To solve the problem, we need to find the sum of the inradius \( r \) and the circumradius \( R \) of an equilateral triangle where one side is given by the equation \( x + y = 3 \) and the centroid is at the origin \( (0,0) \).
### Step-by-Step Solution:
1. **Identify the Equation of the Side**:
The equation of one side of the equilateral triangle is given as \( x + y = 3 \). This line can be rewritten in slope-intercept form as \( y = -x + 3 \).
2. **Find the Vertices of the Triangle**:
Since the centroid of the triangle is at the origin \( (0,0) \), we can denote the vertices of the triangle as \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \). The centroid \( G \) of a triangle formed by these vertices is given by:
\[
G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
\]
Setting this equal to \( (0,0) \), we have:
\[
x_1 + x_2 + x_3 = 0 \quad \text{and} \quad y_1 + y_2 + y_3 = 0
\]
3. **Determine the Coordinates of the Vertices**:
Since one side is \( x + y = 3 \), we can assume two points on this line, say \( B(3,0) \) and \( C(0,3) \). To maintain symmetry and ensure the centroid is at the origin, the third vertex \( A \) must be at \( (-3, -3) \).
4. **Calculate the Length of the Side**:
The distance between points \( B(3,0) \) and \( C(0,3) \) is:
\[
BC = \sqrt{(3-0)^2 + (0-3)^2} = \sqrt{9 + 9} = 3\sqrt{2}
\]
5. **Find the Inradius \( r \)**:
The formula for the inradius \( r \) of an equilateral triangle with side length \( a \) is:
\[
r = \frac{a \sqrt{3}}{6}
\]
Substituting \( a = 3\sqrt{2} \):
\[
r = \frac{3\sqrt{2} \cdot \sqrt{3}}{6} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2}
\]
6. **Find the Circumradius \( R \)**:
The formula for the circumradius \( R \) of an equilateral triangle with side length \( a \) is:
\[
R = \frac{a}{\sqrt{3}}
\]
Substituting \( a = 3\sqrt{2} \):
\[
R = \frac{3\sqrt{2}}{\sqrt{3}} = \sqrt{6}
\]
7. **Calculate \( r + R \)**:
Now, we can find \( r + R \):
\[
r + R = \frac{\sqrt{6}}{2} + \sqrt{6} = \frac{\sqrt{6}}{2} + \frac{2\sqrt{6}}{2} = \frac{3\sqrt{6}}{2}
\]
### Final Answer:
Thus, the value of \( r + R \) is \( \frac{3\sqrt{6}}{2} \).
