Belarusian Mathematical Olympiad

Answer: $S=\frac{(m+n{)}^2}{\sqrt{m-n}}$.

Let $x_R,y_R$ denote the coordinates of point $R$. Let $y=kx+a$ and $y=kx+b$ be the equations of the lines $AD$ and $BC$, respectively. Then $$\{\begin{array}{l}{x}_A^2=y_A=k{x}_A+a,\\ x_D^2=y_D=k{x}_D+a,\end{array}\{\begin{array}{l}{x}_B^2=y_B=k{x}_B+b,\\ x_C^2=y_C=k{x}_C+b.\end{array}$$ So,



$$k=x_A+x_D=x_B+x_C,(1)$$ $$a=-x_A{x}_D,b=-x_B{x}_C.(2)$$ Since $K$ and $L$ are the midpoints of the sides $AD$ and $BC$ respectively, we see that $$x_K=0.5(x_A+x_D),x_L=0.5(x_B+x_C).(3)$$ It is well-known fact that point $M$ of intersection of the diagonals of the trapezoid $ABCD$ lies on the segment joining the midpoints of the trapezoid bases. Taking into account (3) we see that the segment $KL$ containing $M$ is perpendicular to the axis $Ox$, and $x_M=x_K=x_L$. Let $\alpha$ be the angle between the lines $AD$, $BC$ and the positive direction of $Ox$, then $\tan \alpha =k$. If $LN$ is an altitude of the trapezoid $ABCD$, then $\angle NLK=\alpha$ ($LN\perp AD,KL\perp Ox$). Therefore $LN=KL\cos \alpha$. On the other hand, $AD=(y_D-y_A)/\cos \alpha$, $BC=(y_C-y_B)/\cos \alpha$. Therefore, the required area is equal to $$\begin{array}{rl}S& =\frac{1}{2}(AD+BC)LN=\frac{1}{2}(y_D-y_A+y_C-y_B)KL\cdot \frac{\cos \alpha }{\sin \alpha }=[KL=m+n]=\\ & =\frac{1}{2}(y_D-y_A+y_C-y_B)\cdot \frac{m+n}{k}=\frac{1}{2}\left(\frac{y_D-y_A}{k}+\frac{y_C-y_B}{k}\right)(m+n)\stackrel{(1)}{=}\\ & =\frac{1}{2}(x_D-x_A+x_C-x_B)(m+n)\end{array}(4)$$ Let $y=k_{1}x+c$ be the equation of the line $BD$. Then $$\{\begin{array}{l}{x}_B^2=y_B=k_1{x}_B+c,\\ x_D^2=y_D=k_1{x}_D+c,\end{array}\Rightarrow k_1=x_B+x_D,c=-x_B{x}_D.(5)$$ So, $y_M=0.5k_1(x_A+x_D)+c=0.5k_1(x_B+x_C)+c$. Therefore, from (1) - (3), and (5) it follows that $$\begin{array}{rl}m& =y_K-y_M=\frac{1}{2}(x_A+x_D)(x_A+x_D)-x_A{x}_D-\frac{1}{2}(x_B+x_D)(x_A+x_D)+x_B{x}_D=\\ & =\frac{1}{2}(x_A+x_D)(x_A-x_B)-x_D(x_A-x_B)=\frac{1}{2}(x_A-x_B)(x_A-x_D).\end{array}$$ $$\begin{array}{rl}n& =y_M-y_L\stackrel{(1)}{=}\frac{1}{2}(x_B+x_D)(x_A+x_D)-x_B{x}_D-\frac{1}{2}(x_A+x_D)(x_A+x_D)+x_B{x}_C=\\ & =\frac{1}{2}(x_A+x_D)(x_B-x_A)-x_B(x_D-x_C)=[x_D-x_C\stackrel{(1)}{=}{x}_B-x_A]=\\ & =\frac{1}{2}(x_B-x_A)(x_A+x_D-2x_B)\stackrel{(1)}{=}\frac{1}{2}(x_B-x_A)(x_C-x_B).\end{array}$$ Then $m+n=\frac{1}{2}(x_B-x_A)(x_C-x_B+x_D-x_A)$. Hence, (4) can be presented in the form $$S=\frac{1}{2}\cdot \frac{2(m+n)}{x_B-x_A}(m+n)=\frac{(m+n{)}^2}{x_B-x_A}.$$ Note that $$m-n=\frac{1}{2}(x_B-x_A)(x_D-x_A-x_C+x_B)=[x_D-x_C\stackrel{(1)}{=}{x}_B-x_A]=(x_B-x_A{)}^2.$$

Therefore, finally, we have $S=\frac{(m+n{)}^2}{x_B-x_A}=\frac{(m+n{)}^2}{\sqrt{m-n}}$.