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Question Number 189484 by mr W last updated on 17/Mar/23

Commented by mr W last updated on 17/Mar/23

$$anunsolvedoldquestion$$

Answered by mr W last updated on 17/Mar/23

Commented by mr W last updated on 17/Mar/23

$$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$$$$\overrightarrow{AY}=\overrightarrow{a}+\frac{3}{4}\overrightarrow{b}$$$$\overrightarrow{BZ}=-\overrightarrow{a}-\frac{1}{4}\overrightarrow{c}$$$$\overrightarrow{CX}=\frac{3}{4}\overrightarrow{a}+\overrightarrow{c}$$$$sayAP=\xi \times AY,BP=\eta \times BZ$$$$\xi \overrightarrow{AY}=\overrightarrow{a}+\eta \overrightarrow{BZ}$$$$\xi (\overrightarrow{a}+\frac{3}{4}\overrightarrow{b})=\overrightarrow{a}+\eta (-\overrightarrow{a}-\frac{1}{4}\overrightarrow{c})$$$$(1-\frac{\xi }{4}-\eta )\overrightarrow{a}+(\frac{3\xi }{4}-\frac{\eta }{4})\overrightarrow{c}=0$$$$\Rightarrow \eta =3\xi$$$$\Rightarrow 1-\frac{\xi }{4}-3\xi =0$$$$\Rightarrow \xi =\frac{4}{13},\eta =\frac{12}{13}$$$$\Rightarrow AP:PR:RY=4:8:1$$$$\left[ABC\right]=\mathrm{\Delta }=52$$$$\left[AYC\right]=\frac{\mathrm{\Delta }}{4}$$$$\left[APZ\right]=\frac{4}{13}\times \frac{1}{4}\times \left[AYC\right]=\frac{\mathrm{\Delta }}{52}$$$$\left[CRY\right]=\frac{\left[AYC\right]}{13}=\frac{\mathrm{\Delta }}{52}$$$$\left[PZCR\right]=\frac{\mathrm{\Delta }}{4}-2\times \frac{\mathrm{\Delta }}{52}=\frac{11\mathrm{\Delta }}{52}$$$$\left[PQR\right]=\mathrm{\Delta }-3\times \frac{\mathrm{\Delta }}{52}-3\times \frac{11\mathrm{\Delta }}{52}=\frac{4}{13}\left[ABC\right]$$$$similary$$$$\left[APQX\right]=\left[BYRQ\right]=\frac{11\mathrm{\Delta }}{52}$$$$\Rightarrow greenarea=\mathrm{\Delta }-3\times \frac{11\mathrm{\Delta }}{52}=\frac{19\mathrm{\Delta }}{52}=19✓$$$$$$$$\left[PQR\right]=\mathrm{\Delta }-3\times \frac{\mathrm{\Delta }}{52}-3\times \frac{11\mathrm{\Delta }}{52}=\frac{4}{13}\left[ABC\right]$$

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