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2

Part of 2011 Dutch Mathematical Olympiad

Problems(1)

6 concyclic points, sides' trisectors, lead to equilateral triangle

Source: Dutch NMO 2011 p2

9/6/2019
Let ABCABC be a triangle. Points PP and QQ lie on side BCBC and satisfy ∣BP∣=∣PQ∣=∣QC∣=13∣BC∣|BP| =|PQ| = |QC| = \frac13 |BC|. Points RR and SS lie on side CACA and satisfy ∣CR∣=∣RS∣=∣SA∣=13∣CA∣|CR| =|RS| = |SA| = 1 3 |CA|. Finally, points TT and UU lie on side ABAB and satisfy ∣AT∣=∣TU∣=∣UB∣=13∣AB∣|AT| = |TU| = |UB| =\frac13 |AB|. Points P,Q,R,S,TP, Q,R, S, T and UU turn out to lie on a common circle. Prove that ABCABC is an equilateral triangle.
geometryConcyclicEquilateral