*This problem is inspired by the Problem of the Week.*

A closed glass, shaped of a unit cube, consists of water \(\frac{1}{6}\) filled. It is known that if we rotate the cube smoothly (so that the water does not "jump"), the water cross-section changes to one of some known shapes.

What is the volume of space, where the water cross-section will never reach throughout the rotation? For this problem, assume that the cross-section remains flat while changing its form.

**Note:** The answer is not \(\frac{5}{6}\) since it is the volume of space where the water can't reach *without* rotating the cube.

**Important:** Please do not only post the answer. Share solutions!

If you are interested in using or borrowing this for your problem in the community, (as long as the correct answer doesn't appear yet) please let me know!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI don't understand what you are asking.

Log in to reply

If the cube is fully rotated all the way around, you will notice the "empty" solid tangent to the water plane at some points. What is the volume of the space?

Log in to reply

Wouldn't that just be the remaining \(\frac{5}{6}\)?

Log in to reply

I may consider revising the problem in case you interpret the problem incorrectly.

Log in to reply

If it is the volume of empty space that u r interested in knowing, it shall be nothing other than 5/6 cubic units. As the space taken by the 'emptiness' or water itself won't change at all, be it stationary or in motion

Log in to reply

Well, I would assume that the space would have 4/6 or 2/3 of a unit in diameter, so a 1/3 radius. When you do 4/3pi(r)^3 you get about 4/25. This is because the space within the cube is a sphere, correct?

Log in to reply

The solid within the cube is actually not the sphere since the slope of the cross-section changes relative to the total volume of the water. Taking the volume of space about the origin in the rectangular coordinate system, the radius is \(\frac{1}{3}\) if the point touching it is \(\frac{1}{6}\) units away from one of the cube's faces. It is intuitive that the radius of the space is \(\frac{1}{3}\) as if the water cross-section is perpendicular throughout the rotation; however, this is not entirely true. Triangular pyramid of equilateral triangle is the popular example that disproves this.

Consider (1) the point closest to the centroid, and (2) the "sliced" cross-sections formed within the water. Two base cross-sections and a midsection should be helpful to answer the problem.

Log in to reply