Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.2 National olympiad Prove it Greece

Let $\triangle ABC$ a triangle with $90^\circ \neq \hat{A} \neq 135^\circ$ . Let $D$ and $E$ be external points to the triangle such that $DAB$ and $EAC$ are isosceles triangles with right angles in $D$ and $E$ , respectively. Let $F = BE \cap CD$ , and $M, N$ the midpoints of $BC, DE$ , respectively.
Prove that, if three of the points $A, F, M, N$ are collinear, then the four points are collinear.

Solution

a) If $M, N, F$ are collinear, we must have $DE \parallel BC$ , hence the distances from $D$ and $E$ to $BC$ be equal, and this is equivalent to $b = c$ .

b) If $A, M, F$ are collinear, $\tan \vec{BAM} = \frac{b \sin A}{c - b \cos A}$ , $\tan \vec{DAN} = \frac{b \cos A}{c - b \sin A}$ , and $\vec{DAN} = 45^\circ + \vec{BAM} \Leftrightarrow b = c$ .

c) If $A, M, N$ are collinear, we call $H = CD \cap AB$ , $G = BE \cap AC$ and computing we have
$\frac{AG}{GC} = \frac{c \sin (A + 45^\circ)}{a \sin (B + 45^\circ)}$
and similarly $\frac{AH}{HB}$ . By Ceva the condition becomes $b = c$ .

d) if $A, F, N$ are collinear, to use Ceva in $DEF$ we consider $X = CD \cap AE$ , $Y = BG \cap AD$ (eventually improper). We get
$\vec{HCG} = \vec{HBG}, \text{ hence HG is antiparallel to } BC, i.e. \ AH \cdot AB = AG \cdot AC \Leftrightarrow b = c.$

e) If $b = c$ , all four considered points are collinear.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.