Today is the last day of summer*. In my opinion, it is a good time to remember what has happened in the last three months.

* Yes, I know that in some regions it’s already the first day of autumn, but I don’t want to talk about time zones here.

## Space exploration

One of the most memorable events was the Pluto flyby. This event had been highly anticipated, not least because some people tend to call this stone a planet. The closest approach happened on July 14th**, and immediately caused a lot of posts in different places. This culminated a series of wonderful achievements in space exploration. In case you didn’t know, the other elements of this series were reaching a comet and throwing a lander on it as well as directing a spacecraft from Vesta to Ceres.

** I believe that Nasa scientists deliberately adjusted the time so that it landed precisely on my birthday.

## Another birthday present

On July 14th, Wolfram Research released Mathematica 10.2. If the knowledge of Mathematica is represented by a number from 0 to 1, then my knowledge will be somewhere near 1/Graham’s number. As a consequence, I cannot express an educated opinion about the new release. I can, however, tell you about a less recent project by Wolfram Research, namely this website. The site tries to guess what is contained within a given image, and it does its job pretty well, sometimes even telling the name of a depicted person. I encourage you to try it; it’s a lot of fun.

## Pentaquark

Scientists at the LHC have discovered a brand new particle. If you want to read a post about it, go here.

Personally, I have completely ran out of interesting news. If you think you know some interesting recent science-related event, leave a comment below.

Anyway, this article would seem incomplete without an unrelated Math problem. Here it is:

**The problem.** *Does there exist a polyhedron P satisfying these conditions:*

- Every face of P is a polygon with odd number of sides (triangle, pentagon, etc.).
- P has odd number of faces.

And finally, I’m sorry that the article is somewhat rushed, but the summer is coming to end really fast!

In the southern hemisphere it is the last day of winter and in some regions it’s already spring, so it is one of those rare times when it is each of the four seasons somewhere on the earth.

If a polyhedron’s faces all have an odd number of sides, its dual has an odd number of faces meeting at every vertex. If you explode the faces of the dual outward, each vertex touches only one face and thus one vertex becomes N vertices, where N is the number of faces that the vertex touched.

Since N is always odd, each vertex becomes an odd number of vertices. So if the number of vertices was odd, it would still be odd after the faces of the dual got exploded. But as each vertex and each edge touch only one face now, the number of vertices must be the same as the number of edges, because every polygon has the same number of vertices as edges.

Since two polygons met at every edge, after explosion each edge becomes two edges. So the number of edges must be an even number. But this means the number of vertices must be an even number, and therefore the number of vertices in the un-exploded dual polyhedron (and by definition, the number of faces in the original polyhedron) must be even.

So there is no polyhedron with an odd number of faces, where each face has an odd number of sides.

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Thanks for publishing your solution.

Actually, taking the dual was not necessary. You could just write something like this:

Explode the polyhedron. Every edge became two edges, so the number of edges is now even. The number of vertices is now the sum of an odd number of odd numbers (we just sum the number of vertices in a face over all faces), so it must be odd. But the number of edges must now equal the number of vertices, resulting in a contradiction.I thought along the same lines when I first encountered this problem. Similar solutions appear in many textbooks.

By the way, I must assure you that your solution is

stunningly incorrect, and indeed it is possible to find a polyhedron satisfying the conditions.LikeLike

Does the henagonal henahedron count? It can exist only as a spherical projection and has 1 face and 1 vertex. The face has one vertex, but the shape has no edges. However as it has a schlafli symbol of {1,1} we could still call this face a henagon.

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No, it doesn’t count. It’s cheating.

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If “number of sides on a face” is defined to be

“number of distinct edges of the polyhedron that belong to the face”

rather than

“number of edges traced around the face to get back to the starting point”

this works: http://i.imgur.com/oF4rE9K.png

It has seven faces. Six of them are triangles, and the seventh is a heptagon that touches itself, or an octagon if the edge at which it meets itself is counted as two sides.

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Well… When I published this problem, I wasn’t expecting that seemingly simple terms like “polyhedron” and “side” would cause

thatmuch controversy. Now, however, this discussion is quickly approaching the scope of this board. For that is not the best possible outcome, I would now like to present the solution I was hoping to hear and thus end the argument.So, you start by taking a tetrahedron and dividing one of its faces into two triangles with a median (or any other line connecting the vertex and the opposite edge). One of these triangles then serves as a base for another tetrahedron. That’s all – it can be verified that the shape has 7 triangular faces.

To reiterate, I do not want to engage in a religious war, and so I am not going to claim that my solution is better or worse than yours. On the contrary, I will peacefully put all solutions into one small picture:

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