The Magic Tetrahedron

Here's a neat trick involving all the numbers from 1 to 26, and three of the five Platonic Solids, the most basic polyhedral shapes. Let's start with the tetrahedron. A tetrahedron, sometimes called a triangular pyramid, is a shape made up of four corners, four equilateral triangular faces, and six edges. Since each face is a triangle, it also has a total of 12 angles on its four faces. So the tetrahedron has a total of 26 components, (4 corners + 4 faces + 6 edges + 12 angles = 26). These components can be numbered from 1 to 26 in a special way.

The basis of the trick is to use three pairs of numbers in order to create a value for every part of the tetrahedron. The three pairs are: 1 & 2, 3 & 6, 9 & 18. These six numbers will be placed on the six edges of the tetrahedron. Each edge is always directly opposite another edge; that is, if you draw a line through the center of the object starting from one edge, you will always reach another edge on the opposite side.

Select any edge and place the number 1 on it. On the opposite edge place the number 2. Select any of the remaining edges and place the number 3 on it, and then place the number 6 on the opposite edge. On the last two edges place the numbers 9 and 18. Now you're ready to determine all the rest of the numbers, and where they go on the tetrahedron.

Every corner has three edges that meet at it. Add up the value of the three edges and give that number to the corner. Every side has three edges surrounding it. Add up the value of the three edges and give that number to the face. Every angle on the faces is formed by two edges meeting there. Add the value of these two edges and give that number to the angle.

When you are done, you will discover that you have used all the numbers from 1 to 26 without repeating any numbers! There are two possible tetrahedra that can be made this way. Can you find them both?

This trick can also be done on a cube, but without using the angles of the faces. A cube has six faces, eight corners, and twelve edges; 6 + 8 + 12 = 26. Using the same three pairs of numbers we started with above, we can give values to the six faces. Place the number 1 on a face, and the number 2 on the opposite face. Place the number 3 on any of the remaining faces, and the number 6 on the opposite face. Then place the numbers 9 and 18 on the remaining faces.

Just like before, we will add these six numbers together to create all the others. Every edge of a cube is the meeting place of two faces. Add the numbers of these two faces and give that total to the edge. Every corner of a cube is the meeting place of three faces. Add the numbers of these faces together and give that total to the corner. Do this for all parts of the cube, and you will use all the numbers from 1 to 26 without any repetitions. Unlike the tetrahedron, there is only one way to accomplish this feat on a cube.

The technique will also work on the octahedron, which has six corners, eight sides, and twelve edges; 6 + 8 + 12 = 26. In this case the three pairs of opposite numbers are placed on the six corners of the octahedron. As before, place 1 and 2 on opposite corners, 3 and 6 on opposite corners, and 9 and 18 on the remaining corners. Each of the twelve edges of the octahedron connects two corners. Add these two corners to find the number for that edge. Each of the eight triangular sides of the octahedron is surrounded by three corners. Add those three corners together to get the value for that side. When completed, you will again have used all the numbers from 1 to 26 with no repetitions. As with the cube, there is only one way to accomplish this.

The sharp observer will have noticed that the special set of 3 opposite pairs are used for the edges of a tetrahedron, the faces of a cube, and the corners of an octahedron, which may reveal something important about the relationships of points in space. Also, for the die-hard puzzle solver, the solutions to the above figures may be written in Base 3 numerals, with very interesting results.

Guess Your Birthday !

Here's a fun trick to show a friend, a group, or an entire class of people. I have used this fun, mathematical trick on thousands of people since 1963 when I learned it. Tell the person (or class) to think of their birthday...and that you are going to guess it.

Step 1) Have them take the month number from their birthday: January = 1, Feb = 2 etc.
Step 2) Multiply that by 5
Step 3) Then add 6
Step 4) Then multiply that total by four
Step 5) Then add 9
Step 6) Then multiply this total by 5 once again
Step 7) Finally, have them add to that total the day they were born on. If they were born on the 18th, they add 18, etc.

Have them give you the total. In your head, subtract 165, and you will have the month and day they were born on!

How It Works: Let M be the month number and D will be the day number. After the seven steps the expression for their calculation is:

5 (4 (5 M + 6 ) + 9 ) + D = 100 M + D + 165

Thus, if you subtract off the 165, what will remain will be the month in hundreds plus the day!

Technorati Tags: trick , mathematical , birthday

Card Trick

Here's a real clean card trick that is bound to amaze and surprise your friends.

Have a friend shuffle a standard 52-card deck to his satisfaction. The ask him/her to turn over, face up, a pile of twenty-five cards. As they count out the cards to twenty-five, act like you are intensely memorizing every single one of them in order. In truth you are only interested in the 17th card in the pile. Memorize that 17th card..

Turn the twenty-five card pile over, now face down, and set aside.

Take the remaining cards and do the following:

If it is a two through nine card, place it face up, and count out cards up to the number ten. For instance, if you turn up a six, you would place it face up, and then count out four cards coming down, saying "Seven, eight, nine, ten." If you turned up a three, then seven cards, saying "Four, five, six, seven, eight, nine, ten."

If the card you turn up is an ace, ten, or picture, tell your friend that you must have cards 2 through nine for this trick and place the card on the bottom of the deck, face down like the rest of the cards. (Someone told me that aces and pictures work too, but I haven't verified it yet.)

Repeat this procedure until you have completed four columns of "ten counts."

Take whatever remaining cards you have after making the four columns, and place them face down on TOP of the twenty-five card pile you set aside earlier.

Now ask your friend to total up the numbers at the top of the four columns.

Say the total was 23. Counting from the very top of the set-aside-pile-plus-remainder-cards, state you are interested in the 23rd card. Just before turning over the 23rd card, state that it is the 17th card that you memorized at the beginning of the trick!

