In `DeltaABC, R, r, r_(1), r_(2), r_(3)` denote the circumradius, inradius, the exradii opposite to the vertices A,B, C respectively. Given that `r_(1) :r_(2): r_(3) = 1: 2 : 3`
The sides of the triangle are in the ratio

Options

A. `1 : 2 : 3`
B. `3 : 5 : 7`
C. `1 : 5 : 9`
D. `5 : 8 : 9`

Explanation

To solve the problem, we will follow these steps systematically:

### Step 1: Understand the given ratios of exradii
We are given that the exradii \( r_1, r_2, r_3 \) opposite to the vertices \( A, B, C \) of triangle \( ABC \) are in the ratio:
\[ r_1 : r_2 : r_3 = 1 : 2 : 3 \]

### Step 2: Write the relationship between exradii and sides
The exradii are related to the area \( \Delta \) of the triangle and the semi-perimeter \( s \) as follows:
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]
where \( a, b, c \) are the lengths of the sides opposite to vertices \( A, B, C \) respectively.

### Step 3: Set up the equations based on the given ratio
From the ratio \( r_1 : r_2 : r_3 = 1 : 2 : 3 \), we can write:
\[ \frac{\Delta}{s - a} : \frac{\Delta}{s - b} : \frac{\Delta}{s - c} = 1 : 2 : 3 \]
This implies:
\[ \frac{\Delta}{s - a} = k, \quad \frac{\Delta}{s - b} = 2k, \quad \frac{\Delta}{s - c} = 3k \]
for some constant \( k \).

### Step 4: Express \( s - a, s - b, s - c \) in terms of \( k \)
From the above equations, we can express:
\[ s - a = \frac{\Delta}{k}, \quad s - b = \frac{\Delta}{2k}, \quad s - c = \frac{\Delta}{3k} \]

### Step 5: Rearranging to find \( a, b, c \)
Rearranging gives:
\[ a = s - \frac{\Delta}{k}, \quad b = s - \frac{\Delta}{2k}, \quad c = s - \frac{\Delta}{3k} \]

### Step 6: Find \( s \) in terms of \( k \)
Now, we can express \( s \) in terms of \( k \):
Adding \( a, b, c \):
\[ a + b + c = 3s - \left(\frac{\Delta}{k} + \frac{\Delta}{2k} + \frac{\Delta}{3k}\right) \]
The sum of the fractions can be simplified:
\[ \frac{\Delta}{k} + \frac{\Delta}{2k} + \frac{\Delta}{3k} = \Delta \left(\frac{1}{k} + \frac{1}{2k} + \frac{1}{3k}\right) = \Delta \left(\frac{6 + 3 + 2}{6k}\right) = \frac{11\Delta}{6k} \]
Thus:
\[ a + b + c = 3s - \frac{11\Delta}{6k} \]

### Step 7: Relate \( s \) to \( a, b, c \)
Since \( s = \frac{a + b + c}{2} \), we can equate:
\[ 2s = 3s - \frac{11\Delta}{6k} \implies s = \frac{11\Delta}{6k} \]

### Step 8: Substitute \( s \) back to find \( a, b, c \)
Now substituting \( s \) back into the equations for \( a, b, c \):
\[ a = s - \frac{\Delta}{k} = \frac{11\Delta}{6k} - \frac{\Delta}{k} = \frac{11\Delta - 6\Delta}{6k} = \frac{5\Delta}{6k} \]
\[ b = s - \frac{\Delta}{2k} = \frac{11\Delta}{6k} - \frac{\Delta}{2k} = \frac{11\Delta - 3\Delta}{6k} = \frac{8\Delta}{6k} \]
\[ c = s - \frac{\Delta}{3k} = \frac{11\Delta}{6k} - \frac{\Delta}{3k} = \frac{11\Delta - 2\Delta}{6k} = \frac{9\Delta}{6k} \]

### Step 9: Ratio of sides
Thus, the sides \( a, b, c \) are in the ratio:
\[ a : b : c = 5 : 8 : 9 \]

### Conclusion
The sides of triangle \( ABC \) are in the ratio \( 5 : 8 : 9 \).

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