A Pre-RMO question! -27

We don't need to find the altitude $AH$ to solve the problem.

Because $\triangle ABC$ and $\triangle ADM$ have the same height or altitude the ratio of their areas is the ratio of their base lengths or $\dfrac {[ADM]}{[ABC]} = \dfrac {DM}{BC} \implies [ADM] = \dfrac {DM}{18}[ABC]$ .

To find $DM$ , we note that $DM = BD - BM = BD - 9$ . To find $BD$ , we use angle bisector theorem as follows:

$\begin{aligned} \frac {BD}{AB} & = \frac {DC}{AC} \\ \frac {BD}{14} & = \frac {18-BD}{10} \\ \implies BD & = \frac {21}2 \\ \implies DM & = BD - 9 = \frac 32 \end{aligned}$

To find $[ABC]$ , we use Heron's formula as follows:

$\begin{aligned} [ABC] & = \sqrt{s(s-a)(s-b)(s-c)} & \small \blue{\text{where }s = \frac {a+b+c}2} \\ & = \sqrt{21(7)(11)(3)} = 21\sqrt{11} \end{aligned}$

Therefore, $[ADM] = \dfrac {DM}{18}[ABC] = \dfrac 32 \times \dfrac 1{18} \times 21\sqrt{11} = \dfrac {7\sqrt{11}}4$ . $\implies k = \boxed{11}$ .

$\dfrac {|\overline {BD}|}{18-|\overline {BD}|}=\dfrac {14}{10}$

$\implies |\overline {BD}|=10.5\implies |\overline {MD}|=10.5-9=1.5$

Area of $\triangle {ABC}$ is $21\sqrt {11}$ (Heron's Formula)

$\implies |\overline {AH}|=\dfrac {7\sqrt {11}}{3}$

Area of $\triangle {ADM}$ is

$\dfrac 12\times |\overline {MD}|\times |\overline {AH}|=\dfrac {7\sqrt {11}}{4}$

Hence $k=\boxed {11}$ .