How can you intersect a cube at a single point?

Geometry Level 4

In the $(x, y, z, w)$ coordinate system, there is a line through the points $(1, 3, -8, 4)$ and $(9, 8, 5, 6)$ . If the point of intersection of the line with the $zwx$ 3D space/cube/whatever you call it (point, line, plane, ??? ) can be represented as the point $\left( -\dfrac{a}{b}, \dfrac{c}{d}, -\dfrac{e}{f}, \dfrac{g}{h} \right)$ , where each pair in parentheses $(a, b), (e, f), (g, h)$ contain only coprime positive integers, find the minimum sum of $a+b+c+d+e+f+g+h$ .

Bonus: Make sense of the problem from a geometric point of view.

Notes and Assumptions:

$e$ here is a variable and isn't equal to $\displaystyle \lim_{n \to \infty} \left( 1 + \dfrac{1}{n} \right)^n$ or $\displaystyle \sum_{k=0}^{\infty} \dfrac{1}{k!}$

The answer is 128.

1 solution

Hobart Pao
Nov 12, 2016

The equation of the line is:

$\vec{r}(t) = t \left \langle 9 - 1, 8-3, 5+8, 6-4 \right \rangle + \left \langle 1, 3, -8, 4 \right \rangle$

The four equations generated are:

$x = 8t + 1 \\ y = 5t+3 \\ z = 13t - 8 \\ w = 2t + 4$

The point of intersection with the $zwx$ 3D space/whatever you call it occurs when $y= 0$ . Then $t = \dfrac{-3}{5}$ . Substitute that value of $t$ into other three equations to generate the $x, z, w$ coordinates, namely producing $\left( -\dfrac{19}{5}, \dfrac{0}{1}, -\dfrac{79}{5}, \dfrac{14}{5} \right)$ . $19 + 5 + 0 + 1 + 79 + 5 + 14 + 5 = \boxed{128}$

How can you intersect a cube at a single point?

The answer is 128.

1 solution

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