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Question Number 169815 by mr W last updated on 09/May/22

Commented by mr W last updated on 10/May/22

$$canyoupleasegiveaexplanation?$$

Commented by mr W last updated on 09/May/22

$$Givenatriangle\mathrm{\Delta }ABCwithsides$$$$a,b,c.$$$$Findthemaximumareaofthe$$$$inscribedtrianglewhosesidesare$$$$perpendiculartothesidesof\mathrm{\Delta }ABC$$$$asshown.$$

Commented by Rasheed.Sindhi last updated on 10/May/22

$$Ithinkatmostonlyonesuchtraingle$$$$ispossible.$$

Commented by mr W last updated on 10/May/22

$$youarerightsir.$$

Commented by Rasheed.Sindhi last updated on 10/May/22

$$\mathit{s}\mathit{i}\mathit{r},Icansayonlyintuitively.All$$$$suchtrianglesshouldbe$$$$paralleltoeachother.Imeantheir$$$$sidesmustbeparallel.Isawonly$$$$onesuchtriangle\left(intuitively\right),maybe$$$$I^{\prime }mwrong!$$

Commented by Rasheed.Sindhi last updated on 10/May/22

$$\mathcal{TH}an\mathcal{X}\mathcal{S}ir!$$

Commented by mr W last updated on 10/May/22

Commented by mr W last updated on 10/May/22

$$ifsuchatriangleexists,saythered$$$$one,thenitexistsonlyonetime.$$$$wecangettwosides\left(blue\right)parallel$$$$tothesidesofredtriangle,butthe$$$$thirdside\left(green\right)isnolonger$$$$paralleltothecorrespondingred$$$$triangleside.i.e.theredtriangle$$$$isunique.$$

Answered by mr W last updated on 10/May/22

Commented by mr W last updated on 10/May/22

$$wecanseethat$$$$\mathrm{\Delta }QRP\sim \mathrm{\Delta }ABC$$$$\Rightarrow \frac{p}{c}=\frac{q}{a}=\frac{r}{b}=k,say$$$$a=\frac{q}{\tan B}+\frac{r}{\sin C}$$$$a=\frac{ka}{\tan B}+\frac{kb}{\sin C}$$$$a(\frac{1}{k}-\frac{1}{\tan B})=\frac{b}{\sin C}$$$$\Rightarrow \frac{a(\sin B-k\cos B)}{k\sin B}=\frac{b}{\sin C}...\left(i\right)$$$$similarly$$$$\Rightarrow \frac{b(\sin C-k\cos C)}{k\sin C}=\frac{c}{\sin A}...\left(ii\right)$$$$\Rightarrow \frac{c(\sin A-k\cos A)}{k\sin A}=\frac{a}{\sin B}...\left(iii\right)$$$$\left(i\right)\times \left(ii\right)\times \left(iii\right):$$$$\frac{(\sin A-k\cos A)(\sin B-k\cos B)(\sin C-k\cos C)}{k^3}=1$$$$(1+\cos A\cos B\cos C)k^3-\left(\sin A\sin B\sin C\right)k^2+(1+\cos A\cos B\cos C)k-\sin A\sin B\sin C=0$$$$(k-\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C})(k^2+1)=0$$$$\Rightarrow k=\frac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C}$$$$\Rightarrow k=\frac{2\sin A\sin B\sin C}{{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C}$$$$\sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}$$$$\Rightarrow \sin A\sin B\sin C=\frac{abc}{8R^3}$$$$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{a^2+b^2+c^2}{4R^2}$$$$k=\frac{abc}{8R^3}\times \frac{8R^2}{a^2+b^2+c^2}$$$$k=\frac{abc}{(a^2+b^2+c^2)R}$$$$\Rightarrow k=\frac{4{\mathrm{\Delta }}_{ABC}}{a^2+b^2+c^2}$$$${\mathrm{\Delta }}_{PQR}=k^2{\mathrm{\Delta }}_{ABC}=\frac{16{\mathrm{\Delta }}_{ABC}^3}{{(a^2+b^2+c^2)}^2}✓$$

Commented by Tawa11 last updated on 08/Oct/22

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

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