Jump to content
BrainDen.com - Brain Teasers
  • 0

Going through the tetrahedron


BMAD
 Share

Question

Suppose you are in a contest with another player. The gameshow created a holographic regular tetrahedron with each edge at 1 inch. Your job is to make the largest possible tetrahedron and throw it threw the hologram. The catch is that the tetrahedron you make must also be regular and must completely go through the hologram, so that every part of your shape must make it through the hologram. You pick the orientation of the holographic tetrahedron. The largest tetrahedron to make it through the holographic tetrahedron wins!

Assume that your shape does not rotate when it leaves your hands and that the hologram also does not rotate once you set its orientation.

Link to comment
Share on other sites

8 answers to this question

Recommended Posts

  • 0

Do you have a measurement?

I hope CAD measure are ok but what i thought in above figure is not right .

What i did is draw a 1" triangle ABC, then inside draw the front view of that

tetrahedron DEC-EFshowing the height E that is = sqrt 2/3 or 0.816497 where its top

vertex E is touching the AC side of the 1" triangle (no edge along z-axis).

A

G

E

B D H F C

This position is the optimal to go through the Hologram, so i made a parallel line

GB to make a bigger BGC-GH tetrahedron where the height =0.82 w/ side =1.00429"

Link to comment
Share on other sites

  • 0

If I understand this correctly, the solution is as follows:

The orientation of the hologram doesn't matter. Your tetrahedron is oriented the same as the hologram, and its edges are infinitesimally shorter than those of the hologram. Every point in your tetrahedron will pass through at least one point in the hologram.

Link to comment
Share on other sites

  • 0

If I understand this correctly, the solution is as follows:

The orientation of the hologram doesn't matter. Your tetrahedron is oriented the same as the hologram, and its edges are infinitesimally shorter than those of the hologram. Every point in your tetrahedron will pass through at least one point in the hologram.

In looking at TimeSpaceLightForce's question on Particles, I was reminded of Prince Rupert's cube. In this problem, it is shown that a cube larger than a unit cube can in fact successfully pass through a unit cube. This problem is now being put forth to a tetrahedron to see if the same applies.

http://en.wikipedia.org/wiki/Prince_Rupert's_cube

Edited by BMAD
Link to comment
Share on other sites

  • 0

Though you know my feelings towards CAD

Do you have a measurement?

I hope CAD measure are ok but what i thought in above figure is not right .

What i did is draw a 1" triangle ABC, then inside draw the front view of that

tetrahedron DEC-EFshowing the height E that is = sqrt 2/3 or 0.816497 where its top

vertex E is touching the AC side of the 1" triangle (no edge along z-axis).

A

G

E

B D H F C

This position is the optimal to go through the Hologram, so i made a parallel line

GB to make a bigger BGC-GH tetrahedron where the height =0.82 w/ side =1.00429"

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...