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2

Part of 1969 Kurschak Competition

Problems(1)

ABC is equilateral if a(1 - 2 cos A) + b(1 - 2 cos B) + c(1 - 2 cos C) = 0

Source: 1969 Hungary - Kürschák Competition p2

10/11/2022
A triangle has side lengths a,b,ca, b, c and angles A,B,CA, B, C as usual (with bb opposite BB etc). Show that if a(1−2cos⁡A)+b(1−2cos⁡B)+c(1−2cos⁡C)=0a(1 - 2 \cos A) + b(1 - 2 \cos B) + c(1 - 2 \cos C) = 0 then the triangle is equilateral.
trigonometrygeometryEquilateral