Belorusija 2012

b) Answer: $S=27/4$. We find the coordinates $(x_T;y_T)$ of the point of intersection of the parabola $y=a{x}^2$ and the hyperbola $y=1/x$. We have $a{x}_T^2=1/x_T$, so $x_T=1/\sqrt{a}=1/t$ and $y_T=1/x_T=t$. The equation of the tangent to the parabola at point $P(x_P;y_P)$ has the form $y-y_P=2a{x}_P(x-x_P)$, and the equation of the tangent to the hyperbola at point $Q(x_Q;y_Q)$ has the form $y-y_Q=(-1/x_Q^2)(x-x_Q)$. Since the tangent is a common tangent for both the graphs, we have $$2a{x}_P=-1/x_Q^2,(1)$$ $$-2a{x}_P^2+y_P=1/x_Q+y_Q.(2)$$ Since $y_P=a{x}_P^2$ and $y_Q=1/x_Q$, the obtained equality has the form $$-a{x}_P^2=2/x_Q.(2)$$ From equations (1), (2) it follows that $x_Q=-1/(2t)$, $x_P=-2/t$, and so $y_Q=-2t$, $y_P=4t$. Therefore, $$T(\frac{1}{t};t),P(-\frac{2}{t};4t),Q(-\frac{1}{2t};-2t).(3)$$ Let $M,K,L$ be the midpoints of the sides $PQ,PT,TQ$ respectively. Then $$M(-\frac{5}{4t};t),K(-\frac{1}{2t};\frac{5}{2}t);L(\frac{1}{4t};-\frac{t}{2}).$$ Comparing the obtained values with (3), we note that $MT$ parallel to the axis $Ox$, since $y_M=y_T$, and $QK$ parallel to $Oy$, since $y_Q=y_K$. Therefore, medians $MT$ and $QK$ in the triangle $PQT$ are perpendicular, as required. Let $H$ be a centroid of the triangle $PQT$. Then $$\begin{gathered} S(PQT)=2S(QKT)=[TH\perp QK]=2\cdot \frac{1}{2}QK\cdot TH= \\ =[TH=\frac{2}{3}MT]=\frac{2}{3}QK\cdot TM. \end{gathered}$$ Since $TM\parallel Ox$, we have $TM=1/t-(-5/(4t))=9/(4t)$, Similarly, since $QK\parallel Oy$, we have $QK=5t/2-(-2t)=9t/2$. Therefore, $$S(PQT)=\frac{2}{3}\cdot \frac{9}{4t}\cdot \frac{9t}{2}=\frac{27}{4}.$$