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We have to prove $(r_1− r) (r_2− r)(r_3− r) = 4 Rr^2$

I know the following formula , But I could not understand how to use them

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Fawad
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Koolman
  • 2,898
  • Use $r=\frac{\Delta}{S}$ – Fawad Dec 29 '16 at 08:35
  • Heron's formula (which I don't see here) would be helpful: $\Delta^2 = s(s-a)(s-b)(s-c)$. Otherwise, if you're supposed to just use a bunch of formulas, then plugging in the formulas in VIII(a) and IX(a) would give a series of straightforward computations (e.g. $r_1-r = \frac{\Delta a}{s(s-a)}$, and similarly for the others). – Joey Zou Dec 29 '16 at 08:37

2 Answers2

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We have \begin{align} (r_1-r)(r_2-r)(r_3-r)&= \left(\frac{\Delta}{s-a} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s-b} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s-c}-\frac{\Delta}{s}\right)\\ &={\Delta}^3 \left(\frac{ abc}{s^3(s-a)(s-b)(s-c)}\right)\\ &=\frac{abc{\Delta}^2}{s^2\Delta}\\ &=4R{\left(\frac{\Delta}{s}\right)}^2\\ &=4Rr^2. \end{align}

Fawad
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We know $$r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$ Thus, $$r_1-r = 4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} -4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = 4R\sin ^2 \frac{A}{2}$$ Similarly it can be shown that: $$r_2-r = 4R\sin ^2 \frac{B}{2} \text{ and that } r_3-r = 4R\sin ^2\frac{C}{2}$$ Thus, $$(r_1-r)(r_2-r)(r_3-r) = 64R^3\sin^2\frac{A}{2}\sin ^2\frac{B}{2}\sin ^2\frac{C}{2} = 4R(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})^2 = 4Rr^2$$ Hope it helps.