Question:
In a triangle \( ABC \), the identity holds:
\[
s \left( \frac{r_1 - r}{a} + \frac{r_2 - r}{b} + \frac{r_3 - r}{c} \right) = ?
\]
In a triangle \( ABC \), the identity holds:
\[
s \left( \frac{r_1 - r}{a} + \frac{r_2 - r}{b} + \frac{r_3 - r}{c} \right) = ?
\]
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Familiarity with inradius and exradius identities is essential for advanced triangle geometry problems.
Updated On: May 17, 2025
- \( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \)
- \( r_1 + r_2 + r_3 \)
- \( r_1 r_2 r_3 \)
- \( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \quad \text{(written again)} \)
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The Correct Option is B
Solution and Explanation
This identity relates the inradius \( r \), the exradii \( r_1, r_2, r_3 \), and the triangle's side lengths.
From known triangle geometry identities:
\[
s \left( \frac{r_1 - r}{a} + \frac{r_2 - r}{b} + \frac{r_3 - r}{c} \right) = r_1 + r_2 + r_3 - 3r
\]
But it turns out the original expression actually simplifies as:
\[
s \left( \frac{r_1 - r}{a} + \frac{r_2 - r}{b} + \frac{r_3 - r}{c} \right) = r_1 + r_2 + r_3
\]
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