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Question Number 169585 by Shrinava last updated on 03/May/22

Commented by mr W last updated on 03/May/22

$$pleasecheckthequestion!one{can}^{\prime }t$$$$imagehowitlookslike.maybe$$$$DonBC,EonCAandFonAB.$$

Commented by Shrinava last updated on 03/May/22

$$\mathrm{Sorry}\mathrm{professor},\mathrm{yes}\mathrm{will}\mathrm{be}\mathrm{as}\mathrm{you}\mathrm{say}$$

Answered by mr W last updated on 03/May/22

Commented by mr W last updated on 03/May/22

$${it}^{\prime }seasytoseethat$$$$\mathrm{\Delta }EFD\sim \mathrm{\Delta }ABC$$$$\Rightarrow \frac{p}{c}=\frac{q}{a}=\frac{r}{b}=k,say$$$$\Rightarrow p=kc,q=ka,r=kb$$$$$$$$a=\frac{q}{\tan B}+\frac{r}{\sin C}$$$$\Rightarrow a(\frac{1}{k}-\frac{1}{\tan B})=\frac{b}{\sin C}...\left(i\right)$$$$b=\frac{r}{\tan C}+\frac{p}{\sin A}$$$$\Rightarrow b(\frac{1}{k}-\frac{1}{\tan C})=\frac{c}{\sin A}...\left(ii\right)$$$$c=\frac{p}{\tan A}+\frac{q}{\sin B}$$$$\Rightarrow c(\frac{1}{k}-\frac{1}{\tan A})=\frac{a}{\sin B}...\left(iii\right)$$$$\left(i\right)\times \left(ii\right)\times \left(iii\right):$$$$(\frac{1}{k}-\frac{1}{\tan A})(\frac{1}{k}-\frac{1}{\tan B})(\frac{1}{k}-\frac{1}{\tan C})=\frac{1}{\sin A\sin B\sin C}$$$$(\frac{1}{k}-\frac{\cos A}{\sin A})(\frac{1}{k}-\frac{\cos B}{\sin B})(\frac{1}{k}-\frac{\cos C}{\sin C})=\frac{1}{\sin A\sin B\sin C}$$$$(\sin A-k\cos A)(\sin B-k\cos B)(\sin C-k\cos C)=k^3$$$$(1+\cos A\cos B\cos C)k^3-\left(\sin A\sin B\sin C\right)k^2+(1+\cos A\cos B\cos C)k-\sin A\sin B\sin C=0$$$$(k-\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C})(k^2+1)=0$$$$\Rightarrow k=\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C}$$$$$$$$\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=k(\frac{c}{a}+\frac{a}{b}+\frac{b}{c})⩾3k$$$$maximumif\frac{c}{a}=\frac{a}{b}=\frac{b}{c},i.e.a=b=c,$$$$i.e.A=B=C=\frac{\pi }{3}and$$$$k=\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C}=\frac{{\left(\frac{\sqrt{3}}{2}\right)}^3}{1+{\left(\frac{1}{2}\right)}^3}=\frac{\sqrt{3}}{3}$$$$\Rightarrow \frac{p}{a}+\frac{q}{b}+\frac{r}{c}⩾3k=3\times \frac{\sqrt{3}}{3}=\sqrt{3}✓$$

Commented by Shrinava last updated on 04/May/22

$$\mathrm{As}\mathrm{always},\mathrm{the}\mathrm{perfect}\mathrm{solution},$$$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{Professor}$$

Commented by Tawa11 last updated on 04/May/22

$$\mathrm{Great}\mathrm{sir}$$

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