Going through the tetrahedron

[BMAD]

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.

[TimeSpaceLightForce]

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"

[Morningstar]

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.

[BMAD]

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 April 19, 2013 by BMAD

[TimeSpaceLightForce]

the perpendicular triangle side of the halographic

tetrahedron is the greatest area of projection to go

through and the least area of projection of your solid

is when one of its edge is along z-axis forward orientation..

[BMAD]

and so how big is it?

[TimeSpaceLightForce]

and so how big is it?

[BMAD]

Do you have a measurement?

[BMAD]

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"