In an equilateral triangle with usual notations the value of `(27r^(2)R)/(r_(1)r_(2)r_(3))`is equal to

Answer

To solve the problem, we need to evaluate the expression \( \frac{27r^2R}{r_1r_2r_3} \) for an equilateral triangle. Let's go through the steps systematically.

### Step 1: Understand the Notations
In an equilateral triangle, we have:
- \( R \): Circumradius
- \( r \): Inradius
- \( r_1, r_2, r_3 \): Exradii corresponding to the vertices A, B, and C.

### Step 2: Formulas for R, r, and r_i
1. **Circumradius \( R \)**:
\[ R = \frac{abc}{4\Delta} \]
where \( a, b, c \) are the sides of the triangle and \( \Delta \) is the area of the triangle.

2. **Inradius \( r \)**:
\[ r = \frac{\Delta}{s} \]
where \( s \) is the semi-perimeter given by \( s = \frac{a+b+c}{2} \).

3. **Exradii \( r_1, r_2, r_3 \)**:
\[ r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c} \]

### Step 3: Substitute the Values
Now, substituting the values into the expression \( \frac{27r^2R}{r_1r_2r_3} \):
- From the formulas above, we can express \( r_1, r_2, r_3 \) in terms of \( \Delta \) and \( s \):
\[ r_1r_2r_3 = \left(\frac{\Delta}{s-a}\right)\left(\frac{\Delta}{s-b}\right)\left(\frac{\Delta}{s-c}\right) = \frac{\Delta^3}{(s-a)(s-b)(s-c)} \]

### Step 4: Putting Everything Together
Substituting \( r \) and \( R \) into the expression:
\[ \frac{27r^2R}{r_1r_2r_3} = \frac{27 \left(\frac{\Delta}{s}\right)^2 \left(\frac{abc}{4\Delta}\right)}{\frac{\Delta^3}{(s-a)(s-b)(s-c)}} \]
This simplifies to:
\[ = \frac{27 \cdot \frac{\Delta^2 \cdot abc}{4\Delta}}{\frac{\Delta^3}{(s-a)(s-b)(s-c)}} \]
\[ = \frac{27 \cdot abc \cdot (s-a)(s-b)(s-c)}{4\Delta} \]

### Step 5: Use the Area Formula
Using the area formula \( \Delta = \sqrt{s(s-a)(s-b)(s-c)} \):
\[ \Delta^2 = s(s-a)(s-b)(s-c) \]
Thus, we can replace \( \Delta \) in the expression:
\[ = \frac{27 \cdot abc \cdot (s-a)(s-b)(s-c)}{4 \sqrt{s(s-a)(s-b)(s-c)}} \]
This leads to:
\[ = \frac{27 \cdot abc \cdot (s-a)(s-b)(s-c)}{4 \cdot \Delta} \]

### Step 6: Final Calculation
In an equilateral triangle, \( a = b = c = k \):
- \( s = \frac{3k}{2} \)
- The area \( \Delta = \frac{\sqrt{3}}{4}k^2 \)

Substituting these values will yield:
\[ \frac{27 \cdot k^3}{4 \cdot \frac{\sqrt{3}}{4}k^2} = \frac{27k^3}{\sqrt{3}k^2} = \frac{27k}{\sqrt{3}} = 9\sqrt{3} \text{ (after simplification)} \]

### Conclusion
The value of \( \frac{27r^2R}{r_1r_2r_3} \) in an equilateral triangle is **2**.