Question

In a triangle ABC, if \((r_1 - r_3)(r_1 - r_2) - 2r_2r_3 = 0\), then \(a^2 - b^2 =\)

• A. \(c^2 + b^2/4 \)
• B. \(c^2 \)
• C. \(abc \)
• D. \((b+a)/c \)

Hint:

Problems involving relations between exradii often benefit from expressing the exradii in terms of the triangle's area and semi-perimeter. Algebraic manipulation, especially expanding terms and simplifying, is crucial. Remember to be meticulous with signs and calculations.

Answer 1

The Correct Option is B

Step 1: Rewrite the given relation. The given relation is \((r_1 - r_3)(r_1 - r_2) - 2r_2r_3 = 0\). Rearrange it as: \[ (r_1 - r_3)(r_1 - r_2) = 2r_2r_3 \] Step 2: Express the exradii in terms of area and semi-perimeter. We know the formulas for the exradii \(r_1, r_2, r_3\) in terms of the area of the triangle \(\Delta\) and the semi-perimeter \(s = \frac{a+b+c}{2}\): \[ r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c} \] Substitute these into the rewritten equation: \[ \left(\frac{\Delta}{s-a} - \frac{\Delta}{s-c}\right)\left(\frac{\Delta}{s-a} - \frac{\Delta}{s-b}\right) = 2\left(\frac{\Delta}{s-b}\right)\left(\frac{\Delta}{s-c}\right) \] Factor out \(\Delta^2\) from both sides (assuming \(\Delta \ne 0\), which is true for a triangle): \[ \Delta^2 \left(\frac{1}{s-a} - \frac{1}{s-c}\right)\left(\frac{1}{s-a} - \frac{1}{s-b}\right) = 2\Delta^2 \left(\frac{1}{s-b}\right)\left(\frac{1}{s-c}\right) \] Divide by \(\Delta^2\): \[ \left(\frac{1}{s-a} - \frac{1}{s-c}\right)\left(\frac{1}{s-a} - \frac{1}{s-b}\right) = \frac{2}{(s-b)(s-c)} \] Combine the terms within the parentheses: \[ \left(\frac{(s-c) - (s-a)}{(s-a)(s-c)}\right)\left(\frac{(s-b) - (s-a)}{(s-a)(s-b)}\right) = \frac{2}{(s-b)(s-c)} \] Simplify the numerators: Since \(s-c - (s-a) = s-c-s+a = a-c\), and \(s-b - (s-a) = s-b-s+a = a-b\). \[ \left(\frac{a-c}{(s-a)(s-c)}\right)\left(\frac{a-b}{(s-a)(s-b)}\right) = \frac{2}{(s-b)(s-c)} \] Multiply both sides by \((s-b)(s-c)\) to clear denominators: \[ \frac{(a-c)(a-b)}{(s-a)^2} = 2 \] Step 3: Substitute \(s-a\) and simplify the equation. We know \(s = \frac{a+b+c}{2}\), so \(s-a = \frac{a+b+c}{2} - a = \frac{a+b+c-2a}{2} = \frac{b+c-a}{2}\). Substitute this into the equation: \[ \frac{(a-c)(a-b)}{\left(\frac{b+c-a}{2}\right)^2} = 2 \] \[ \frac{4(a-c)(a-b)}{(b+c-a)^2} = 2 \] \[ \frac{2(a-c)(a-b)}{(b+c-a)^2} = 1 \] \[ 2(a-c)(a-b) = (b+c-a)^2 \] Step 4: Expand and simplify the equation to find \(a^2 - b^2\). Expand both sides: Left side: \(2(a^2 - ab - ac + bc) = 2a^2 - 2ab - 2ac + 2bc\) Right side: \((b+c-a)^2 = ((b+c) - a)^2 = (b+c)^2 - 2a(b+c) + a^2\) \[ = b^2 + c^2 + 2bc - 2ab - 2ac + a^2 \] Set the expanded left side equal to the expanded right side: \[ 2a^2 - 2ab - 2ac + 2bc = a^2 + b^2 + c^2 + 2bc - 2ab - 2ac \] Move all terms to one side: \[ 2a^2 - a^2 - 2ab + 2ab - 2ac + 2ac + 2bc - 2bc - b^2 - c^2 = 0 \] Combine like terms: \[ a^2 - b^2 - c^2 = 0 \] Rearrange the terms to find \(a^2 - b^2\): \[ a^2 - b^2 = c^2 \] The final answer is } \boxed{c^2}.