $LP_3$ of $\angle A_1LA_2.$ By symmetry then, arcs $QP_1$ and $P_2R$ are equal, implying $\angle P_1P_3P_2=\angle PP_3R$ which, as has been shown, equals $60^{\circ}.$ It follows that $\angle P_1P_3P_2=60^{\circ}.$ The same obviously holds for the other two angles in $\Delta P_1P_2P_3.$

### Acknowledgment

The proof first appeared in the oft-quoted Taylor/Marr paper (Proceedings of Edinburgh Math. Society, 1914) which attributed it to W. E. Philip. It was also included in Johnson's book, Advanced Euclidean Geometry (Modern Geometry), pp. 253-254.

I am grateful to Roger Smyth for bringing this proof to my attention.