## Tetrahedron Problems

My blog is getting more famous. Now I don’t need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov’s Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:

Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?

Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:

Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?

Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.

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## Christ Schlacta:

the second is 7.

19 February 2011, 12:31 am## Vishesh Kumar:

Second question: 6 faces!

19 February 2011, 2:07 amFirst question: If the segments are equal, I (believe, I) can.

## Vishesh Kumar:

oops. qn 2 is 7 faces.*sheepish head shake* ðŸ™‚

19 February 2011, 2:07 am## Austin:

Second question: 5 faces.

19 February 2011, 10:48 am## Tanya Khovanova:

Austin,

Nice. I have an octahedron die and a tetrahedron die. I actually physically attached them to each other to see what is going on.

19 February 2011, 2:25 pm## Austin:

Good idea! I got my answer the dirty way (coordinatization). Is there a more insightful solution?

A dozen years ago when I was preparing for the AIME, I ran across this problem from an earlier exam: “If the cube with corners at (+/-1, +/-1, +/-1) is cut along the planes x=y, y=z, and z=x, how many pieces result?” Though I came up with multiple reasons that the answer must be 6, I couldn’t convince my practical-minded aunt until we settled it by slicing up a block of cheese. I was considering applying the same method to this problem until you suggested the dice. (I just happen to have a lot of 4-sided dice left over from the thesis defense of Beth Chen, who encouraged her audience to build parts of her record-breaking tetrahedron packing — the subject of her thesis — out of d4’s and poster putty which she passed out.)

(To digress further, the cube-slicing problem above leads to a nice non-calculus explanation for the volume formula for a pyramid or cone.)

19 February 2011, 3:15 pm## Jonathan:

Consider five equal edges, length 1. They will join into two equilateral triangles, sharing a common base. That base serves as a sort of hinge. If we have a tiny 6th side, we almost fold it closed. If we have a bigger sixth side, we open it wider. Opened flat, it is a 60/120 rhombus (lozenge, I think it is called) with long diagonal sqr(3)

If we take any 3, of course we can form a triangle. For the sixth edge, we must not equal or exceed 2x. We can get as small as we like.

But if we chose our last side y such that sqr(3) â‰¤ y â‰¤ 2, then we meet the condition of being able to form a triangle with any 3 segments, but we are too long to be the last edge of a tetrahedron.

The first statement is false.

19 February 2011, 8:20 pm## Jonathan:

that “2x” should have been “2”

Integral counterexample: 4, 4, 4, 4, 4, 7

19 February 2011, 8:36 pm## Vishesh Kumar:

How, is question 2: 5?

20 February 2011, 7:14 am## Tanya Khovanova:

Vishesh,

Two out of new three faces become continuations of two existing faces.

20 February 2011, 10:45 am## Austin:

Yeah, in the same way that two squares of equal size make a quadrilateral rather than a hexagon. In this case, the tetrahedron and the square pyramid make a (non-right) triangular prism with three parallelogram faces, I think. It’s far from intuitive.

20 February 2011, 1:50 pm## JBL:

Jonathan, very nice example. I was working toward something where the sixth side would be too small (e.g., start with 4, 4, 4, sqrt(13), sqrt(21), something between sqrt(21) – sqrt(13) and 1) but since you don’t have to assemble the 4s into an equilateral triangle it’s harder to rule out a valid tetrahedron.

@Austin, for a more conceptual explanation: consider a cube. One can choose four vertices so that the segments joining them are face diagonals of the cube, i.e., so that they form a regular tetrahedron. Cutting this tetrahedron out, we are left with four pyramids, each with three isosceles right triangular faces and one equilateral triangular face. Note that two such pyramids share an edge with the tetrahedron, and the three angles sum to make the flat side of the cube. Now, glue these four pyramids together to form the square pyramid in the question; when you attach the tetrahedron, the same angles are reconstructed.

Alternative viewpoint: chop space up with the planes x + y + z = 2n, x + y – z = 2n, x – y + z = 2n and -x + y + z = 2n for all integers n. The regions they determine are regular tetrahedra and regular octahedra.

20 February 2011, 2:32 pm## Jonathan:

JBL,

I wonder if 4,4,4,4,4,7 is the smallest integral counterexample. Certainly this is the best for five congruent and one large. Four congruent and two congruent can’t produce a counterexample. I am not sure how to check something on the order of 9, 8, 8, 8, 7, 3. It seems that it would need to fail three cases – equilateral triangle, 8’s emanating from a single vertex, and 8’s lined end to end, but not closed. And that third case looks like 3 cases.

20 February 2011, 3:51 pm## JBL:

Jonathan, it wouldn’t surprise me (at least for reasonable definitions of the word “smallest”).

Your 9,8,8,8,7,3 example is the sort of thing I was looking for, based on the assumption that the 8s would form a triangle (although in your example I think the 3 is just large enough that we *can* form a triangle in this case — something like 2.01 is small enough, though, and would surely break the situation when the 8s all have a common vertex, too). The last case (or the last 3 cases, as you note), are trickier.

Perhaps it’s easier to classify workable situations if we also use some tool like Ptolemy’s Inequality: http://mathworld.wolfram.com/PtolemyInequality.html — in a tetrahedron ABCD, we have AB * CD < AC * BD + AD * BC. Unfortunately, this isn’t even sensitive enough to find the 4s and 7 example; I wonder if it can find any example that doesn’t have a violation of the triangle inequality?

21 February 2011, 10:35 am## BB:

The second question is a nice one! It’s easy to understand what’s going on if you imagine an octahedron inscribed in a tetrahedron. (Just connect the midpoints of the tetrahedron’s edges).

21 February 2011, 1:11 pmThen dissect this construction by a plane containing the 4 vertices of the octahedron. What’s left is a square-based pyramid with 2 tetrahedrons attached to it.

## David:

BB,

22 February 2011, 5:45 amThanks! I understood the idea, but was having great trouble lining up the planes in my mind. Your construction clarified the situation very nicely.

## Geneal:

The second question made the front page of the New York Times many years ago when it appeared on a math SAT test with the intended answer of 7. One person who gave the right answer appealed, with the result that ETS accepted all answers of 5 *as well as* all answers of 7. Go figure (sorry, I couldn’t resist).

Another easy way to visualize this is to place two square pyramids next to each other on the sand, sharing one square edge. Then join their apices with another edge. It’s easy to see that (1) you’ve just created a regular tetrahedron, and (2) the new faces you’ve just created are coplanar with the already existing square pyramid faces.

22 February 2011, 12:29 pm## Christ Schlacta:

perhaps in the mathematical world an answer of 5 works, but you try joining two coplanar surfaces into a single face in the physical world..

23 February 2011, 12:58 am## Tim Chu:

You can also envision the picture of second question by imagining a triangular tent with wind blowing into a triangular face ðŸ™‚

24 February 2011, 4:48 pm## mashplum:

In the second question, we are not told that the faces of the tetrahedron are the same size as the triangular faces of the pyramid. We are also not told how to orient the faces when we glue them together. I imagine you could argue for any answer between 5 and 13.

3 April 2011, 9:17 pm