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Question Number 16441 by mrW1 last updated on 22/Jun/17

Commented by mrW1 last updated on 22/Jun/17

$$\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{question}\mathrm{to}\mathrm{this}\mathrm{diagram}?$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

$$r=\sqrt{r_a.r_b}+\sqrt{r_a.r_c}+\sqrt{r_c.r_b}$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/Jun/17

$$cos\frac{A}{2}=\frac{2\sqrt{r.r_a}}{r+r_a}=m$$$$m=\frac{2\sqrt{r.r_a}}{r+r_a}\Rightarrow m^2=\frac{4r.r_a}{{(r+r_a)}^2}$$$$\Rightarrow m^2{(r+r_a)}^2-4r.r_a=0$$$$\Rightarrow r_a^2+2r(1-\frac{2}{m^2})r_a+r^2=0$$$$\Rightarrow r_a=\frac{-2r(1-\frac{2}{m^2})+\sqrt{4r^2{(1-\frac{2}{m^2})}^2-4r^2}}{2}=$$$$r_a=r(\frac{2\sqrt{1-m^2}}{m^2}-1+\frac{2}{m^2})=r.\frac{2\sqrt{1-m^2}-m^2+2}{m^2}=$$$$=r.\frac{1-m^2+2\sqrt{1-m^2}+1}{m^2}=r.\frac{{(\sqrt{1-m^2}+1)}^2}{m^2}$$$$1-m^2=1-{cos}^2\frac{A}{2}={sin}^2\frac{A}{2}$$$$\Rightarrow r_a=r.\frac{{(1+sin\frac{A}{2})}^2}{{cos}^2\frac{A}{2}}=r.{\left(\frac{1+sin\frac{A}{2}}{cos\frac{A}{2}}\right)}^2$$$$r_b=r.{\left(\frac{1+sin\frac{B}{2}}{cos\frac{B}{2}}\right)}^2,r_c=r.{\left(\frac{1+sin\frac{C}{2}}{cos\frac{C}{2}}\right)}^2$$$$\Rightarrow \sqrt{r_a.r_b}+\sqrt{r_b.r_c}+\sqrt{r_c.r_a}=$$$$=r.[\frac{(1+sin\frac{A}{2})(1+sin\frac{B}{2})}{cos\frac{A}{2}.cos\frac{B}{2}}+\frac{(1+sin\frac{B}{2})(1+sin\frac{C}{2})}{cos\frac{B}{2}.cos\frac{C}{2}}+$$$$+\frac{(1+sin\frac{A}{2})(1+sin\frac{C}{2})}{cos\frac{A}{2}.cos\frac{C}{2}}]=$$$$=r.[\frac{cos\frac{C}{2}+cos\frac{C}{2}sin\frac{A}{2}+cos\frac{C}{2}sin\frac{B}{2}+cos\frac{C}{2}sin\frac{A}{2}sin\frac{B}{2}}{cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}+$$$$+\frac{cos\frac{A}{2}+cos\frac{A}{2}sin\frac{B}{2}+cos\frac{A}{2}sin\frac{C}{2}+cos\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}}{cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}+$$$$+\frac{cos\frac{B}{2}+cos\frac{B}{2}sin\frac{A}{2}+cos\frac{B}{2}sin\frac{C}{2}+cos\frac{B}{2}sin\frac{A}{2}sin\frac{C}{2}}{cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}]=$$$$=r.\left[\frac{\mathrm{\Sigma }cos\frac{A}{2}+\mathrm{\Sigma }sin\frac{A+B}{2}+\mathrm{\Sigma }cos\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}}{\mathrm{\Pi }cos\frac{A}{2}}\right]=$$$$=r.\left[\frac{2\mathrm{\Sigma }cos\frac{A}{2}+\mathrm{\Sigma }cos\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}}{\mathrm{\Pi }cos\frac{A}{2}}\right]=$$$$\mathrm{\Sigma }cos\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\mathrm{\Sigma }\sqrt{\frac{p(p-a)}{bc}}\sqrt{\frac{(p-a)(p-c)(p-a)(p-b)}{ac.ab}}=$$$$=\mathrm{\Sigma }\frac{(p-a).S}{abc}=\frac{(3p-a-b-c).S}{abc}=\frac{p.S}{4R.S}=\frac{p}{4R}$$$$\mathrm{\Pi }cos\frac{A}{2}=\mathrm{\Pi }\sqrt{\frac{p(p-a)}{bc}}=\frac{p.S}{abc}=\frac{p.S}{4R.S}=\frac{p}{4R}$$$$\Rightarrow \mathrm{\Sigma }\sqrt{r_a.r_b}=r.(1+\frac{8R\mathrm{\Sigma }cos\frac{A}{2}}{p})$$$$ithinkthereissomethingwrong.$$$$pleasesomeonerecheckmysolution.$$

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