Question

In a triangle \(ABC\), if \[ r_1 r_2 + rr_3 = 35, \quad r_2 r_3 + rr_1 = 63, \quad r_3 r_1 + rr_2 = 45 \] then \(2s\) is:

• A. \(28\)
• B. \(25\)
• C. \(21\)
• D. \(36\)

Hint:

Remember the formulas for the inradius $r$ and the exradii $r_1$, $r_2$, $r_3$.

Answer 1

The Correct Option is C

We know that

\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c}, \quad r = \frac{\Delta}{s}, \]

where \( \Delta \) is the area of triangle \( ABC \), \( s \) is the semiperimeter, and \( a, b, c \) are the side lengths. Then

\[ r_1 r_2 + rr_3 = \frac{\Delta}{s - a} \cdot \frac{\Delta}{s - b} + \frac{\Delta}{s} \cdot \frac{\Delta}{s - c} \]

\[ = \frac{\Delta^2}{(s - a)(s - b)} + \frac{\Delta^2}{s(s - c)} \]

\[ = \Delta^2 \cdot \frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)} \]

\[ = \Delta^2 \cdot \frac{s^2 - cs + s^2 - as - bs + ab}{s(s - a)(s - b)(s - c)} \]

\[ = \Delta^2 \cdot \frac{2s^2 - (a + b + c)s + ab}{s(s - a)(s - b)(s - c)} \]

\[ = \Delta^2 \cdot \frac{2s^2 - 2s^2 + ab}{s(s - a)(s - b)(s - c)} \]

\[ = \frac{ab \Delta^2}{s(s - a)(s - b)(s - c)}. \]

By Heron's formula,

\[ \Delta = \sqrt{s(s - a)(s - b)(s - c)}, \]

so

\[ \Delta^2 = s(s - a)(s - b)(s - c). \]

Hence,

\[ r_1 r_2 + rr_3 = \frac{ab \Delta^2}{s(s - a)(s - b)(s - c)} = ab. \]

Similarly,

\[ r_2 r_3 + rr_1 = bc, \quad r_3 r_1 + rr_2 = ca, \]

so we can write the given equations as

\[ ab = 35, \quad bc = 63, \quad ca = 45. \]

Multiplying all these equations, we get

\[ (ab)(bc)(ca) = 35 \cdot 63 \cdot 45 = 99225, \]

so

\[ a^2 b^2 c^2 = 99225. \]

Then

\[ abc = \sqrt{99225} = 315. \]

Dividing \( abc = 315 \) by \( ab = 35 \), we get

\[ c = \frac{315}{35} = 9. \]

Dividing \( abc = 315 \) by \( bc = 63 \), we get

\[ a = \frac{315}{63} = 5. \]

Dividing \( abc = 315 \) by \( ca = 45 \), we get

\[ b = \frac{315}{45} = 7. \]

Then

\[ 2s = a + b + c = 5 + 7 + 9 = 21, \]

so the answer is \( \boxed{21} \).