Question

The graph of the function f(x) = x10 + 9x9 + 7x8 + ... + a1x + a0 intersect the line y = b at the points B1, B2, B3, ......, B10 (from left to right) and the line y = c at the points C1, C2, C3,......C10(from left to right). Let P be a point on the line y = c to the right of the point C10. If b = 5 and c = 3 then find the sum

$\sum_{n=1}^{10} cot (B_nC_nP)$

Answer 1

0

Answer 2

Let the function be $f(x) = x^{10} + 9x^9 + 7x^8 + \dots + a_1x + a_0$.

The graph of $y = f(x)$ intersects the line $y = b$ at points $B_1, B_2, \dots, B_{10}$. Let the x-coordinates of these points be $x_{B_1}, x_{B_2}, \dots, x_{B_{10}}$. These x-coordinates are the roots of the equation $f(x) = b$, which is $x^{10} + 9x^9 + 7x^8 + \dots + a_1x + (a_0 - b) = 0$. By Vieta's formulas, the sum of the roots is $\sum_{n=1}^{10} x_{B_n} = -\frac{\text{coefficient of } x^9}{\text{coefficient of } x^{10}} = -\frac{9}{1} = -9$.

The graph of $y = f(x)$ intersects the line $y = c$ at points $C_1, C_2, \dots, C_{10}$. Let the x-coordinates of these points be $x_{C_1}, x_{C_2}, \dots, x_{C_{10}}$. These x-coordinates are the roots of the equation $f(x) = c$, which is $x^{10} + 9x^9 + 7x^8 + \dots + a_1x + (a_0 - c) = 0$. By Vieta's formulas, the sum of the roots is $\sum_{n=1}^{10} x_{C_n} = -\frac{\text{coefficient of } x^9}{\text{coefficient of } x^{10}} = -\frac{9}{1} = -9$.

We are given $b=5$ and $c=3$. The points are $B_n = (x_{B_n}, b)$ and $C_n = (x_{C_n}, c)$. The point $P$ is on the line $y = c$ to the right of $C_{10}$. Let $P = (x_P, c)$. Since $P$ is to the right of $C_{10}$, and $C_1, \dots, C_{10}$ are ordered from left to right, we have $x_P > x_{C_{10}} \ge x_{C_n}$ for all $n=1, \dots, 10$.

We want to calculate $\sum_{n=1}^{10} \cot(B_n C_n P)$. Let $\theta_n = B_n C_n P$. Consider the points $C_n = (x_{C_n}, c)$, $B_n = (x_{B_n}, b)$, and $P = (x_P, c)$. The line segment $C_n P$ lies on the horizontal line $y=c$. Since $x_P > x_{C_n}$, the vector $\vec{C_n P} = (x_P - x_{C_n}, 0)$ points in the positive x-direction. The vector $\vec{C_n B_n} = (x_{B_n} - x_{C_n}, b - c)$. The angle $\theta_n$ is the angle between the vector $\vec{C_n B_n}$ and the positive x-axis (which is the direction of $\vec{C_n P}$). Let $\Delta x_n = x_{B_n} - x_{C_n}$ and $\Delta y_n = b - c$. The vector $\vec{C_n B_n}$ is $(\Delta x_n, \Delta y_n)$. The angle $\theta_n$ is the angle whose tangent is $\frac{\Delta y_n}{\Delta x_n}$, provided $\Delta x_n \neq 0$. If $\Delta x_n = 0$, then $x_{B_n} = x_{C_n}$, which implies $f(x_{B_n}) = b$ and $f(x_{C_n}) = c$ with the same x-coordinate. This would mean $b=c$, which is not true ($b=5, c=3$). So $x_{B_n} \neq x_{C_n}$, and $\Delta x_n \neq 0$.

The slope of the line segment $C_n B_n$ is $m_n = \frac{b - c}{x_{B_n} - x_{C_n}}$. The angle $\theta_n$ is the angle that the segment $C_n B_n$ makes with the horizontal line $y=c$ at $C_n$, measured towards $P$. Since $P$ is to the right of $C_n$, the direction of $C_n P$ is along the positive x-axis. The tangent of the angle that $C_n B_n$ makes with the positive x-axis is $\frac{b-c}{x_{B_n}-x_{C_n}}$. The angle $\theta_n = B_n C_n P$ is either this angle or this angle plus/minus $\pi$. However, in the triangle $B_n C_n P$, the angle $\theta_n$ is between 0 and $\pi$. Let's consider the coordinates relative to $C_n$ as the origin. $C_n = (0,0)$, $B_n = (x_{B_n} - x_{C_n}, b - c)$, $P = (x_P - x_{C_n}, 0)$. Since $x_P - x_{C_n} > 0$, the point $P$ is on the positive x-axis relative to $C_n$. The angle $\theta_n$ is the angle between the vector $(x_{B_n} - x_{C_n}, b - c)$ and the positive x-axis. The cotangent of this angle is given by $\cot \theta_n = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{x_{B_n} - x_{C_n}}{b - c}$.

We are given $b = 5$ and $c = 3$, so $b - c = 5 - 3 = 2$. Thus, $\cot \theta_n = \frac{x_{B_n} - x_{C_n}}{2}$.

We need to find the sum $\sum_{n=1}^{10} \cot(B_n C_n P) = \sum_{n=1}^{10} \cot \theta_n = \sum_{n=1}^{10} \frac{x_{B_n} - x_{C_n}}{2}$. This sum can be written as: $\frac{1}{2} \sum_{n=1}^{10} (x_{B_n} - x_{C_n}) = \frac{1}{2} \left( \sum_{n=1}^{10} x_{B_n} - \sum_{n=1}^{10} x_{C_n} \right)$.

Using the sums of the roots we found earlier: $\sum_{n=1}^{10} x_{B_n} = -9$ $\sum_{n=1}^{10} x_{C_n} = -9$

Substituting these values into the sum: $\frac{1}{2} ((-9) - (-9)) = \frac{1}{2} (-9 + 9) = \frac{1}{2} (0) = 0$.

The sum $\sum_{n=1}^{10} \cot(B_n C_n P) = 0$.