Find the area of the region in the coordinate plane satisfying the three conditions

$\star$  $x \le 2y$
$\star$ $y \le 2x$
$\star$ $x + y \le 60.$
1. **Identify the inequalities and their corresponding lines:**
   - The inequality \( x \leq 2y \) can be rewritten as \( y \geq \frac{x}{2} \). This represents the region above the line \( y = \frac{x}{2} \).
   - The inequality \( y \leq 2x \) represents the region below the line \( y = 2x \).
   - The inequality \( x + y \leq 60 \) represents the region below the line \( x + y = 60 \).

2. **Find the intersection points of the lines:**
   - Intersection of \( y = \frac{x}{2} \) and \( y = 2x \):
     \[
     \frac{x}{2} = 2x \implies x = 0 \implies y = 0
     \]
     So, one intersection point is \((0,0)\).

   - Intersection of \( y = \frac{x}{2} \) and \( x + y = 60 \):
     \[
     y = \frac{x}{2} \implies x + \frac{x}{2} = 60 \implies \frac{3x}{2} = 60 \implies x = 40 \implies y = \frac{40}{2} = 20
     \]
     So, another intersection point is \((40,20)\).

   - Intersection of \( y = 2x \) and \( x + y = 60 \):
     \[
     y = 2x \implies x + 2x = 60 \implies 3x = 60 \implies x = 20 \implies y = 2 \cdot 20 = 40
     \]
     So, the third intersection point is \((20,40)\).

3. **Verify the vertices of the triangle:**
   The vertices of the triangle formed by the intersection of the lines are \((0,0)\), \((40,20)\), and \((20,40)\).

4. **Calculate the area using the Shoelace formula:**
   The Shoelace formula for the area of a polygon with vertices \((x_1, y_1)\), \((x_2, y_2)\), ..., \((x_n, y_n)\) is given by:
   \[
   \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
   \]
   For the vertices \((0,0)\), \((40,20)\), and \((20,40)\):
   \[
   \text{Area} = \frac{1}{2} \left| 0 \cdot 20 + 40 \cdot 40 + 20 \cdot 0 - (0 \cdot 40 + 20 \cdot 20 + 40 \cdot 0) \right|
   \]
   \[
   = \frac{1}{2} \left| 0 + 1600 + 0 - (0 + 400 + 0) \right|
   \]
   \[
   = \frac{1}{2} \left| 1600 - 400 \right|
   \]
   \[
   = \frac{1}{2} \left| 1200 \right|
   \]
   \[
   = \frac{1}{2} \cdot 1200 = 600
   \]

The final answer is \(\boxed{600}\).