In a triangle the sum of two sides is x and the product of the same is y. If `x^2 - c^2 = y` where c is the third side. Determine the ration of the in-radius and circum-radius

Options

A. `(3y)/(2x(x+c))`
B. `(3y)/(2c(x+c))`
C. `(3y)/(4x(x+c))`
D. `(3y)/(4c(x+c))`

Answer

To solve the problem, we need to determine the ratio of the in-radius (r) and circum-radius (R) of a triangle given certain conditions. Let's break down the solution step by step.

### Step 1: Define the sides of the triangle
Let the sides of the triangle be \( a \), \( b \), and \( c \). According to the problem, we have:
- The sum of two sides \( x = a + b \)
- The product of the same two sides \( y = ab \)

### Step 2: Use the given equation
We are given the equation:
\[ x^2 - c^2 = y \]
Substituting for \( x \) and \( y \):
\[ (a + b)^2 - c^2 = ab \]

### Step 3: Expand and rearrange the equation
Expanding \( (a + b)^2 \):
\[ a^2 + 2ab + b^2 - c^2 = ab \]
Rearranging gives:
\[ a^2 + b^2 - c^2 + ab = 0 \]
This can be rearranged to:
\[ a^2 + b^2 - c^2 = -ab \]

### Step 4: Relate to cosine of angle C
Using the cosine rule, we know:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
From our previous equation, we can substitute \( c^2 \):
\[ a^2 + b^2 - (a^2 + b^2 - 2ab \cos C) = -ab \]
This simplifies to:
\[ 2ab \cos C = ab \]
Thus,
\[ \cos C = \frac{1}{2} \]
This means that \( C = 60^\circ \) or \( C = \frac{\pi}{3} \).

### Step 5: Calculate the area (Δ) of the triangle
Using the formula for the area of a triangle:
\[ \Delta = \frac{1}{2}ab \sin C \]
Substituting \( C = 60^\circ \):
\[ \Delta = \frac{1}{2}ab \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}ab \]

### Step 6: Calculate the semi-perimeter (s)
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} = \frac{x + c}{2} \]

### Step 7: Calculate the in-radius (r)
The in-radius \( r \) is given by:
\[ r = \frac{\Delta}{s} \]
Substituting \( \Delta \) and \( s \):
\[ r = \frac{\frac{\sqrt{3}}{4}ab}{\frac{x + c}{2}} = \frac{\sqrt{3}ab}{2(x + c)} \]

### Step 8: Calculate the circum-radius (R)
The circum-radius \( R \) is given by:
\[ R = \frac{abc}{4\Delta} \]
Substituting \( \Delta \):
\[ R = \frac{abc}{4 \cdot \frac{\sqrt{3}}{4}ab} = \frac{c}{\sqrt{3}} \]

### Step 9: Find the ratio \( \frac{r}{R} \)
Now we can find the ratio:
\[ \frac{r}{R} = \frac{\frac{\sqrt{3}ab}{2(x + c)}}{\frac{c}{\sqrt{3}}} = \frac{3ab}{2c(x + c)} \]

### Conclusion
Thus, the ratio of the in-radius to the circum-radius is:
\[ \frac{r}{R} = \frac{3y}{2c(x + c)} \]
This matches with one of the options provided in the question.