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
$${let}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} ....{p}_{\mathrm{60}} \:{be}\:\mathrm{60}\:{points}\:{on}\:{BC} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{60}} {\sum}}\left({AP}_{{i}} ^{\:\mathrm{2}} +{P}_{{i}} {B}×{P}_{{i}} {C}\right)=? \\ $$
Answered by 1549442205 last updated on 02/Jun/20
$$\mathrm{Applying}\:\mathrm{Stewart}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{AB}^{\mathrm{2}} ×\mathrm{P}_{\mathrm{i}} \mathrm{C}+\mathrm{AC}^{\mathrm{2}} ×\mathrm{P}_{\mathrm{i}} \mathrm{B}−\mathrm{AP}_{\mathrm{i}} ^{\mathrm{2}} \mathrm{BC}= \\ $$$$\mathrm{BC}×\mathrm{P}_{\mathrm{i}} \mathrm{B}×\mathrm{P}_{\mathrm{i}} \mathrm{C}\Rightarrow\mathrm{100}×\mathrm{BC}−\mathrm{AP}_{\mathrm{i}} ^{\mathrm{2}} \mathrm{BC} \\ $$$$=\mathrm{BC}×\mathrm{P}_{\mathrm{i}} \mathrm{B}×\mathrm{P}_{\mathrm{i}} \mathrm{C}\Rightarrow\mathrm{AP}_{\mathrm{i}} ^{\mathrm{2}} +\mathrm{P}_{\mathrm{i}} \mathrm{B}×\mathrm{P}_{\mathrm{i}} \mathrm{C}=\mathrm{100} \\ $$$$\mathrm{Hence},\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{60}} {\Sigma}}\left(\mathrm{AP}_{\mathrm{i}} ^{\mathrm{2}} +\mathrm{P}_{\mathrm{i}} \mathrm{B}×\mathrm{P}_{\mathrm{i}} \mathrm{C}\right)=\mathrm{60}×\mathrm{100} \\ $$$$=\mathrm{6000} \\ $$
Commented by M±th+et+s last updated on 02/Jun/20
$${thank}\:{you}\:{sir}\:. \\ $$$${correct}\:{solution}\: \\ $$