Library / Clock-tower School Juniors Competition /1 of 2

Clock-tower School Juniors Competition, 2011

Geometry Difficulty 4.0 AMC 10/12 Prove it Romania

The triangle $ABC$ and the points $M \in (BC)$ , $N \in (AC)$ , $P \in (AB)$ fulfill the conditions $\angle BMP \equiv \angle CNM \equiv \angle APN$ and $BM = CN = AP$ .

Prove that the triangle $ABC$ is equilateral.

Solution

We start noticing that $m(\overline{PMN}) = 180^\circ - \overline{BMP} - \overline{NMC} = 180^\circ - \overline{CNM} - \overline{NMC} = \overline{NCM}$ and, in the same way, $\overline{MNP} = \overline{NAP}$ , so
$\Delta ABC \sim \Delta NPM. \qquad (1)$

Suppose now, without loss of generality, that $\overline{C} \ge \overline{A} \ge \overline{B}$ . (2)
Then $AB \ge BC \ge AC$ , whence, from (1), $NP \ge PM \ge MN$ . These, together with $\angle BMP \equiv \angle CNM \equiv \angle APN$ and $BM = CN = AP$ , lead to $\overline{A} \ge \overline{B} \ge \overline{C}$ . (3)

Relations (2) and (3) show that $\overline{A} = \overline{B} = \overline{C}$ , whence the conclusion.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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