Math CentralQuandaries & Queries


Question from Joel, a teacher:

We've been working on this problem diligently and can't seem to come up with the answer book's answer. We think it may be wrong, yet want to check it with an expert. Here goes.
The school's new soccer balls are covered with 64 regular hexagonal panels. Each hexagon measures 2 inches between opposite corners and 1.5 inches between opposite sides. What is the total surface area of the soccer ball?

We have two responses for you

Who in the name of Euclid came up with THAT question? From the existence of an answer book, it sounds as if it was in an official textbook. Say it ain't so, Joel...

Number one, those dimensions are incompatible. If a regular hexagon is 2" between corners, it is exactly sqrt(3) ~ 1.732... inches between opposite sides. 1.5 inches isn't even close.

Number two, no soccer ball can possibly be made up of only regular hexagonal panels. There is a theorem of Euler's that says that for any convex polyhedron, F+V = E+2 where there are F faces, V vertices, and E edges. (You can find a nice proof of this, easy enough for your class to follow, on Wikipedia:

Now, if the soccer ball had only regular hexagon panels, there would be 3 edges meeting at every vertex and (always!) 2 vertices on each edge. Thus E = 3/2 V.
Also, there would be three hexagons at a vertex (more than that goes saddle-shaped, less goes flat!) and six vertices per hexagon; so V = 2F, and E = 3F. But then F+2F = 3F + 2
which is impossible.

So not only is the answer book's answer wrong, the textbook's question is wrong - twice! Perhaps there are not many mistakes like this in the textbook. If there are many, the book should not be in use in any school - and you should tell the school board this.

Good hunting!


In a regular hexagon (all sides and angles the same), all interior angles are 120 degrees (see Wikipedia). Draw a regular hexagon with a diagonal joining the left corner of the bottom with the opposite corner. Let this diagonal have length two as suggested below. Draw a line from the top end of the diagonal to the other corner of the base below it, forming a right triangle with angles 90,60,60. Call the height of this triangle d. The sine of 60 degrees is equal to d/2. Solving this gives d=sqrt(3). Thus a regular hexagon with a diagonal length of 2 has distance sqrt(3) between opposite sides. It is not possible to have a regular hexagon with diagonal length 2 and height 1.5. The problem below states that the distance between any two opposite corners is 2, the same for all pairs of opposite corners which would make it a regular hexagon. This is an impossibility. No such soccer ball exists!

Thx, L. Dame

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