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Step-by-Step Solution
Step 1: Recognize the special properties of an equilateral triangle
In an equilateral triangle, the centroid, incenter, and circumcenter all coincide at a single point. This means:
• The incenter is the point from which the incircle (circle touching each side) is drawn.
• The circumcenter is the point from which the circumcircle (circle passing through each vertex) is drawn.
• Therefore, the distance from the center to a side is the inradius $r$, and the distance from the center to a vertex is the circumradius $R$.
Moreover, for an equilateral triangle of side length $a$, the inradius $r$ and the circumradius $R$ satisfy $R = 2r.$
Step 2: Find the inradius using distance from the centroid to the given side
The side is given by the line $x + y = 3.$ The centroid of the triangle is at the origin $(0, 0).$
The distance $d$ from a point $(x_{0}, y_{0})$ to the line $Ax + By + C = 0$ is given by:
$$d = \frac{|A x_{0} + B y_{0} + C|}{\sqrt{A^2 + B^2}}.$$
Here, $A = 1$, $B = 1$, and $C = -3$. Substituting $(x_{0}, y_{0}) = (0,0)$, we get
$$r = \frac{|1\cdot 0 + 1\cdot 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}.$$
Hence, the inradius of the triangle is $r = \frac{3}{\sqrt{2}}.$
Step 3: Relate the circumradius to the inradius
For an equilateral triangle, $R = 2r.$ Therefore,
$$R = 2 \times \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}.$$
Step 4: Compute $R + r$
Summing the circumradius and the inradius:
$$R + r = 3\sqrt{2} + \frac{3}{\sqrt{2}}.$$
To combine them into a single fraction, note that $3\sqrt{2} = \frac{6}{\sqrt{2}}$:
$$3\sqrt{2} + \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}.$$
Thus,
$$R + r = \frac{9}{\sqrt{2}}.$$
Step 5: Final Answer
The value of $(R + r)$ is
$$\frac{9}{\sqrt{2}}.$$
