Collinearity

We will now show that any 3 points are collinear.

Thus, we have to show $(a_k, b_k)$, $(a_m, b_m)$ and $(a_n, b_n)$ are collinear.

$\iff$ Line formed by $(a_k, b_k)$ and $(a_m, b_m)$ is parallel to line formed by $(a_m, b_m)$ and $(a_n, b_n)$

Using the fact of arithmetic progressions, we can let $a_k=a_0+k \times d_a$ and $b_k=b_0+k \times d_b$, for some constants $a_0, d_a, b_0, d_b$

Therefore we can show that the gradients of the 2 lines are equal.

Gradient of line formed by $(a_k, b_k)$ and $(a_m, b_m)=\frac{b_m-b_k}{a_m-a_k}=\frac{d_b}{d_a}$

Gradient of line formed by $(a_m, b_m)$ and $(a_n, b_n)=\frac{b_m-b_n}{a_m-a_n}=\frac{d_b}{d_a}$

Since gradients are equal, it follows that the lines are parallel and thus $(a_k, b_k)$, $(a_m, b_m)$ and $(a_n, b_n)$ are collinear.

Now we can start with base case n=3 and induct by taking points $(a_1, b_1), (a_2, b_2)$ and $(a_{n+1}, b_{n+1})$. Hence proven.

Let $a_i = \lambda_a i+\mu_a$, $b_i = \lambda_b i+\mu_b$ for some real numbers $\lambda_a,\lambda_b,\mu_a,\mu_b$.

Then $i=\frac{a_i-\mu_a}{\lambda_a}$, and thus $b_i = \lambda_b\frac{a_i-\mu_a}{\lambda_a}+\mu_b = \frac{\lambda_b}{\lambda_a}a_i+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a})$.

Hence all the points $(a_i,b_i)$ lie on the same line $y=\frac{\lambda_b}{\lambda_a}x+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a})$, as desired.

This prob came in C.M.I....................... Do you know how to prove that f(x)=1/(x+2cos(x)) is onto rigorously...I mean intuitively it can be seen, but is there any othere way to do it.............I mean using 'hardcore math'.

there's a easy method using determinant criteria, for 3 points prove that the determinant of area equal to 0