## Origamic Tetrahedron

```Date:           Sun, 22 Oct 2000 12:06:07 +0900
From:           Toshi Kato <junk.kato@nifty.ne.jp>
To:             eppstein@ics.uci.edu
Subject:        Origamic tetrahedron
```

```Dear David Eppstein,

Thank you,

Toshi Junk Kato
```

Here is a triangle which lengths are 2, 3 and 4. Make a tetrahedron by folding three lines.

```Date:           Sat, 21 Oct 2000 21:46:38 -0700
From:           David Eppstein <eppstein@ics.uci.edu>
To:             Toshi Kato <junk.kato@nifty.ne.jp>
Subject:        Re: Origamic tetrahedron
```

```On 10/22/00 12:06 PM +0900, Toshi Kato wrote:
> Here is a triangle which lengths are 2, 3 and 4.
> Make a tetrahedron by folding three lines.

I thought one could always fold any triangle into a tetrahedron by making
four congruent triangles using folds through the midpoints of the edges.  Is
there something special about 2-3-4 that allows another solution?
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```

```Date:           Sun, 22 Oct 2000 17:12:47 +0900
From:           Toshi Kato <junk.kato@nifty.ne.jp>
To:             eppstein@ics.uci.edu
Subject:        Re: Origamic tetrahedron
```

```> > Here is a triangle which lengths are 2, 3 and 4.
> > Make a tetrahedron by folding three lines.
>                                 ~~~~~~~~~~~
> I thought one could always fold any triangle into a tetrahedron by making
> four congruent triangles using folds through the midpoints of the edges.  Is

> there something special about 2-3-4 that allows another solution?

Can you do it by folding only three lines about 2-3-4 triangle?

Best wishes,

Junk Kato
```

```Date:           Sun, 29 Oct 2000 11:30:25 -0800
From:           David Eppstein <eppstein@ics.uci.edu>
To:             Toshi Kato <junk.kato@nifty.ne.jp>
Subject:        Re: Origamic tetrahedron
```

```On 10/22/00 5:12 PM +0900, Toshi Kato wrote:
> > I thought one could always fold any triangle into a tetrahedron by making
> > four congruent triangles using folds through the midpoints of the edges.
> > Is
>
> > there something special about 2-3-4 that allows another solution?
>
> Can you do it by folding only three lines about 2-3-4 triangle?

You're right, this is a good puzzle.  I had to actually try folding some
paper in order to understand it.

The four congruent triangle "solution" (top part of attached GIF image)
doesn't work.  It is possible to get a tetrahedron in which each of the
original triangle edges is doubled back along its midpoint, but it takes more
than three folds (middle of GIF).

My solution is the bottom of the GIF.  The fold points are at the middle of
the long edge, and 1/6 and 1/3 of the way along the middle edge.
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```

```Date:           Mon, 30 Oct 2000 11:01:55 +0900
From:           Toshi Kato <junk.kato@nifty.ne.jp>
To:             eppstein@ics.uci.edu
Subject:        Re: Origamic tetrahedron
```

```Dear Prof. David Eppstein,

> You're right, this is a good puzzle.  I had to actually try folding some
> paper in order to understand it.
>
> The four congruent triangle "solution" (top part of attached GIF image)
> doesn't work.  It is possible to get a tetrahedron in which each of the
> original triangle edges is doubled back along its midpoint, but it takes mor
e
> than three folds (middle of GIF).
>
> My solution is the bottom of the GIF.  The fold points are at the middle of
> the long edge, and 1/6 and 1/3 of the way along the middle edge.

Very good. You are right.
You are first solver.
Thank you very much.
If you have time, please find other solution.

Best wishes,

Junk Kato
```

```Date:           Sun, 29 Oct 2000 23:08:16 -0800
From:           David Eppstein <eppstein@ics.uci.edu>
To:             Toshi Kato <junk.kato@nifty.ne.jp>
Subject:        Re: Origamic tetrahedron
```

```On 10/30/00 11:01 AM +0900, Toshi Kato wrote:
> If you have time, please find other solution.

There's more than one three-fold tetrahedron?

I did try folding it some more, and found other tetrahedra, but they need
five, seven, or even nine folds -- see attached image.
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```

```Date:           Mon, 30 Oct 2000 23:54:01 +0900
From:           Toshi Kato <junk.kato@nifty.ne.jp>
To:             eppstein@ics.uci.edu
Subject:        Re: Origamic tetrahedron
```

```Dear David Eppstein,

> > If you have time, please find other solution.
>
> There's more than one three-fold tetrahedron?

Yes, I think there is one more.

> I did try folding it some more, and found other tetrahedra, but they need
> five, seven, or even nine folds -- see attached image.

