China Western Mathematical Olympiad

Let $d(P,l)$ denote the distance from point $P$ to line $l$. We first prove the following lemma.

Lemma Let $\angle SAT=\alpha$ be a given angle and $P$ a moving point in $\angle SAT$. If the sum of distances from $P$ to lines $AS$ and $AT$ is a constant number $m$, then the trace of $P$ is a segment of $BC$ with points $B$ and $C$ on $AS$ and $AT$ respectively, and $AB=AC=\frac{m}{\sin \alpha }$. Further, if a point $Q$ lies inside $△ABC$, then the sum of distances from $Q$ to lines $AS$ and $AT$ is less than $m$, and it is greater than $m$ if $Q$ lies outside $△ABC$.



Proof of Lemma For $AB=AC=\frac{m}{\sin \alpha }$ and $P$ on $BC$, we have $S_{△PAB}+S_{△PAC}=S_{△ABC}$, i.e. $\frac{1}{2}AB\cdot d(P,AB)+\frac{1}{2}AC\cdot d(P,AC)=\frac{1}{2}\cdot AB\cdot AC\cdot \sin \alpha$. Then $d(P,AB)+d(P,AC)=m$. If point $Q$ lies inside $△ABC$, $S_{△QAB}+S_{△QAC}<S_{△ABC}$, then $d(Q,AB)+d(Q,AC)<m$. If $Q$ lies outside $△ABC$, $S_{△QAB}+S_{△QAC}>S_{△ABC}$, then $d(Q,AB)+d(Q,AC)>m$. The proof of Lemma is complete.

We now consider the following two cases:

(1) Neither pair of the opposite sides of the quadrilateral $ABCD$ is parallel. We may assume that sides $BC$ and $AD$ meet at point $F$ and sides $BA$ and $CD$ meet at point $E$. Through point $P$ draw segments $l_1$ and $l_2$ such that the sum of distances from any point on $l_1$ to $AB$ and $CD$ is constant and the sum of distances from any point on $l_2$ to $BC$ and $AD$ is also constant. As seen in the second figure, for any point $Q$ in area $S$, using the given condition and the lemma, we have

$$\begin{array}{rl}d(P,AB)+d(P,BC)+d(P,CD)+d(P,DA)& \\ & =d(Q,AB)+d(Q,BC)+d(Q,CD)+d(Q,DA)\\ & =[d(Q,AB)+d(Q,CD)]+[d(Q,BC)+d(Q,DA)]\\ & >[d(P,AB)+d(P,CD)]+[d(P,BC)+d(P,DA)].\end{array}$$

It leads to a contradiction.

(2) The quadrilateral $ABCD$ is a trapezoid. In the same way, we can also show that it will lead to a contradiction.

This completes the proof.