## My Favorite Math Party Trick

I have a minor personal triumph to relate today.  So, I try to have several cute, quick math puzzles/facts on hand to bust out at parties, because I am cool like that.  The most popular of these is a participation-based trick that goes as follows (feel free to play along at home):

Take a pen and paper and draw a quadrilateral.  There are no restrictions (it can be concave or self-intersecting), but don’t make it too close to the sides of the paper.  Now, for each edge, draw the square containing that edge that is outside the quadrilateral.  Put a dot in the center of each of the four squares, and draw a line connecting opposite dots, ie, those that came from opposite edges.

The Punchline: The lines you just drew are the same length, and perpendicular.

(If you lack pen and paper, theres an applet here)

It works pretty well because it isn’t very sensitive to sloppy geometry on the part of the artist; so you will pretty consistantly get ‘perpendicular-looking’ lines.  Also, I often bend the truth a bit and attribute it to Napolean Bonaparte, even though he proved something different but closely related.  The usual attribution of this result is to van Aubel, who to my knowledge conquered very little of Europe.

I’ve been using this for a couple of years now, and periodically I attempt to find a nice geometric proof without passing to coordinates.  Such a proof eluded me up to last night, when I came up with a reasonably nice vector-based proof.  Its a little cheap since using vectors isn’t totally different from passing to coordinates, but in my mind a geometric proof is one which can be done only with pictures (though I will use words for laziness’ sake).  For the record, I was unaware of van Aubel’s proof until this morning, which is a more traditional geometric proof, but a little bit more indirect.

The first step is to notice that the vector connecting the midpoints of opposite sides is the average of the vectors composing the other two sides:

So, in the picture, the red vector is the average of the two purple vectors.  Of course, instead of averaging them, we can cut each of them in half and then add them.

Now, the red vector is the sum of the two purple ones.

Next, lets rotate the purple vectors 90 degrees to the right, and position them so they go from the midpoint of the squares to the midpoints of the edges:

We haven’t changed the length of the vectors at all, and we’ve rotated them both the same amount, so their sum is still the same length as the red vector but perpedicular.

We can apply identical logic to the other vector connecting opposite midpoints:

We have that the sum of the purple vectors is the same length and perpendicular to the red vector, and that the sum of the pink vectors is the same length and perpendicular to the blue vector.  Hence, the purple vectors plus the blue vector is the same length and perpendicular to the pink vectors plus the red vector.  QED

Its not hard to see the coordinate-based version of this proof beneath the surface, especially if one use complex coordinates, where rotation by 90 degrees is multiplication by $i$.  In fact, I think this theorem is a pretty good illustration of the power and usefulness of cartesian coordinates.

(Note: The images above are based on a picture I stole from Mathworld)

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### 32 Responses to “My Favorite Math Party Trick”

1. leuffi Says:

A very simple proof can be given using affine transformations of the plane: a 90 degree rotation which exchanges the line segments is given by the product of two affine transformations – first rotate by 45 degrees and homotet (is that a word?) by sqrt(2) at a(n appropriate) vertex, second rotate by 45 degrees and homotet by sqrt(2)^(-1) at the diagonal vertex.

2. Greg Muller Says:

Leuffi,
I’m not totally sure I follow your proof. Which vertex is the diagonal vertex? Also, I think the word you are looking for is ‘homothet’, as in what you do to make something into a homothetic thing, but why use that when you can say ‘scale’?

• steve gray Says:

Napoleon and Van Aubel can both be proved the same way. Without overdoing the details, the concept is to displace one vertex of the original n-gon (n=3 or 4) and note that the resulting displacements of two vertices of the first constructed n-gon are related to each other by being equal in length and at a certain angle g to each other where g=360/n. Use this displacement n-2 times and you can get a regular n-gon. If this is not clear, I can probably rewrite it to be even less clear.

3. Greg Muller Says:

Oh, random question for smart people who might be reading this. The resources online frequently mention that both this and Napolean’s theorem are special cases of the Petr-Douglas-Neumann Theorem. However, the internet seems to be lacking in facts about this theorem. Anyone know it?

• steve gray Says:

Greg – See the Math Monthly, March 2003, where I published a big generalization of the Petr-Douglas-Neumann theorem. I have an even further generalization which is not published.

Know it? I love the Petr-Douglas-Neumann Trio! That one guy was gooned. Oh… theorem? No, don’t know it. Are they good?

5. Heinrich Says:

That’s a very nice proof. Despite being with vectors, it’s not with coordinates: you choose vectors that arise naturally and not an arbitrary and ill-fitting basis for monstrous calculations.

But you’re too quick at the last step, it’s not clear that

purple * red = 0 /\ pink * blue = 0 /\ |purple|=|red| /\ |pink|=|blue|
=> (purple + blue) * (pink + red) = 0

In fact, this is wrong in 3 dimensions (pink=purple=e1, red=e2, blue=e3) , so some more insight is required here. The insight is of course that

i*purple = red /\ pink = i * blue
=> i * (purple + blue) = red + pink

Note that orientation matters, pink = -i* blue is not enough.

PS: How do I use LaTeX in comments?

6. missanderson08 Says:

that is cool how did you come up with that text me back please

7. Greg Muller Says:

Heinrich,
Thats a good point, that the way I say it seems a bit quick. I wrote ‘is perpendicular’, but in my head it was always ‘is a 90 degree rotation’, and specifically, the same 90 degree rotation each time. While perpendicularity isn’t necessarily additive, linear maps like rotation most certainly are. Of course, the quickest way to label this rotation is as i, as you point out.

