Geometry · Circumcircle · Inscribed angles · Cevians · Angle chasing · Equilateral triangle
A point \(P\) inside a triangle \(ABC\) satisfies
\[ \angle BPC = \angle BAC + 60^\circ, \qquad \angle CPA = \angle CBA + 60^\circ, \qquad \angle APB = \angle ACB + 60^\circ. \]
The lines \(AP\), \(BP\), \(CP\) meet the circle circumscribed about \(ABC\) a second time at \(A_1\), \(B_1\), \(C_1\) respectively. Prove that the triangle \(A_1B_1C_1\) is equilateral.
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Serbian Republic Competition (Republicko takmicenje) 1995, high school grade I, problem 3. Organized by the Mathematical Society of Serbia (DMS). Source