geometry

Extend point P so it is perpendicular to AB. Label the intersection point F. 

 

\([AMP] = {2 \over3} \times PF \div 2\)

 

\(\triangle APF \) is similar to \(\triangle ABN\). Because of this, \(PF = {1 \over3}\)

 

Thus, \([AMP] = {2 \over 3} \times {1 \over 3} \div 2 = \color{brown}\boxed{1 \over 9}\)

Here I provide a solution with coordinate geometry.

 

Let D = (0, 0), C = (2, 0), A = (0, 2), B = (2, 2), M = (2/3, 2), N = (2, 1).

 

The equation of DM is:

\(y = \dfrac{2 - 0}{\dfrac23 - 0}x\\\)

i.e., \(y = 3x\)

 

The equation of AN is:

\(y - 1 = \dfrac{2 - 1}{0 - 2}(x - 2)\\ -2(y - 1) = x - 2\\ x + 2y-4=0\)

 

We solve for the coordinates of P.

\(\begin{cases}y = 3x\\x+2y-4=0\end{cases}\\ x + 2(3x) - 4 = 0\\ x= \dfrac47\\ y = 3\left(\dfrac47\right) = \dfrac{12}7\)

Hence P = (4/7, 12/7).

 

Therefore, 

\(\begin{array}{rcl} \text{area of }\triangle AMP &=& \dfrac12\cdot AM \cdot \left(2 - \dfrac{12}7\right) \\ &=& \dfrac12 \cdot \dfrac23 \cdot \dfrac27\\ &=& \dfrac2{21} \end{array}\)

Here's an alternative solution, that may be easier to understand.

 

\(\begin{array}{rcl} \tan \angle BAN &=& \dfrac{1}{2}\\ \angle BAN &=& \arctan \dfrac12 \end{array}\)

\(\begin{array}{rcl} \tan \angle AMD &=& \dfrac{2}{\frac23}\\ &=& 3\\ \angle AMD &=& \arctan 3 \end{array}\)

We now use the sine rule:

\(\begin{array}{rcl} \dfrac{MP}{\sin \angle BAN} &=& \dfrac{AM}{\sin\left(180^\circ - \arctan \dfrac12 - \arctan 3\right)}\\ \sqrt5 MP&=& \dfrac{5 \sqrt 2}7\cdot\dfrac23\\ MP &=& \dfrac{2\sqrt{10}}{21} \end{array}\)

Then, 

\(\begin{array}{rcl} \text{area of }\triangle AMP &=& \dfrac12 \cdot \dfrac23 \cdot MP \cdot \sin \angle AMP\\ &=& \dfrac{2\sqrt{10}}{63}\cdot \dfrac{3\sqrt{10}}{10}\\ &=& \dfrac2{21} \end{array}\)