Problem
GEO-B3-M06-P023 Three Cevians with \(10^\circ\) Angles
In triangle \(ABC\) with angles \(50^\circ,60^\circ,70^\circ\), cevians are drawn from the vertices cutting off angles \(10^\circ,20^\circ,30^\circ\) in cyclic order. Prove that they are concurrent if the order is chosen so that trig Ceva reduces to \(\sin10^\circ\sin20^\circ\sin80^\circ=\sin20^\circ\sin20^\circ\sin30^\circ\).
C. Hint 1. First write trig Ceva instead of trying to draw the point.
D. Hint 2. Prove the identity using \(\sin80^\circ=\cos10^\circ\) and double-angle formulas.
By assumption, the angle order makes trig Ceva reduce to \[ \sin10^\circ\sin20^\circ\sin80^\circ=\sin20^\circ\sin20^\circ\sin30^\circ. \] Cancel \(\sin20^\circ\). We need \(\sin10^\circ\sin80^\circ=\frac12\sin20^\circ\).
Since \(\sin80^\circ=\cos10^\circ\), the left side is \(\sin10^\circ\cos10^\circ=\frac12\sin20^\circ\). Thus trig Ceva holds, and the cevians are concurrent.
Good olympiad practice: the difficulty is the correct angle arrangement, not long algebra.