Latex in comments is done with a $, then the word ‘latex’ with no space or quotation marks, followed by a space, the Latex code, and another$ symbol. I’d write it out, but I don’t know the syntax for a literal environment in wordpress.

8. uthek Says:

Without the pictures I would have just been confused. *has no buisness looking at anything math related* n_n;

9. jd2718 Says:

I didn’t even know about Von Aubel until this summer. Andy Fox had a way cool post on using complex numbers (do I hear “vectors”?) in his proof.

And now, a new party trick! Thanks!

10. jd2718 Says:

van Aubel. OK, I still have a lot to learn. Still very cool.

11. leuffi Says:

I didn’t want to give names to the various vertices, but I guess I cannot get away that easily. Let me be more precise.

Let $T(x, \theta, r)$ be the rotation by the angle $\theta$ and the dilation by $r$ at the point $x$, i.e. the map given by $z \mapsto e^{i\theta}r(z - x) + x$. Let $A_1, \dots, A_4$ be the initial four points. Then $B_i = T(A_{i+1}, \pi/4, \sqrt(2)^{-1})A_i$ are the four points constructed, here $A_5 = A_1$. We need to show that $B_1B_3$ is perpendicular to $B_2B_4$ and equal in length.

The composition $g = T(A_3, \pi/4, \sqrt(2)^{-1}) \circ T(A_1, \pi/4, \sqrt(2))$ maps $B_1 \mapsto A_2 \mapsto B_2$. On the other hand, it is not difficult to see that $g = T(A_1, \pi/4, sqrt(2)^{-1}) \circ T(A_3, \pi/4, \sqrt(2))$ which maps $B_3 \mapsto A_4 \mapsto B_4$. Being an affine transformation with no dilation, $g$ is a rotation by $\pi/4 + \pi/4$ (around its unique fixed point) and it maps $B_1B_3$ to $B_2B_4$. QED.

12. leuffi Says:

Arh, sorry about the messy TeX. Only if I could do a preview before I post..

13. Hallowed Says:

That’s a cool deal. I’m not sure it would go over well at any party I’ve been at…. But still, very cool.

14. Rahul Sharma Says:

Very Cool! Indeed!

15. salobrena Says:

This is cool.. yer a math guy, I have a question.. I know a card trick that I cannot figure out but it works every time..it is obviously math based but I can’t get it. Maybe you can explain it.. if you don’t know it, you will have another party trick up your sleeve.
Take one standard deck of cards. remove the jokers and spares.
Separate the suits and place in order ace high
Stack the suits red black red black so that the deck is back together again.
Now cut the deck 13 times. No more, no less.
Now deal out the cards into 13 piles. In each pile will be 4 of the same from each suit..not only that, they will all be in the same order ie.. hearts,spades,diamonds clubs..or what ever your red black combos were..
Why does this work?

16. John Armstrong Says:

@salobrena: what do you mean by “cut the deck”? Because using what I usually think of as “cutting” (pick a random number of cards from the top and put them on the bottom) it clearly doesn’t work.

17. mousomer Says:

“The first step is to notice that the vector connecting the midpoints of opposite sides is the average of the vectors composing the other two sides”

Is it? Sure, if we’re speaking about the line in the middle between the parallel edges of a paralelogram. But for a general quadrangle this isn’t true.
A counterexample: take a paralelogram ABCD (AB || CD). Let G be the middle point of AB, H the middle point of CD. Now you claim that GH is the average of BC and AD. This isn’t true. One can lengthen AB indefinitely, and as long as it’s symmetric about point G, GH stays the same length, while both BC and AD go to infinity.

Am I missing something trivial?

18. Greg Muller Says:

Mousomer,
Saying a vector is the average of two other vectors is different than the length of a vector being the average length of two other vectors. For instance, if I have two vectors of the same length and pointing opposite directions, their average is zero since their sum is zero. However, their average length is not zero.

19. James Lefevre Says:

sabobrena: The thirteen cuts is a red herring. This trick works because two cuts are the same as one. You can see this by thinking of a cut as adding x (mod 52) to the position of each card, or alternatively by noting that the two cards separated by the first cut are put together again by the second cut. Thus 13 or any other number of cuts are equivalent to just one. Then when you deal them out it is just like dealing the original deck, but starting from a different position. Each pile will have 4 of the same number. The suits will be in the same cyclic order, but probably rotated – eg if the original order was HSDC, a pile could be SDCH or CHSD.

20. John Armstrong Says:

I think James is right. I read the “deal” in a different way. He’s saying deal thirteen piles of four, one at a time. I was thinking to deal four cards, then another four, then.. and so on.

21. mousomer Says:

Oh, silly me. I should have read this more carefully. So perpendicularity of the vectors is the crucial element in proving the lines have same lengths. Yes, this proof is insightful. Thanks.

You mentioned that you copy the image from Mathworld. If you have knowledge of Metapost see

were you can find a place on the net that will allow you to create images.

Also here is a link to a nice article on the Van Aubel theorem

http://mysite.mweb.co.za/residents/profmd/aubel2.pdf

for a complex number proof of Van Aubel’s theorem see

http://foxmath.blogspot.com/2007/08/van-aubels-theorem-with-complex-numbers.html

24. amca01 Says:

The Petr-Douglas-Neumann Theorem. is described at Mathworld as the Petr-Neumann-Douglas Theorem: http://mathworld.wolfram.com/Petr-Neumann-DouglasTheorem.html.

Very nice proof of van Aubel’s theorem! – elegant, visually appealing, and intuitive – just how I like my geometry.

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