Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it North Macedonia

Let $B$ be an interior point of the segment $AC$ . The equilateral triangles $\triangle ABM$ and $\triangle BCN$ are constructed in same half plane determined by the line $AC$ . The lines $AN$ and $CM$ intersect in $L$ . Find the angle $\angle CLN$ .

Solution

Notice that the angle $\angle MBN = 60^\circ$ . Hence $\angle ABN = \angle MBC = 120^\circ$ . Because the triangles $ABM$ and $BCN$ are equilateral we have $\overline{AB} = \overline{BM}$ and $\overline{BC} = \overline{BN}$ . Hence $\triangle ABN \cong \triangle MBC$ , from where $\angle ANB = \angle MCB = \alpha$ .

Let $K$ be the intersection point of the lines $MC$ and $BN$ . Because $\angle BKC = \angle MKN = \beta$ (opposite angles), from $\triangle BKC$ we have $\alpha + \beta = 180^\circ - 60^\circ = 120^\circ$ .

Finally from $\triangle LKN$ we obtain $\angle KLN = 180^\circ - (\alpha + \beta) = 60^\circ$ , hence $\angle CLN = 60^\circ$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.