Mathematical Olympiad questions: Plane Geometry-7

Set $AC, ce$ is a positive hexagon $ABCDEF $ two diagonal, point $M, n$ respectively within the $AC, ce$ ratio: $\displaystyle{am\over AC} = {Cn\over CE} = R $. If $B, M, n$ three points collinear.

Try to find $r $.

(IMO 23.5)

Analysis:

The known proportional formula shows that $\triangle{abc}$ and $\triangle{cde}$ are corresponding triangles, and $\triangle{abm}$ and $\triangle{cdn}$ rotate congruent. This is a breakthrough and the analysis is made by combining the properties of the hexagonal.

Answer:

Nexus $DN $, easy to know $\triangle{abm}$ can be around point $O $ counterclockwise to $\triangle{cdn} \rightarrow BM = dn$ and $\angle{bnd} = 120^{\circ}$.

Nexus $OD, ob$, $\angle{bod} = 120^{\circ}\rightarrow O, B, D, n$ four points round.

Nexus $OC $, easy to know $CO = CD = Cb\rightarrow C $ is $\triangle{obd}$ outside the heart $\rightarrow CN = co$.

therefore $r = \displaystyle{cn \over CE} = {CO \over ce} = {\sqrt{3}\over 3}$.

Q$\cdot$e$\cdot$d

Mathematical Olympiad questions: Plane Geometry-7