每日一题_190919

在\(\triangle ABC\)中,点\(D\)在线段\(BC\)上,且满足\(\overrightarrow{BD}=\dfrac{1}{2}\overrightarrow{DC}\),过点\(D\)的直线分别交直线\(AB\),\(AC\)于不同的两点\(M\),\(N\),\(\overrightarrow {AM}=m\overrightarrow{AB}\),\(\overrightarrow{AN}=n\overrightarrow{AC}\),则\((\qquad)\)

\(\mathrm{A.}\quad m+n=2\quad\) \(\mathrm{B.}\quad2m+n=3\quad\) \(\mathrm{C.}\quad\dfrac{1}{m}+\dfrac{1}{n}=2\quad\) \(\mathrm{D.}\quad\dfrac{2}{m}+\dfrac{1}{n}=3\)
解析:
法一 由题

\[\overrightarrow{AD}=\dfrac{2}{3}\cdot\overrightarrow{AB}+\dfrac{1}{3}\cdot \overrightarrow{AC}=\dfrac{2}{3m}\cdot \overrightarrow{AM}+\dfrac{1}{3n}\cdot\overrightarrow{AC}.$$由于$M,D,N$三点共线,故 \]

1=\dfrac{2}{3m}+\dfrac{1}{3n}.$$于是选项$\mathrm{D}$正确.
 法二 仅需考察直线$MN$与线段$AB$相交时的情形,如图,过点$D$作$DE\parallel CA$,

\(DE\)直线交\(AB\)于点\(E\),不妨设\(|AE|=2\),则一方面$$\overrightarrow{AM}=m\overrightarrow{AB},m\neq \dfrac{2}{3}\quad (\ast)$$另一方面,我们考虑用\(n\)将\(\overrightarrow{AM}\)与\(\overrightarrow{AB}\)联系起来.由于$$|DE|=\dfrac{1}{3}\cdot|AC|=\dfrac{1} {3n}\cdot|AN|.$$所以$$\dfrac{|ME|}{|AE|}=\dfrac{|DE|}{|AN|-|DE|}=\dfrac{1}{3n-1},n\neq \dfrac{1}{3}.$$进而$$\dfrac{|AM|}{|AB|}=\dfrac{|AE|+|ME|}{\frac{3}{2}\cdot |AE|}=\dfrac{n}{\frac{3n}{2}-\frac{1}{2}}.$$结合\((\ast)\)我们可得$$
\dfrac{n}{\frac{3n}{2}-\frac{1}{2}}=m\Longleftrightarrow \dfrac{2}{m}+\dfrac{1}{n}=3 .$$
所以\(\mathrm D\)选项正确.
法三 仅考察直线\(MN\)与线段\(AB\)交于点\(M\)的情形,

由梅涅劳斯定理我们有$$\dfrac{|AM|}{|MB|}\cdot \dfrac{|BD|}{|DC|}\cdot \dfrac{|CN|}{|NA|}=1.$$也即$$\dfrac{m}{1-m}\cdot \dfrac{1}{2}\cdot \dfrac{n-1}{n}=1.$$整理得$$
\dfrac 2m+\dfrac 1n=3.$$所以选项\(\mathrm D\)正确.
备注 此题亦可用极限思想分析,当\(M\)无限接近线段\(AB\)上靠近\(B\)点的三等分点时,此时\(m\to \dfrac{2}{3}\),\(n\to +\infty\).验证仅\(\mathrm D\)选项满足题意.