If you did everything correctly, it works every time! Amazing!

A Fractal Guide to Tic-Tac-Toe

A fractal is a shape that can be divided into parts that are smaller versions of the whole. A genuine fractal such as Sierpinski�s gasket has detailed structure on all scales of magnification: any piece of it, no matter how small, will resemble the whole. A quasi-fractal, in contrast, is an approximation of a true fractal�it has detailed structure over a large but finite range of magnification scales. The patterns of a quasi-fractal do not continue to infinitely fine scales, but because the human eye cannot distinguish such small details, quasi-fractals look convincingly fractal. One of the accomplishments of
Grim and St. Denis was to devise a quasifractal diagram that represents all the possible games of tic-tac-toe. As everyone knows, tic-tac-toe is played on a 3-by-3 grid of squares by two players, X and O. Each player takes turns marking squares, and the first to get three in a row (across, down or diagonally) wins. Traditionally, X goes first, and optimal play always results in a draw. But exactly how many games are possible? At X�s first turn, he chooses among nine squares; then O chooses among eight, and so on. So the
total number of games is 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880.
Here�s how Grim and St. Denis built their quasi-fractal. Start with a big 3-by-3
square grid, and divide each square into a 3-by-3 subgrid [see illustration on opposite
page]. Player X has nine opening moves, corresponding to the positions in the
larger grid. One possible move is that X chooses to mark the top left corner. Find
the 3-by-3 subgrid in the top left corner of the big grid and draw an X in the subgrid�s
top left corner. The subgrid is now a picture of the game after this opening move. Another possibility is that X opens with the bottom center square; to represent this move, find the subgrid in the bottom center square of the big grid and draw an X in the subgrid�s bottom center square. In this way, each of the nine subgrids receives an X in a different subsquare. Now concentrate on the subgrid in the top left corner of the big grid. X�s first
move is already drawn in the top left corner; the other eight subsquares represent possible moves for O. If we just put O�s in each of those subsquares, though, we would have nowhere to put X�s second move. Instead we repeat the trick already used for the opening move: We divide each of the eight unmarked subsquares into a 3-by-3 grid of sub-subsquares, getting eight small tic-tac-toe boards. We put an X in the top left corner of each, to represent X�s opening move. Then we put one of O�s eight possible moves into each of the small tic-tac-toe boards. We can continue in this fashion, recording all the possible moves in subgrids of ever smaller size. At every stage, all the unoccupied squares are subdivided into 3-by-3 grids, and all moves previous to that stage are copied into the cells of those grids. The final figure has a quasi-fractal structure because the rules of the game are recursive: the possible moves at each stage are determined by the moves made before. The geometry of fractals is also recursive: similar shapes repeat on ever smaller scales. The tic-tac-toe figure is a quasi-fractal rather than a true fractal because the game ends after a finite number of moves. Now we turn to logic. The simplest area of conventional mathematical logic, propositional calculus, is concerned with statements whose �truth-value� is either 1, representing true, or 0, representing false. For example, the statement P = �pigs can fly� has a truth-value of 0, whereas Q = �Africa is a continent� has a truth-value of 1. Statements can be combined using various logical operators, such as AND and OR. If P and Q are as above, the statement P AND Q is �pigs can fly, and Africa is a continent.� This statement is false, so the truth-value of P AND Q is 0. The results of applying AND to statements can be summed up in a truth table:
P Q P AND Q
0 0 0
0 1 0
1 0 0
It is also possible to change 0 to 1 and 1 to 0 by applying the operator NOT: that is, NOT P is true if P is false, and vice versa. There are 16 possible truth tables for two statements, representing all the possible ways to put 0�s and 1�s in the table�s final column. We can denote them with successive four-digit binary numbers: 0000, 0001, 0010, 0011 and so on, up to 1111. (In decimal notation, these numbers are 0, 1, 2, 3, ... , 15.) This list leads to another quasi-fractal. To draw it, sketch a 16-by-16 array of squares and add a border above the top row that identifies each column with one of the binary numbers [see illustration on page 86]. Then add a similar border down the left side of the array to enumerate the rows. Choose 16 different colors to correspond to the 16 binary numbers and color the border squares accordingly. Next, choose a logical operator: for example, the Sheffer stroke, which is represented by the symbol |. In computer engineering, the Sheffer stroke is known as NAND, because P | Q = NOT (P AND Q). Its truth table is:
P Q P | Q
0 0 1
0 1 1
1 0 1
1 1 0
Now, for each of the squares in the 16-by-16 array, put the square�s four-digit row number in the first column of the P | Q truth table and put the square�s column number in the table�s second column. Then perform the NAND operations and put the resulting truth-values in the table�s final column. This yields another four-digit binary number. Find the color that corresponds to this number and use it to mark the square in the 16-by-16 array. For instance, consider the square in row 5, column 11. In binary notation, these numbers are 0101 and 1011. Plugging them into the truth table for P | Q yields:
P Q P | Q
0 1 1
1 0 1
0 1 1
1 1 0
The number in the final column is 1110, or 14 in decimal notation. So the square in row 5, column 11 is given the color corresponding to 14. The final product of this laborious process is shown in the illustration. Notice that the green squares, corresponding to the binary number 1111, form a shape very similar to Sierpinski�s gasket! Instead of color-coding the picture, one can also graph the value of each square in a third dimension, as a height given by its decimal number divided by 16. For example, the height of the square in row 5, column 11 would be 14/16 =0.875. These graphs are called value solids. In the value solid for the Sheffer stroke, a gasketlike shape can clearly be seen. The explanation is simple: any formal logical system that involves recursion�whether a game or a truth table�can provide a recipe for drawing quasi-fractals.