Can a 10 inch long cylinder an inch in diameter fit into a hollow cube with six inch edges?

Question

llort · Answer

this is what I have thus far:

$$\sqrt((6^2+6^2)+6^2)\approx10.39$$

anonymous · Answer

then figure out the diagonal of the cylinder and see if it's shorter or longer.

anonymous · Answer

actually, that doesn't work.

amistre64 · Answer

can 1 inch wide fit into 6 inches wide?  yes

anonymous · Answer

if you draw the cross section, so you have the cube diagonal represented as a rectangle that is 10sqrt(2) wide and 10 tall, and the cylinder as a 10 inch x 1 inch rectangle, you should be able to figure it out.

amistre64, I am assuming he means will it fit completely inside the volume bounded by the cube.

amistre64 · Answer

attachment

amistre64 · Answer

a hollow cube ... i assume acts like a sleeve

llort · Answer

nope, the cube is closed.

amistre64 · Answer

then 10 inches doesnt fit into 6 inches ...

anonymous · Answer

will it fit in the diagonal though?

amistre64 · Answer

it might :)  if thats the question being asked; which now seems more reasonable lol

llort · Answer

yeah, I already proved with trig, that if you use all three dimensions it will fit, with about 4/10 of an inch to spare.

anonymous · Answer

but it's an inch wide so it might not fit.

llort · Answer

but what about the diameter of the cylinder, will that fit into the cube?

so that is the question

anonymous · Answer

right. draw the diagram I suggested.

amistre64 · Answer

does:

(6-(1/sqrt(2)))^2+(6-(1/sqrt(2)))^2<=10^2

anonymous · Answer

that assumes the cylinder is at 45 degrees, which it is not.

amistre64 · Answer

attachment

amistre64 · Answer

6 tall; and the diagonal wide ... hmmm

amistre64 · Answer

well, if it fits in that way, you know it fits in the other way lol

anonymous · Answer

that's true!