Sliding Lines

Calculus Level 4

Two lines have the following equations (assuming $p = (x, y)$ ):

$n_1 \cdot (p - p_1(t) ) = 0$

with $n_1 = (1, 2) , p_1(t) = (1, 1) + (1, 3) t$ , and

$n_2 \cdot (p - p_2(t) ) = 0$

with $n_2 = (1, -1), p_2(t) = (-1,1) + (1, -1) t$

$t$ is the time parameter. The intersection of these two lines traces a line (shown in red in the animation). If the slope of the traced line is $\dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers , find $a + b$

4 21 16 19

1 solution

Hosam Hajjir
Aug 10, 2016

Consider sliding lines $n_1 \cdot ( r - (r_1 + v_1 t) ) = 0$ and $n_2.( r - (r_2 + v_2 t ) ) = 0$

The solution for a given $t$ is a single point. (assuming $n_1$ is not a multiple of $n_2$ ) We have the following linear system:

$A r = b$

with

$A = \begin{bmatrix} n_1^T \\ n_2^T \end{bmatrix}$

and

$b = \begin{bmatrix} n_1^T (r_1 + v_1 t) \\ n_2^T (r_2 + v_2 t) \end{bmatrix}$

The solution is

$r = A^{-1} b = A^-1 \begin{bmatrix} n_1^T r_1 \\ n_2^T r_2 \end{bmatrix} + A^{-1} \begin{bmatrix} n_1^T v_1 \\ n_2^T v_2 \end{bmatrix} t$

which is of the form,

$r = r_0 + d t$

From which, the slope is $m = d(2) / d(1)$ .

In this problem,

$n_1 = [1, 2]^T, n_2 = [1, -1]^T, r_1 = [1, 1]^T, r_2 = [-1, 1]^T, v_1 = [1, 3], v_2 = [1, -1]$

Substituting these values in the above formulas, we get m = $\frac{5}{11}$ , making the answer $5+11=\boxed{16}$

Sliding Lines

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