Presented on TheProblemSite.com, September 2004

In our three-dimensional world, we recognize various degrees of "flatness".  A figure that lies in a plane is "flat" to us, because it is two-dimensional.  A figure that lies completely within a line is much "flatter" because it is one-dimensional.  The one-dimensional analog of an equilateral triangle with unit sides is a line segment.  Its height is 1.  The height of an equilateral triangle with unit sides is sqrt(3)/2.  The three-dimensional analog of this triangle is a tetrahedron with unit edges.  Its height is sqrt(6)/3.

For the sake of this puzzle, I'll invent a new word: a unit "n-equilateroid" is an analog in n dimensions of the equilateral triangle (or of the tetrahedron) whose sides (or edges) are one unit long.

(A good way to visualize the height of a tetrahedron is to imagine the triangular base with a point P in the "center" (incenter, circumcenter, or centroid -- they're all the same).  From this point, an altitude is raised until it reaches the fourth vertex, V.  Draw a line segment from P to any of the vertices, B, on the base, and then complete the triangle using an edge of the tetrahedron.  Can you see it?  The triangle is PVB.  You may know that a point P in the center of an equilateral triangle is twice as far from a vertex as it is from a side, and since the height of a triangle is sqrt(3)/2, the length of PB is sqrt(3)/3.  The length of VB, the hypotenuse of PVB, is 1.  So the length of PV, which is the height of the tetrahedron is sqrt(1-PB^2) = sqrt(6)/3)

If you've mastered that trick of visualization in three dimensions, then you're ready for this month's puzzle.  Imagine that you live in an n-dimensional world, where n is an arbitrarily large number.  In this world, there are many more different degrees of "flatness".  A three-dimensional object would seem very flat to you, if, say, you live in an eleven-dimensional world.  In this world, it would be natural for you to not only imagine, but build and use four-, five-, and six-dimensional equilateroids.  Don't you think a nine-dimensional platonic decahedron would make a lovely ornament on your keychain?

Are you still with me?  Good.

As you go up in dimensions, I'm sure you've noticed that the height of the equilateroid gets smaller.  A one-equilateroid (a line segment) with unit edges has a height of 1.  In two dimensions, the height is sqrt(3)/2.  In three dimensions (an actual tetrahedron) the height is sqrt(6)/3, which (you can check) is a smaller number.

Here's the puzzle:

Imagine you're an n-dimensional creature who really enjoys rational numbers.  As such, you've noticed that a one-equilateroid with unit edges has a rational height.  As you explore your world, finding and building larger-dimensional equilateroids, you serendipitously discover another one with a rational height.  Are these the only two?  Prove it!

Answer: Yes

Solution:

First, a lemma: The distance from the "center" of an n-equilateroid to any vertex (the center-vertex distance) is nh/(n+1), where h is the height of the figure, measured along a line perpendicular to any (n-1)-equilateroidal base to the opposite vertex.

For example, a 2-equlateroid (an equilateral triangle) has a center that is a distance of 2h/3 from any vertex, measured along a line perpendicular to any of the 1-dimensional bases that passes through the opposite vertex.

The proof of this, which I will only sketch here, is to use the definition of the "centroid" (center of balance), which is the ratio of two integrals.  If you integrate along the x-axis perpendicular to the base, and position the vertex at the origin, then (noting that the n-1-dimensional "area" inside the integral is proportional to the n-1st power of x), you'll find the integral works out so very simply that no recollection of even trigonometry is required!)

Second, an interesting phenomenon, which I'll bet many of you have already noticed.  If you write the lengths of the legs of a right triangle as sqrt(a), sqrt(b), and of the hypotenuse as sqrt(c), then you'll see that a+b=c.  For example, consider a 3-4-5 triangle.  Write the lengths of the sides as sqrt(9), sqrt(16), and sqrt(25).  Then you'll see that 9+16=25.

Maybe that's a bit simple.  Now consider triangles in which the hypotenuse is 1.  These come up naturally in all sorts of contexts, and you'll see they come up in the discussion that will follow.  If one leg is sqrt(11/32) then it's obvious using this trick that the other leg is sqrt(21/32).  Just so it doesn't catch you off-guard, I'll warn you now that I will use this quick trick to find the missing leg in the steps that follow.

Now, let's try for some induction, shall we?

A 1-equilateroid has a height of 1, which I will write as sqrt(2/2).  The center-vertex distance (the distance from the centroid to a vertex) is nh/(n+1) = 1/2, which I will write as sqrt(1/4).

In a 2-equilateroid (an equilateral triangle), we draw a right triangle that has as one leg the center-vertex distance of the base, and has an edge of unit length as the hypotenuse.  The center-vertex distance of the base is sqrt(1/4), so the height is sqrt(3/4).  The center-vertex distance is (2/3)sqrt(3/4) = sqrt(2/6)

In a 3-equilateroid (a tetrahedron), we draw a right triangle that has one leg as the center-vertex distance of the base, and has an edge of unit length as the hypotenuse.  The center-vertex distance of the base is sqrt(2/6), so the height is sqrt(4/6).  The center-vertex distance is (3/4)sqrt(4/6) = sqrt(3/8)

Maybe you see the pattern here.

Assume the height of an n-equilateroid is sqrt((n+1)/(2n)).  That's certainly true for n=1,2,3.
The center-vertex distance = (n/(n+1))sqrt((n+1)/(2n))
center-vertex distance = sqrt(n / (2n+2))

So the height of an n+1-equilateroid is sqrt((n+2)/(2n+2))

Now we know the height of an n-equilateroid is sqrt((n+1)/(2n)), so when is this rational?   Whenever n+1 and 2n are both perfect squares.

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