In square $ABCD$ with side length $2$, let $M$ be the midpoint of $AB$. Let $N$ be a point on $AD$ such that $AN = 2ND$. Let point $P$ be the intersection of segment $MN$ and diagonal $AC$. Find the area of triangle $BPM$.

[i]Proposed by Jacob Xu[/i]
1. **Assign coordinates to the vertices of the square:**
   \[
   A = (0, 0), \quad B = (0, 2), \quad C = (2, 2), \quad D = (2, 0)
   \]

2. **Find the coordinates of point \( M \), the midpoint of \( AB \):**
   \[
   M = \left( \frac{0+0}{2}, \frac{0+2}{2} \right) = (0, 1)
   \]

3. **Find the coordinates of point \( N \) on \( AD \) such that \( AN = 2ND \):**
   Since \( AN = 2ND \), we can write:
   \[
   AN = 2x \quad \text{and} \quad ND = x \quad \text{with} \quad AN + ND = AD = 2
   \]
   Solving for \( x \):
   \[
   2x + x = 2 \implies 3x = 2 \implies x = \frac{2}{3}
   \]
   Therefore, \( AN = 2x = \frac{4}{3} \). Thus, the coordinates of \( N \) are:
   \[
   N = \left( \frac{4}{3}, 0 \right)
   \]

4. **Find the equation of line \( MN \):**
   The slope of \( MN \) is:
   \[
   \text{slope of } MN = \frac{0 - 1}{\frac{4}{3} - 0} = -\frac{3}{4}
   \]
   Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \) with point \( M(0, 1) \):
   \[
   y - 1 = -\frac{3}{4}(x - 0) \implies y = -\frac{3}{4}x + 1
   \]

5. **Find the intersection of line \( MN \) and diagonal \( AC \):**
   The equation of diagonal \( AC \) is \( y = x \). Setting \( y = x \) in the equation of \( MN \):
   \[
   x = -\frac{3}{4}x + 1 \implies x + \frac{3}{4}x = 1 \implies \frac{7}{4}x = 1 \implies x = \frac{4}{7}
   \]
   Therefore, the coordinates of point \( P \) are:
   \[
   P = \left( \frac{4}{7}, \frac{4}{7} \right)
   \]

6. **Calculate the area of triangle \( BPM \):**
   The base \( \overline{BM} \) has length \( 1 \) (since \( B = (0, 2) \) and \( M = (0, 1) \)).
   The height from point \( P \) to line \( BM \) (which is vertical) is the \( x \)-coordinate of \( P \), which is \( \frac{4}{7} \).

   The area of triangle \( BPM \) is:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times \frac{4}{7} = \frac{2}{7}
   \]

The final answer is \(\boxed{\frac{2}{7}}\).