slacking

[fireyice]

I'm having troubles concentrating this afternoon. So instead of working I've been thinking about a question that someone asked in my graphics class earlier this afternoon. Assume that you have a box and there is a triangle that partially sticks outside of the box. If you clip off the parts of the triangle sticking outsid, what is the maximum

Comments

[thatgirljj]

Is your box a cube?

[fireyice]

Hmm, I think it can be three dimensional but all sides do not have to be the same length. The triangle is two dimensional.

[denoue]

The length of the sides of the box shouldn't matter. If it's a 2d box then 7's pretty simple, can you do 8?

[fireyice]

I can only get 7.

[mrsmalkav]

i have 9 on a cube, 7 on a square

[fireyice]

do you have a pic?

[drspin]

you can only get 7 on a square, i'm pretty confident. to get 7, you have two triangle vertices outside two edges of the square, and the last vertex outside the one of the corners. uh, for square on (0,0) (10,10), coords (-2,5), (5,12), (15,-5). hmm, shit, i can't explain why you can't get 8 without a diagram.

[denoue]

Yea. This became pretty clear once I tried to do it.

[drspin]

oh, duh, and here's an easy proof. you have three lines for the triangle and four lines for the square - you can't have a concave polygon that uses any of those segments more than once, so at most you can have 7 segments.

and the proof extends to the 3d case - you have 6 faces and 3 triangle edges, hence max of 9 sided polygon.

[drspin]

er, convex. i always get those backward.

i'm gonna pump up your comment number a bit more now!

[drspin]

[fireyice]

heh, I'm pretty sure that I know the answer if the box is 2d. I think I know what the 3d answer should be, but I can't picture how to get it.

[denoue]

I think I see how to get 9 for the 3d case.

[fireyice]

Draw it! That's what I think the answer should be... but I can't picture it. I suck at 3d.

[denoue]

I lied. I have 8, and I think that might actually be the answer.

[mrsmalkav]