IMC 2000 · Day 2 · P10

a) Without loss of generality, we can assume that the point $A_2$ is the origin of system of coordinates. Then the polynomial can be presented in the form $$y = \bigl( a_0 x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4 \bigr) x^2 + a_5 x,$$ where the equation $y = a_5 x$ determines the straight line $A_1 A_3$. The abscissas of the points $A_1$ and $A_3$ are $-a$ and $a$, $a > 0$, respectively. Since $-a$ and $a$ are points of tangency, the numbers $-a$ and $a$ must be double roots of the polynomial $a_0 x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4$. It follows that the polynomial is of the form $$y = a_0 \bigl( x^2 - a^2 \bigr)^2 + a_5 x.$$ The equality follows from the equality of the integrals $$\int_{-a}^{0} a_0 \bigl( x^2 - a^2 \bigr) x^2\,dx = \int_{0}^{a} a_0 \bigl( x^2 - a^2 \bigr) x^2\,dx$$ due to the fact that the function $y = a_0 (x^2 - a^2)$ is even.

b) Without loss of generality, we can assume that $a_0 = 1$. Then the function is of the form $$y = (x + a)^2 (x - b)^2 x^2 + a_5 x,$$ where $a$ and $b$ are positive numbers and $b = ka$, $0 < k < \infty$. The areas of the figures at the segments $A_1 A_2$ and $A_2 A_3$ are equal respectively to $$\int_{-a}^{0} (x + a)^2 (x - b)^2 x^2\,dx = \frac{a^7}{210} (7k^2 + 7k + 2)$$ and $$\int_{0}^{b} (x + a)^2 (x - b)^2 x^2\,dx = \frac{a^7}{210} (2k^2 + 7k + 7)$$ Then $$K = k^5\, \frac{2k^2 + 7k + 7}{7k^2 + 7k + 2}.$$ The derivative of the function $f(k) = \dfrac{2k^2 + 7k + 7}{7k^2 + 7k + 2}$ is negative for $0 < k < \infty$. Therefore $f(k)$ decreases from $\frac{7}{2}$ to $\frac{2}{7}$ when $k$ increases from 0 to $\infty$. Inequalities $\frac{2}{7} < \frac{2k^2 + 7k + 7}{7k^2 + 7k + 2} < \frac{7}{2}$ imply the desired inequalities.