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Question Number 206704 by mr W last updated on 22/Apr/24

Commented by mr W last updated on 22/Apr/24

$$find\frac{{\mathrm{\Delta }}_{XYZ}}{{\mathrm{\Delta }}_{ABC}}=?$$

Answered by A5T last updated on 22/Apr/24

$$BD=x;DC=y$$$$\frac{AE}{EC}\times \frac{CZ}{ZF}\times \frac{FB}{BA}=1\Rightarrow \frac{AE}{EC}=\frac{BA}{FB}=\frac{x+2y}{x+y}$$$$\frac{CD}{DB}\times \frac{BY}{YE}\times \frac{EA}{AC}=1\Rightarrow \frac{y}{x}=\frac{CD}{DB}=\frac{AC}{EA}$$$$\frac{BF}{FA}\times \frac{AX}{XD}\times \frac{DC}{CB}=1\Rightarrow \frac{BF}{FA}=\frac{BC}{DC}=\frac{x+y}{y}$$$$\Rightarrow \frac{AE}{EC}\times \frac{CD}{DB}\times \frac{BF}{FA}=\frac{x+2y}{x}$$$$\Rightarrow \frac{\left[XYZ\right]}{\left[ABC\right]}=\frac{{(1-\frac{x+2y}{x})}^2}{(1+\frac{x+2y}{x+y}+\frac{y(x+2y)}{x(x+y)})(1+\frac{y}{x}+\frac{x+y}{x})(1+\frac{x+y}{y}+\frac{x+2y}{y})}$$$$=\frac{y^3}{2(x+2y){(x+y)}^2}$$

Commented by A5T last updated on 23/Apr/24

$$\frac{AC}{EA}=\frac{y}{x}\wedge \frac{AC}{EA}=\frac{2x+3y}{x+2y}$$$$\Rightarrow \frac{y}{x}=\frac{2x+3y}{x+2y}\Rightarrow y=\frac{x(\sqrt{5}+1)}{2}$$$$\Rightarrow \frac{\left[XYZ\right]}{\left[ABC\right]}=\frac{7-3\sqrt{5}}{4}\approx 0.07295$$

Commented by mr W last updated on 23/Apr/24

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Answered by mr W last updated on 23/Apr/24

Commented by mr W last updated on 23/Apr/24

$$\frac{EY}{YB}\times \frac{a}{b}\times \frac{c+d}{d}=1$$$$\Rightarrow \frac{c}{d}=\frac{b}{a}-1=\lambda -1with\lambda =\frac{b}{a}$$$$\frac{FZ}{ZC}\times \frac{c}{d}\times \frac{p+q}{q}=1$$$$\Rightarrow (\lambda -1)(1+\frac{p}{q})=1\Rightarrow \frac{p}{q}=\frac{1}{\lambda -1}-1$$$$\frac{DX}{XA}\times \frac{p}{q}\times \frac{a+b}{b}=1$$$$\Rightarrow (\frac{1}{\lambda -1}-1)(1+\frac{1}{\lambda })=1$$$$\Rightarrow {\lambda }^2-\lambda -1=0$$$$\Rightarrow \lambda =\frac{1+\sqrt{5}}{2}=\phi \left(goldenratio\right)$$$$(similarly\frac{d}{c}=\frac{q}{p}=\lambda )$$$$let\mu =\frac{ZY}{YB}=\frac{XZ}{ZC}=\frac{YX}{XA}$$$$\frac{ZY}{YB}\times \frac{a}{b}\times \frac{CX}{XZ}=1$$$$\Rightarrow \mu \times \frac{1}{\lambda }\times (\frac{1}{\mu }+1)=1$$$$\Rightarrow \mu =\lambda -1=\frac{\sqrt{5}-1}{2}=\frac{1}{\lambda }$$$$\left[\mathrm{\Delta }ADC\right]=\frac{b}{a+b}\left[\mathrm{\Delta }ABC\right]=\frac{\lambda }{1+\lambda }\left[\mathrm{\Delta }ABC\right]$$$$\left[\mathrm{\Delta }CXD\right]=\frac{\left[\mathrm{\Delta }ADC\right]}{2}=\frac{\lambda }{2(1+\lambda )}\left[\mathrm{\Delta }ABC\right]$$$$\left[\mathrm{\Delta }CYX\right]=\frac{XY}{YD}\left[\mathrm{\Delta }CDX\right]=\mu \left[\mathrm{\Delta }CDX\right]=\frac{\mu \lambda }{2(1+\lambda )}\left[\mathrm{\Delta }ABC\right]$$$$\left[\mathrm{\Delta }XYZ\right]=\frac{XZ}{XC}\left[\mathrm{\Delta }CYX\right]=\frac{\mu }{1+\mu }\left[\mathrm{\Delta }CYX\right]=\frac{{\mu }^2\lambda }{2(1+\mu )(1+\lambda )}\left[\mathrm{\Delta }ABC\right]$$$$\Rightarrow \frac{\left[\mathrm{\Delta }XYZ\right]}{\left[\mathrm{\Delta }ABC\right]}=\frac{{\mu }^2\lambda }{2(1+\mu )(1+\lambda )}=\frac{1}{2{\lambda }^4}$$$$=\frac{7-3\sqrt{5}}{4}\approx 0.07295$$

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