Search:

Easy First Steps

Paths Do End: Paths from the corner of an integral box always end at a corner.

To help visualize the spectrum of box shapes, we introduce what we call the "box triangle".

Every box has three dimensions, so the universe of box shapes would seem to be three-dimensional. But if our interest is in where the path from a corner leads, then we can shed one dimension. The reason is that the scale of the box is irrelevant. That is, for example, the picture for the {$3\times 5\times6$} box is the same as the picture for the {$9\times 15\times18$} box.

So if we want to work with all possible box shapes, we can simply look at all boxes where the dimensions add up to 1. A nice way to visualize all triples, {$a,b,c$} with {$a+b+c=1$}, is to look at all points inside an equilateral triangle with height 1.

For any point in the triangle, the sum of the distances to the sides

adds to 1. (The area of the triangle, {$\frac s2$}, is equal to the areas of the three smaller triangles, {$\frac {as}2+\frac {bs}2+\frac {cs}2$}.)

Thus, boxes of every shape correspond to points in the triangle and vice-versa. We'll use ordered triples, {$(a,b,c)$}, to refer to points in the triangle. The distances {$a$}, {$b$}, and {$c$} are called the "barycentric coordinates" of the point.

To make the picture more vivid, we color the corners.

Then if we notice, for example, that the path on the {$3\times 5\times6$} box ends at the yellow corner

we color the point in the box triangle corresponding to this box (distances {$\frac 3{14}$}, {$\frac 5{14}$}, {$\frac 6{14}$}) yellow.

Square Bottom: If the bottom of the box is square, you end at the yellow corner.

This proposition tells us how to color these points on the box triangle:

Sum of Two: If either {$b=a+c$} or {$a=b+c$} then the path on the {$a\times b\times c$} box ends at the gray corner.

This proposition tells us how to color these points on the box triangle:

By the way, it's not simple if {$c=a+b$}. In the {$3\times 5\times 8$} box, for example, the path has 25 segments.

Just Right: If either {$\ a,c<b<a+c\ $} or {$\ b,c<a<b+c\ $}, then the path on the {$a\times b\times c$} box returns to the start, to the black corner.

This proposition is the secret behind the surprising fact about the {$\pi\times \sqrt2\times e$} box. Note that {$$\sqrt2, e<\pi<\sqrt2+ e.$$}

Putting together the information of these propositions, we have the picture:

Back to Geometry.

Print - Search
Page last modified on June 25, 2013, at 09:13 AM