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Question Number 197834 by ajfour last updated on 30/Sep/23

Answered by mr W last updated on 01/Oct/23

Commented by mr W last updated on 03/Oct/23

$$let\mathrm{\Delta }=areaof\mathrm{\Delta }ABC\left(anytypetriangle\right)$$$$\frac{HD}{EC}=\frac{p}{1}$$$$\Rightarrow HD=pqb$$$$\frac{GA}{DG}=\frac{AE}{HD}=\frac{(1-q)b}{pqb}=\frac{1}{pq}-\frac{1}{p}$$$$\frac{DG}{DA}=\frac{DG}{DG+GA}=\frac{1}{1+\frac{1}{pq}-\frac{1}{p}}$$$$\frac{\mathrm{\Delta }BFG}{\mathrm{\Delta }BDA}=\frac{DG}{DA}=\frac{1}{1+\frac{1}{pq}-\frac{1}{p}}$$$$\frac{\mathrm{\Delta }BDA}{\mathrm{\Delta }}=\frac{p}{1}$$$$\Rightarrow \mathrm{\Delta }BDA=p\mathrm{\Delta }$$$$\Rightarrow \mathrm{\Delta }BFG=\frac{p\mathrm{\Delta }}{1+\frac{1}{pq}-\frac{1}{p}}$$$$similarly$$$$\Rightarrow \mathrm{\Delta }CEK=\frac{q\mathrm{\Delta }}{1+\frac{1}{qr}-\frac{1}{q}}$$$$\Rightarrow \mathrm{\Delta }AFL=\frac{r\mathrm{\Delta }}{1+\frac{1}{rp}-\frac{1}{r}}$$$$bluearea+p\mathrm{\Delta }+q\mathrm{\Delta }+r\mathrm{\Delta }-2(\frac{p\mathrm{\Delta }}{1+\frac{1}{pq}-\frac{1}{p}}+\frac{q\mathrm{\Delta }}{1+\frac{1}{qr}-\frac{1}{q}}+\frac{r\mathrm{\Delta }}{1+\frac{1}{rp}-\frac{1}{r}})=\mathrm{\Delta }$$$$\Rightarrow \frac{bluearea}{\mathrm{\Delta }}=1+2(\frac{p}{1+\frac{1}{pq}-\frac{1}{p}}+\frac{q}{1+\frac{1}{qr}-\frac{1}{q}}+\frac{r}{1+\frac{1}{rp}-\frac{1}{r}})-(p+q+r)$$

Answered by ajfour last updated on 01/Oct/23

Commented by ajfour last updated on 01/Oct/23

$$let△ABChaseachside=2$$$$withAoriginO.$$$$B(\sqrt{3},-1)C(\sqrt{3},1)$$$$linePQ$$$$r_1=\lambda [\sqrt{3}-i(1-q)]$$$$lineQR$$$$r_2=\sqrt{3}-i+\mu [-\frac{r}{\sqrt{3}}+i(2-\frac{r}{2})]$$$$lineRP$$$$r_3=\sqrt{3}+i+ϵ[-(\sqrt{3}-\frac{p}{\sqrt{3}})-i(1+\frac{p}{2})]$$$$Plieson{r}_1\mathrm{\&}{r}_3,hence$$$$\sqrt{3}\lambda =\sqrt{3}-ϵ(\sqrt{3}-\frac{p}{\sqrt{3}})$$$$similarly$$$$\lambda (1-q)=-1+ϵ(1+\frac{p}{2})$$$$\Rightarrow 1-ϵ(1-\frac{p}{3})(1-q)=-1+ϵ(1+\frac{p}{2})$$$$ϵ[(1+\frac{p}{2})+(1-q)(1-\frac{p}{3})=2$$$$\lambda =1-\frac{2(1-\frac{p}{3})}{(1+\frac{p}{2})+(1-q)(1-\frac{p}{3})}$$$$area△PQR=\frac{1}{2}PQ\times PR\left(vector\right)$$$$...prettytediousthisway$$$$Thankyousir.$$$$$$$$$$

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