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