高考试题

\begin{Example}
(2010年广东)设$A(x_1,y_1),B(x_2,y_2)$是平面直角坐标系$xOy$上的两点,现定义由点$A$到点$B$的一种折线距离$\rho (A,B)$为$\rho (A,B)=|x_2-x_1|+|y_2-y_1|$.对于平面$xOy$上给定的不同的两点$A(x_1,y_1),B(x_2,y_2)$.
\begin{enumerate}
\item[(I)] 若点$C(x,y)$是平面$xOy$上的点,试证明: $\rho (A,C)+\rho(C,B)\geqslant \rho(A,B)$;

\item[(II)] 在平面$xOy$上是否存在点$C(x,y)$同时满足

\ding{172} $\rho (A,C)+\rho(C,B)\geq \rho(A,B)$; \qquad \ding{173} $\rho (A,C)=\rho(C,B)$.

若存在,请求出所有符合条件的点;若不存在,请予以证明.
\end{enumerate}
\end{Example}
\begin{proof}
(I)证明:因为$\rho (A,C)=|x-x_1|+|y-y_1|$, $\rho (C,B)=|x_2-x|+|y_2-y|$, $\rho (A,B)=|x_2-x_1|+|y_2-y_1|$,所以
\begin{align*}
\rho \left( A,C \right) +\rho \left( C,B \right) &=\left| x-x_1 \right|+\left| y-y_1 \right|+\left| x_2-x \right|+\left| y_2-y \right|
\\
&=\left( \left| x-x_1 \right|+\left| x_2-x \right| \right) +\left( \left| y-y_1 \right|+\left| y_2-y \right| \right)
\\
&\geqslant \left| \left( x-x_1 \right) +\left( x_2-x \right) \right|+\left| \left( y-y_1 \right) +\left( y_2-y \right) \right|
\\
&=\left| x_2-x_1 \right|+\left| y_2-y_1 \right|=\rho \left( A,B \right).
\end{align*}
(II)解:注意到点$A(x_1,y_1)$与点$B(x_2,y_2)$不同,下面分三种情形讨论.

(i)若$x_1=x_2$,则$y_1\neq y_2$,由条件\ding{173}得
$$
\left| x-x_1 \right|+\left| y-y_1 \right|=\left| x_2-x \right|+\left| y_2-y \right|,
$$
即$\left| y-y_1 \right|=\left| y_2-y \right|$,所以$y=\frac{y_1+y_2}{2}$.

由条件\ding{172}得
$$
\left| x-x_1 \right|+\left| y-y_1 \right|+\left| x_2-x \right|+\left| y_2-y \right|=\left| x_2-x_1 \right|+\left| y_2-y_1 \right|.
$$
所以
$$
2\left| x-x_1 \right|+\frac{1}{2}\left| y_2-y_1 \right|+\frac{1}{2}\left| y_2-y_1 \right|=\left| y_2-y_1 \right|,
$$
所以$\left| x-x_1 \right|=0$,所以$x=x_1$.

因此,所求的点$C$为$\left( x_1,\frac{y_1+y_2}{2}\right)$.

(ii)若$y_1=y_2$,则$x_1\neq x_2$,类似于(i),可得符合条件的点$C$为$\left(\frac{x_1+x_2}{2},y_1\right)$.

(iii)当$x_1\neq x_2$且$y_1\neq y_2$时,不妨设$x_1<x_2$.

(1)若$y_1<y_2$,则由(I)中的证明知,要使条件\ding{172}成立,当且仅当$(x-x_1)(x_2-x)\geqslant 0$与$(y-y_1)(y_2-y)\geqslant 0$同时成立,故$x_1\leqslant x\leqslant x_2$且$y_1\leqslant y\leqslant y_2$.

从而由条件\ding{173},得$x+y=\frac{1}{2}(x_1+x_2+y_1+y_2)$.

此时所求点$C$的全体为
$$
M=\left\{ \left( x,y \right) \left| x+y=\frac{1}{2}\left( x_1+x_2+y_1+y_2 \right) ,x_1\leqslant x\leqslant x_2\,\mathrm{且}\, y_1\leqslant y\leqslant y_2 \right. \right\}.
$$

(2)若$y_1>y_2$,类似地由条件\ding{172}可得$x_1\leqslant x\leqslant x_2$且$y_2\leqslant y\leqslant y_1$.

从而由条件\ding{173}得$x-y=\frac{1}{2}(x_1+x_2-y_1-y_2)$.

此时所求点的全体为
$$
N=\left\{ \left( x,y \right) \left| x-y=\frac{1}{2}\left( x_1+x_2-y_1-y_2 \right) ,x_1\leqslant x\leqslant x_2\,\mathrm{且}\, y_2\leqslant y\leqslant y_1 \right. \right\}.
$$
\end{proof}