Final Round

Let $a(t)$, $b(t)$, $c(t)$ be the abscissae of the first, the second and the third snails at the moment $t$, respectively, and $u_a,u_b,u_c$ be their speeds respectively. First, suppose that all snails move in the same direction. Without loss of generality we assume that the snails move along the positive part of the axis $Ox$ and $u_a\le u_b\le u_c$. Then (see Fig. 1) $$S(ABC)=S(AC{C}_1{A}_1)-S(AB{B}_1{A}_1)-S(BC{C}_1{B}_1)(1)$$ Fig. 1 Fig. 2 Since the quadrilaterals $AC{C}_1{A}_1$, $AB{B}_1{A}_1$, $BC{C}_1{B}_1$ are trapezia and we know the coordinates of their vertices $A_1(a;0)$, $B_1(b;0)$, $C_1(c;0)$ and $A\left(a;\frac{1}{a}\right)$, $B\left(b;\frac{1}{b}\right)$, $C\left(c;\frac{1}{c}\right)$, we can find their areas: $$\begin{array}{rl}S(AC{C}_1{A}_1)& =\frac{1}{2}{A}_1{C}_1\cdot (A{A}_1+C{C}_1)=\frac{1}{2}(c-a)\left(\frac{1}{a}+\frac{1}{c}\right)=\frac{1}{2}\frac{c^2-a^2}{ac}=\\ & =\frac{1}{2}\left(\frac{c}{a}-\frac{a}{c}\right)=\frac{1}{2}\left(\frac{u_{c}t}{u_{a}t}-\frac{u_{a}t}{u_{c}t}\right)=\frac{1}{2}\left(\frac{u_c}{u_a}-\frac{u_a}{u_c}\right).\end{array}$$ Since the speeds of the snails are constant, we see that the area $S(AC{C}_1{A}_1)$ is independent of time. Similarly, we conclude that the areas of the trapezia $AB{B}_1{A}_1$ and $BC{C}_1{B}_1$ are independent of time, too.





Now we consider the case when two snails (say, the second and the third snails) move in the same direction and the first snail moves in the opposite direction. Without loss of generality we assume that the first snail moves along the negative part of the axis $Ox$ (see Fig. 2) and $u_b\le u_c$. Then $$S(ABC)=S(AB{B}_1)+S(BC{C}_1{B}_1)-S(AC{C}_1).$$ We have $$\begin{array}{rl}S(AB{B}_1)& =\frac{1}{2}{B}_{1}A\cdot B{B}_1=\frac{1}{2}(b-a)\left(\frac{1}{b}-\frac{1}{a}\right)=\\ & =\frac{1}{2}\left(2-\frac{b}{a}-\frac{a}{b}\right)=\frac{1}{2}\left(2-\frac{u_{b}t}{-u_{a}t}-\frac{-u_{a}t}{u_{b}t}\right)=1+\frac{1}{2}\left(\frac{u_a}{u_b}+\frac{u_b}{u_a}\right).\end{array}$$ Since the speeds of the snails are constant we see that the areas of the triangles $AB{B}_1$ and $AC{C}_1$ are independent of time. It remains to show that the area of the trapezium $BC{C}_1{B}_1$ is also independent of time. Indeed, $$\begin{array}{rl}S(BC{C}_1{B}_1)& =\frac{1}{2}{C}_1{B}_1\cdot (C{C}_1+B{B}_1)=\frac{1}{2}(c-b)\left(\frac{1}{c}-\frac{1}{a}+\frac{1}{b}-\frac{1}{a}\right)=\\ & =\frac{1}{2}\left(\frac{c}{b}-\frac{b}{c}-2\frac{c}{a}+2\frac{b}{a}\right)=\frac{1}{2}\left(\frac{u_c}{u_b}-\frac{u_b}{u_c}+2\frac{u_c}{u_a}-2\frac{u_b}{u_a}\right).\end{array}$$ Therefore, the area of the triangle ABC is independent of time.