# Proof

• May 24th 2006, 03:07 PM
Natasha1
Proof
I was wondering whether someone could come up with the proof of this exercise I set myself:

We have four different types of triangles to play with (unlimited amounts of each):

one triangle which has 3 sides of length 1
one triangle which has 2 sides of length 1 and one of length square root 2
one triangle which has 2 sides of length square root 2 and one of side length 1
one triangle which has 3 sides of length square root 2

When I assumble 4 triangles together, what is the maximum amout of irregular tetrahedron (3-D shape) I can form. And what is the proof to make sure I have found the maximum?

Using:

4 triangles the same shape ----- you can make 2 irregular tetrahedrons.
3 triangles the same shape----- you can make 2 irregular tetrahedrons.
2 the same + 2 same (but different to the 2 others) ---- you can make 1 irregular tetrahedron.
2 the same + 2 different --- you can make 3 irregular tetrahedrons (I'm not sure on this one)
0 different ---- obviously 0 irregular tetrahedron.

Does this make any sense?
• May 30th 2006, 06:26 AM
Soroban
Hello, Natasha!

A fascinating problem . . . but I don't understand your answers.

Quote:

We have four different types of triangles to play with (unlimited amounts of each):

a triangle which has 3 sides of length 1
a triangle which has 2 sides of length 1 and one of length square root 2
a triangle which has 2 sides of length square root 2 and one of side length 1
a triangle which has 3 sides of length square root 2

When I assemble 4 triangles, what is the maximum number of irregular tetrahedrons I can form?
And what is the proof to make sure I have found the maximum?

Using 4 triangles the same shape ----- you can make 2 irregular tetrahedrons. ??
Are you sure you understand "irregular"?

A regular tetrahedron has all four faces congruent.
. . It is composed of four equilateral triangles.
And these are not counted.

Using 4 triangles of the same shape, you may or may not get a tetrahedron.

Here are the four shapes
Code:

*
*          * *    _
* *          *  * √2
*  * 1      *    *
*  E1 *        *  R  *
* * * * *      * * * * * *
1                1

*                *
* *
* *  _        *  *  _
√2        *    * √2
* I *          *  E2  *
*        *
* * * *      * * * *_* * *
1                √2

I've named the first and last triangles "E" for Equilateral.
The second is "R" for Right triangle.
The third is "I" for Isosceles.

The best approach is to make cardboard triangles and play with them.
. . I have done this . . . I suggest you do the same.

Here's what I've come up with:

two E1's and two R's
one E1 and three I's
one E1, two R's and one I
one R, two I's and one E2
three R's and one E2
two R's and two I's
two I's and 2 E2's
four I's

I hope I didn't miss any . . .

• May 30th 2006, 06:40 AM
Natasha1
Quote:

Originally Posted by Soroban
Hello, Natasha!

A fascinating problem . . . but I don't understand your answers.

Are you sure you understand "irregular"?

A regular tetrahedron has all four faces congruent.
. . It is composed of four equilateral triangles.
And these are not counted.

Using 4 triangles of the same shape, you may or may not get a tetrahedron.

Here are the four shapes
Code:

*
*          * *    _
* *          *  * √2
*  * 1      *    *
*  E1 *        *  R  *
* * * * *      * * * * * *
1                1

*                *
* *
* *  _        *  *  _
√2        *    * √2
* I *          *  E2  *
*        *
* * * *      * * * *_* * *
1                √2

I've named the first and last triangles "E" for Equilateral.
The second is "R" for Right triangle.
The third is "I" for Isosceles.

The best approach is to make cardboard triangles and play with them.
. . I have done this . . . I suggest you do the same.

Here's what I've come up with:

two E1's and two R's
one E1 and three I's
one E1, two R's and one I
one R, two I's and one E2
three R's and one E2
two R's and two I's
two I's and 2 E2's
four I's

I hope I didn't miss any . . .

That's great thanks Soroban. My lecturer told me we can get 10, but I'm sure he is wrong (during the year he got similar problems wrong ;) ) So I bet he is wrong again this time.
• May 31st 2006, 09:51 AM
malaygoel
Quote:

Originally Posted by Soroban
Using 4 triangles of the same shape, you may or may not get a tetrahedron.

