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Question Number 96441 by  M±th+et+s last updated on 01/Jun/20

Commented by  M±th+et+s last updated on 01/Jun/20

$$let{p}_1,p_2,p_3....p_{60}be60pointsonBC$$$$\underset{i=1}{\sum ^{60}}({AP}_i^2+P_{i}B\times P_{i}C)=?$$

Answered by 1549442205 last updated on 02/Jun/20

$$\mathrm{Applying}{\mathrm{Stewart}}^{\prime }\mathrm{s}\mathrm{theorem}\mathrm{we}\mathrm{have}:$$$${\mathrm{AB}}^2\times {\mathrm{P}}_{\mathrm{i}}\mathrm{C}+{\mathrm{AC}}^2\times {\mathrm{P}}_{\mathrm{i}}\mathrm{B}-{\mathrm{AP}}_{\mathrm{i}}^2\mathrm{BC}=$$$$\mathrm{BC}\times {\mathrm{P}}_{\mathrm{i}}\mathrm{B}\times {\mathrm{P}}_{\mathrm{i}}\mathrm{C}\Rightarrow 100\times \mathrm{BC}-{\mathrm{AP}}_{\mathrm{i}}^2\mathrm{BC}$$$$=\mathrm{BC}\times {\mathrm{P}}_{\mathrm{i}}\mathrm{B}\times {\mathrm{P}}_{\mathrm{i}}\mathrm{C}\Rightarrow {\mathrm{AP}}_{\mathrm{i}}^2+{\mathrm{P}}_{\mathrm{i}}\mathrm{B}\times {\mathrm{P}}_{\mathrm{i}}\mathrm{C}=100$$$$\mathrm{Hence},\underset{\mathrm{i}=1}{\stackrel{60}{\mathrm{\Sigma }}}({\mathrm{AP}}_{\mathrm{i}}^2+{\mathrm{P}}_{\mathrm{i}}\mathrm{B}\times {\mathrm{P}}_{\mathrm{i}}\mathrm{C})=60\times 100$$$$=6000$$

Commented by  M±th+et+s last updated on 02/Jun/20

$$thankyousir.$$$$correctsolution$$

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