\(3\overrightarrow{MA}+2\overrightarrow{MB}=\overrightarrow{0}\) suy ra \(\dfrac{MB}{BA}=\dfrac{3}{5}\).
\(CPMQ\) là hình bình hành suy ra \(MP\) song song với \(CQ\), \(MQ\) song song với \(PC\).
Xét tam giác \(ABC\) có \(MP\) song song với \(BC\) suy ra \(\dfrac{AP}{AC}=\dfrac{AM}{AB}=\dfrac{2}{5}\).
Suy ra \(\overrightarrow{AP}=\dfrac{2}{5}\overrightarrow{AC}\).
Tương tự ta cũng suy ra được \(\overrightarrow{BQ}=\dfrac{3}{5}\overrightarrow{BC}\).
\(a\overrightarrow{NA}+b\overrightarrow{NQ}=\overrightarrow{0}\Leftrightarrow a\overrightarrow{AN}=b\overrightarrow{NQ}\) suy ra \(\overrightarrow{AQ}=\overrightarrow{AN}+\overrightarrow{NQ}=\overrightarrow{AN}+\dfrac{a}{b}\overrightarrow{AN}=\dfrac{a+b}{b}\overrightarrow{AN}\)
do đó \(\overrightarrow{AN}=\dfrac{b}{a+b}\overrightarrow{AQ}\).
Để \(B,N,P\) thẳng hàng thì \(\overrightarrow{BN}\) và \(\overrightarrow{BP}\) cùng phương. Ta sẽ phân tích \(\overrightarrow{BN},\overrightarrow{BP}\) theo hai vectơ \(\overrightarrow{BA}\) và \(\overrightarrow{BC}\).
\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=\overrightarrow{BA}+\dfrac{b}{a+b}\overrightarrow{AQ}=\overrightarrow{BA}+\dfrac{b}{a+b}\left(\overrightarrow{AB}+\overrightarrow{BQ}\right)\)
\(=\overrightarrow{BA}-\dfrac{b}{a+b}\overrightarrow{BA}+\dfrac{3b}{5\left(a+b\right)}\overrightarrow{BC}=\dfrac{a}{a+b}\overrightarrow{BA}+\dfrac{3b}{5\left(a+b\right)}\overrightarrow{BC}\).
\(\overrightarrow{BP}=\overrightarrow{BA}+\overrightarrow{AP}=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}=\overrightarrow{BA}+\dfrac{2}{5}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=\dfrac{3}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\).
\(\overrightarrow{BN},\overrightarrow{BP}\) cùng phương nên
\(\dfrac{\dfrac{a}{a+b}}{\dfrac{3}{5}}=\dfrac{\dfrac{3b}{5\left(a+b\right)}}{\dfrac{2}{5}}\Leftrightarrow\dfrac{2a}{5\left(a+b\right)}=\dfrac{9b}{25\left(a+b\right)}\Leftrightarrow10a=9b\).
Vì \(\left(a,b\right)=1\), \(a,b\) nguyên nên \(a=9,b=10\).
Vậy \(a+b=19\).
