Morley's Miracle
Bankoff's proof

This proof has appeared in Mathematics Magazine, 35 (1962) 223-224.

In the diagram,

(1)
sin(∠AQC) = sin(p - (A+C)/3)
  = sin((p - B)/3)
  = sin((2p + B)/3)

Also

(2) sin(3a) = 4sin(a)sin(p/3 + a)sin(p/3 - a)

From the Sine Law,

AQ·sin((p - B)/3) = 2R·sin(B)·sin(C/3),

where R is the circumradius. Therefore, by (2)

AQ = 8R·sin(B/3)·sin(C/3)·sin((p + B)/3).

Similarly, AR = 8R·sin(C/3)·sin(B/3)·sin((p + C)/3). Therefore,

AR/AQ = sin((p + C)/3)/sin((p + B)/3).

But ∠ARQ + ∠AQR = p - A/3 = (p + B)/3 + (p + C)/3. From here,

∠ARQ = (p + C)/3 and ∠AQR = (p + B)/3,

and similarly for triangles BPR and CPQ. It thus follows that the sum of angles around P, excluding ∠QPR is 300°, or ∠QPR = 60°. The other two angles are similarly shown to be 60°.


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