Thank you for other folds.

Best wishes,

Junk Kato
```

```Date:           Mon, 30 Oct 2000 12:57:38 -0800
From:           David Eppstein <eppstein@ics.uci.edu>
To:             Toshi Kato <junk.kato@nifty.ne.jp>
Subject:        Re: Origamic tetrahedron
```

```On 10/30/00 11:54 PM +0900, Toshi Kato wrote:
> > There's more than one three-fold tetrahedron?
>
> Yes, I think there is one more.

Ok, I found it.  It is an N-shape, with vertices one unit along the length-3
side, 1.5 units along the long side, at the obtuse vertex, and 3 units along
the long side.  (Image attached.)

I found this by systematically going through all ways of making an N shape in
the triangle, with one vertex of the N at a triangle vertex.  One way gave my
previous solution, four of them led to the four folds I showed you in my
previous message (all with more than three folds), and the sixth led to this
solution.

By the way, there are also three infinite families of tetrahedra that can be
folded from the triangle. if you fold two of the triangle edges in half, so
that all three vertices meet at a point, you get a shape with one remaining
edge (forming a continuous loop) and two interior vertices at the midpoints
of the original triangle. The other two tetrahedron vertices can be chosen to
be any two opposite points from the loop.  There is one solution in common to
all three such families: the one I first sent you.  However, it seems that
none of these possibilities uses only three folds.
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```

```Dear Prof. Eppstein,

> On 10/30/00 11:54 PM +0900, Toshi Kato wrote:
> > > There's more than one three-fold tetrahedron?
> >
> > Yes, I think there is one more.
>
> Ok, I found it.  It is an N-shape, with vertices one unit along the
> length-3  side, 1.5 units along the long side, at the obtuse vertex,
> and 3 units along the long side.

I have to apologize you for my mistake.
The second one is not a solution because of non-tetrahedoron.
It seems that the problem has unique solution.

Best regards,

Junk Kato
```

```Date:           Mon, 13 Nov 2000 21:38:11 -0800
From:           David Eppstein <eppstein@ics.uci.edu>
To:             Toshi Kato <junk.kato@nifty.ne.jp>
Subject:        Re: Origamic tetrahedron
```

```On 11/14/00 1:41 PM +0900, Toshi Kato wrote:
> I have to apologize you for my mistake.
> The second one is not a solution because of non-tetrahedoron.
> It seems that the problem has unique solution.

It looked like a tetrahedron when I folded it...are you sure?
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```

```Date:           Tue, 14 Nov 2000 19:35:41 +0900
From:           Toshi Kato <junk.kato@nifty.ne.jp>
To:             eppstein@ics.uci.edu
Subject:        Re: Origamic tetrahedron
```

```Dear Prof. Eppstein,
>
> On 11/14/00 1:41 PM +0900, Toshi Kato wrote:
> > I have to apologize you for my mistake.
> > The second one is not a solution because of non-tetrahedoron.
> > It seems that the problem has unique solution.
>
> It looked like a tetrahedron when I folded it...are you sure?

I also looked like a near-flat tetrahedron at first.
Now I think that it is perfect-flat.
How are you feeling?

Best wishes,

Junk Kato
```

```Date:           Tue, 14 Nov 2000 23:26:44 -0800
From:           David Eppstein <eppstein@ics.uci.edu>
To:             Toshi Kato <junk.kato@nifty.ne.jp>
Subject:        Re: Origamic tetrahedron
```

```On 11/14/00 7:35 PM +0900, Toshi Kato wrote:
> I also looked like a near-flat tetrahedron at first.
> Now I think that it is perfect-flat.
> How are you feeling?

You are right, it is flat.
Proof:

Let the triangle be ABC, with AB=2, BC=3, AC=4.
Draw D on AC at distance 1 from A, E = midpoint(D,C), and
F on BC at distance 2 from B.
My proposed fold pattern was along line segments BD, BE, and EF.

Then BCD is isosceles, so 2*angle(BDC)+angle(BCD)=180.
But also angle(ADB)+angle(BDC)=180, so subtracting the two equations
gives angle(BDC))+angle(BCD)=angle(ADB), so these three angles
fold flat rather than making one of the corners of a tetrahedron, QED.

One can also observe that ADB is similar to ABC, BDE is isosceles,
and (since the other tetrahedron corners must also be flat) BEF is a right
triangle. I don't know of any similar nice property for the remaining
triangle, EFC. This construction also makes it clear that the sharp angle of
the 2-3-4 triangle is twice that of a 1-sqrt(15)-4 right triangle, which is
helpful if you want to calculate the angles.
--
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
```