Comments on: My Favorite Math Party Trick http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/ Geometry, Topology, Categories, Groups, Physics, . . . Everything Sat, 19 Jul 2008 10:29:30 +0000 http://wordpress.org/?v=MU By: isallaboutmath http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2639 isallaboutmath Sun, 20 Jan 2008 00:06:00 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2639 for a complex number proof of Van Aubel's theorem see http://foxmath.blogspot.com/2007/08/van-aubels-theorem-with-complex-numbers.html for a complex number proof of Van Aubel’s theorem see

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

]]> By: isallaboutmath http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2638 isallaboutmath Sat, 19 Jan 2008 23:57:48 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2638 You mentioned that you copy the image from Mathworld. If you have knowledge of Metapost see http://blog.isallaboutmath.com/2007/12/01/creating-beautiful-images/ 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 You mentioned that you copy the image from Mathworld. If you have knowledge of Metapost see

http://blog.isallaboutmath.com/2007/12/01/creating-beautiful-images/

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

]]> By: mousomer http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2618 mousomer Tue, 15 Jan 2008 08:50:12 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2618 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. 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.

]]> By: John Armstrong http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2616 John Armstrong Tue, 15 Jan 2008 06:18:52 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2616 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. 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.

]]> By: James Lefevre http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2615 James Lefevre Tue, 15 Jan 2008 05:17:08 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2615 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. 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.

]]> By: Greg Muller http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2611 Greg Muller Mon, 14 Jan 2008 22:07:11 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2611 Mousomer, Saying a vector is the average of two other vectors is different than the <b>length</b> 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. 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.

]]> By: mousomer http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2608 mousomer Mon, 14 Jan 2008 21:36:31 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2608 "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? “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?

]]> By: John Armstrong http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2602 John Armstrong Mon, 14 Jan 2008 14:56:16 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2602 @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. @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.

]]> By: salobrena http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2598 salobrena Mon, 14 Jan 2008 09:12:44 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2598 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? 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?

]]> By: Rahul Sharma http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2597 Rahul Sharma Mon, 14 Jan 2008 03:59:03 +0000 http://cornellmath.wordpress.com/2008/01/12/my-favorite-math-party-trick/#comment-2597 Very Cool! Indeed! Very Cool! Indeed!

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