Without using cardboards, is it possible to tell whether the four given triangles will make a tetrahedron or not?
• May 31st 2006, 03:52 PM
Soroban
Hello, malaygoel!

Quote:

Without using cardboards, is it possible to tell whether the four given triangles will make a tetrahedron or not?
Yes, it is possible . . . but for some configurations, I had to use cardboard triangles.

I'd visualize three triangles together, then I'd try to "see" the fourth one.
. . Most of the time, I couldn't make a good sketch.
Sometimes it was just too much for my poor brain.
Other times, I'd discover that I already had it one my list.
. . I didn't recognize it because it was upside-down.
(But I didn't do too badly for a carbon-based lifeform.)

And I just discovered that I missed a tetradron!
. . We can make one with four R's (right triangles). **

I've found nine now . . . so your lecturer may be right.
. . I'll have to keep looking.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

This means that I was wrong . . . again!

With four of the same type of triangle, we can make a tetrahedron,
. . either regular and irregular.

Even more embarrassing:
I just found that we can use any shape triangle
. . and four of them will make a tetradedron.

This means that you were right . . .
Four of the same shape will form two irregular tetrahedrons.
If that's what you meant, I owe you a big apology.
• May 31st 2006, 05:45 PM
malaygoel
Quote:

Originally Posted by Soroban
Hello, malaygoel!

And I just discovered that I missed a tetradron!
. . We can make one with four R's (right triangles). **

just check, I don't think that a tetrahedron can be made with four R's
• May 31st 2006, 11:30 PM
Natasha1
My lecture did say there was 9 absolute ones an the 10th one is a contrivance (with 5 pieces) could someone draw it? (as i can't picture it)
• Jun 1st 2006, 06:05 AM
Soroban
Hello, malaygoel!

Quote:

just check, I don't think that a tetrahedron can be made with four R's
I didn't think so either . . . but there it was!
Code:

D                  C                  D
*-------------------*-------------------*
*              *  *              *
*    R    *      *    R    *
*      *    R    *      *
*  *              *  *
*-------------------*
A                  B

Place three R's on the table as shown.

Fold up the side triangles along AC and BC, so that DC = CD.

The front "face" is another triangle ADB, which is another R.
. . (Its sides are: $\displaystyle \,DA\,=\,1,\;DB\,=\,1,\;AB\,=\,\sqrt{2}$)

I'm still working on your lecturer's "contrivance".
So far, I've had no luck . . .
• Jun 12th 2006, 09:12 AM
Soroban
Hello, malaygoel!

I came across a fascinating bit of information.

I showed that an irregular tetrahedron can be made with four R's.
Quote:

Code:

D                  C                  D
*-------------------*-------------------*
*              *  *              *
*    R    *      *    R    *
*      *    R    *      *
*  *              *  *
*-------------------*
A                  B

Place three R's on the table as shown.

Fold up the side triangles along AC and BC, so that DC = CD.

The front "face" is another triangle ADB, which is another R.
. . (Its sides are: $\displaystyle \,DA\,=\,1,\;DB\,=\,1,\;AB\,=\,\sqrt{2}$)
Now "double" the R's . . .
Code:

D                  C                  D
*---------*---------*---------*---------*
*      *      * * *      *      *
*    *    *  *  *    *    *_
*  *  *    *    *  *  * √2
* * *      *      * * *
A*---------*---------*B
1        1

Place six R's on the table as shown.
Fold up the side triangles along AC and BC, so DC = CD.

The front "face" is another triangle ADB, which is a large R.
Its sides are: $\displaystyle DA = \sqrt{2},\;DB = \sqrt{2},\;AB = 2$
. . and this can be filled with two R's.

So, an irregular tetrahedron can be made with eight R's.

But we can make larger and larger R's . . . with the same result.

Four R's can be arranged to make an even larger R.
Code:

*
* |
*  |
*    |
*-------*
* |    * |
*  |  *  |
*    | *    |
*-------*-------*

And four of these will form an irregular tetrahedron.

Do you see where this is going?
Since there is "an unlimited supply" of triangles, we can keep going . . .

Of course, if the original problem stipulated that exactly four triangles are used,
. . then none of this is acceptable.
But then your lecturer's "contrived" solution isn't acceptable either,
. . and we have found the nine solutions.