 In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Let $M$ be the midpoint of side $AB$, $G$ be the centroid of $\triangle ABC$, and $E$ be the foot of the altitude from $A$ to $BC$. Compute the area of quadrilateral $GAME$.

[i]Proposed by Evin Liang[/i]

[hide=Solution][i]Solution[/i]. $\boxed{23}$
Use coordinates with $A = (0,12)$, $B = (5,0)$, and $C = (-9,0)$. Then $M = \left(\dfrac{5}{2},6\right)$ and $E = (0,0)$. By shoelace, the area of $GAME$ is $\boxed{23}$.[/hide]
1. **Assign coordinates to the vertices of the triangle:**
   We can place the triangle in the coordinate plane for easier calculation. Let:
   \[
   A = (0, 12), \quad B = (5, 0), \quad C = (-9, 0)
   \]
   These coordinates satisfy the given side lengths:
   \[
   AB = \sqrt{(5-0)^2 + (0-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
   \]
   \[
   BC = \sqrt{(5 - (-9))^2 + (0-0)^2} = \sqrt{14^2} = 14
   \]
   \[
   CA = \sqrt{(0 - (-9))^2 + (12-0)^2} = \sqrt{81 + 144} = \sqrt{225} = 15
   \]

2. **Find the coordinates of the midpoint \( M \) of \( AB \):**
   The midpoint formula gives:
   \[
   M = \left( \frac{0+5}{2}, \frac{12+0}{2} \right) = \left( \frac{5}{2}, 6 \right)
   \]

3. **Find the coordinates of the foot of the altitude \( E \) from \( A \) to \( BC \):**
   Since \( BC \) is horizontal (both \( B \) and \( C \) have the same y-coordinate), the altitude from \( A \) to \( BC \) is vertical. Thus, \( E \) has the same x-coordinate as \( A \) and the same y-coordinate as \( B \) and \( C \):
   \[
   E = (0, 0)
   \]

4. **Find the coordinates of the centroid \( G \) of \( \triangle ABC \):**
   The centroid \( G \) is the average of the coordinates of the vertices:
   \[
   G = \left( \frac{0 + 5 - 9}{3}, \frac{12 + 0 + 0}{3} \right) = \left( \frac{-4}{3}, 4 \right)
   \]

5. **Calculate the area of quadrilateral \( GAME \) using the Shoelace Theorem:**
   The Shoelace Theorem states that for a polygon with vertices \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\), the area is:
   \[
   \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
   \]
   For quadrilateral \( GAME \) with vertices \( G \left( \frac{-4}{3}, 4 \right) \), \( A (0, 12) \), \( M \left( \frac{5}{2}, 6 \right) \), and \( E (0, 0) \), we have:
   \[
   \text{Area} = \frac{1}{2} \left| \left( \frac{-4}{3} \cdot 12 + 0 \cdot 6 + \frac{5}{2} \cdot 0 + 0 \cdot 4 \right) - \left( 4 \cdot 0 + 12 \cdot \frac{5}{2} + 6 \cdot 0 + 0 \cdot \frac{-4}{3} \right) \right|
   \]
   Simplifying inside the absolute value:
   \[
   = \frac{1}{2} \left| \left( -16 + 0 + 0 + 0 \right) - \left( 0 + 30 + 0 + 0 \right) \right|
   \]
   \[
   = \frac{1}{2} \left| -16 - 30 \right|
   \]
   \[
   = \frac{1}{2} \left| -46 \right|
   \]
   \[
   = \frac{1}{2} \times 46 = 23
   \]

The final answer is \(\boxed{23}\).