Step 1: Understanding the Concept:
We are given the ex-radii \( r_1, r_2, r_3 \). We need to find the ratio of the sides \( a:b:c \). The ex-radii are related to the area \( \Delta \) and semi-perimeter \( s \) by \( r_1 = \frac{\Delta}{s-a} \), \( r_2 = \frac{\Delta}{s-b} \), and \( r_3 = \frac{\Delta}{s-c} \). The inradius \( r \) satisfies \( \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \).
Step 2: Detailed Explanation:
First, find the reciprocal of the inradius \( r \):
\[ \frac{1}{r} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6+3+1}{24} = \frac{10}{24} = \frac{5}{12} \]
So, \( r = \frac{12}{5} = 2.4 \).
From the area formulas, we can express \( s-a, s-b, s-c \) in terms of \( \Delta \):
\[ s-a = \frac{\Delta}{r_1} = \frac{\Delta}{4} \]
\[ s-b = \frac{\Delta}{r_2} = \frac{\Delta}{8} \]
\[ s-c = \frac{\Delta}{r_3} = \frac{\Delta}{24} \]
Adding these three equations gives \( s \):
\[ (s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s \]
So,
\[ s = \Delta \left( \frac{1}{4} + \frac{1}{8} + \frac{1}{24} \right) = \Delta \left( \frac{5}{12} \right) \]
Now, find the sides \( a, b, c \) in terms of \( \Delta \):
\[ a = s - (s-a) = \frac{5\Delta}{12} - \frac{\Delta}{4} = \frac{5\Delta - 3\Delta}{12} = \frac{2\Delta}{12} = \frac{\Delta}{6} \]
\[ b = s - (s-b) = \frac{5\Delta}{12} - \frac{\Delta}{8} = \frac{10\Delta - 3\Delta}{24} = \frac{7\Delta}{24} \]
\[ c = s - (s-c) = \frac{5\Delta}{12} - \frac{\Delta}{24} = \frac{10\Delta - \Delta}{24} = \frac{9\Delta}{24} \]
The ratio \( a:b:c \) is:
\[ \frac{\Delta}{6} : \frac{7\Delta}{24} : \frac{9\Delta}{24} \]
Multiply by 24 to clear denominators:
\[ 4 : 7 : 9 \]
Step 4: Final Answer:
The ratio is 4:7:9.