Olimpiadi di Matematica - Febbraio

Solution:

We observe that triangle $MBC$ is isosceles with base $MB$: indeed $\widehat{CBM}=\widehat{MBA}$, by hypothesis, and $\widehat{ABM}=\widehat{BMC}$, since they are alternate interior angles with respect to the parallels $DC \parallel AB$. Hence
$$\overline{AB}=\overline{DC}=2\,\overline{MC}=2\,\overline{CB}=2 \cdot 6=12$$
Let us call $K$ the projection of $C$ onto $AB$ and set $h=\overline{CK}$ and $d=\overline{BK}$. We find $h$ and $d$ through the system
$$\left\{\begin{array}{l} \overline{BC}^2-h^2=d^2 \\ \overline{AC}^2-h^2=(\overline{AB}+d)^2 \end{array}\right.$$
from which, subtracting the equations, we get $d=2/3$ and $h=8\sqrt{5}/3$.

Calling $H$ the projection of $M$ onto $AB$, we have
$$\overline{AH}=\overline{AB}+\overline{BK}-\overline{HK},$$
where $\overline{HK}=\overline{MC}$ (because $HKCM$ is a rectangle by construction) and $\overline{AB}=2\,\overline{MC}$. It follows that $\overline{AH}=\overline{MC}+d=20/3$ and $\overline{AM}^2=(\overline{AH})^2+h^2=80$. Consequently $\overline{AM}=4\sqrt{5}$.

Second solution. As before we have $\overline{AB}=2\,\overline{BC}$. We observe that triangle $ABM$ is right-angled at $M$: indeed, calling $N$ the midpoint of $\overline{AB}$, we have $\overline{NM} \parallel \overline{BC}$, so $\overline{NM}=\overline{AN}=\overline{NB}=6$, hence $M$ lies on a circle of radius $6$ centered at $N$ and in particular the angle at $M$ is right. Calling $\beta$ the angle $\widehat{ABM}$, we have $\overline{AM}=\overline{AB}\sin\beta=12\sin\beta$. Knowing the sides of triangle $ABC$ we can obtain $\cos(2\beta)$ by means of the law of cosines:
$$\overline{AB}^2+\overline{BC}^2-2\overline{AB}\cdot\overline{BC}\cos(2\beta)=\overline{AC}^2,$$
from which $\cos(2\beta)=-1/9$. Thanks to the cosine addition formula we can obtain $\sin\beta$ from the equation $\cos(2\beta)=1-2\sin^2\beta$, from which $\sin\beta=\sqrt{5}/3$. Thus $\overline{AM}=12\cdot\sqrt{5}/3=4\sqrt{5}$.

Third solution. From Heron's formula we obtain that the area of triangle $ABC$ is $16\sqrt{5}$. This area is also equal to the area of $ABM$, which in turn is twice the area of $ANM$. Calling $2x$ the length of $\overline{AM}$, we must have $1/2 \cdot 2x \cdot \sqrt{6^2-x^2}=1/2 \cdot 16\sqrt{5}=8\sqrt{5}$. Squaring both sides and solving for $x^2$ we find $2x=4\sqrt{5}$.