7394
/www/nrich/html/content/id/7394/
1
2011-02-01T00:00:01
<mdoxml version="1.0">In the problem <a href="http://nrich.maths.org/7346&part=">Dicey
Decisions</a>, we encouraged you to consider the possible edge
totals by adding up the numbers that meet on the different edges of
a six-sided die.<br></br>
<span style="font-style: italic;">If you haven't already done
this, why not try now?</span><br></br>
<br></br>
Imagine that instead of a six-sided die we had a dodecahedron
numbered 1-12.<br></br>
There are different ways to arrange the numbers from 1-12. A
standard six-sided die has opposite faces that sum to 7, so perhaps
our dodecahedral die should have opposite faces that sum to
13.<br></br>
<br></br>
<span style="font-weight: bold;">Can you create a net for a
dodecahedral die whose opposite faces sum to 13?</span><br></br>
<br></br>
For the six-sided die, the edge totals were distributed like
this:<br></br>
<table border="1">
<tr>
<td style="text-align: center;">Edge total</td>
<td style="text-align: center;"> 3 </td>
<td style="text-align: center;"> 4 </td>
<td style="text-align: center;"> 5 </td>
<td style="text-align: center;"> 6 </td>
<td style="text-align: center;"> 7 </td>
<td style="text-align: center;"> 8 </td>
<td style="text-align: center;"> 9 </td>
<td style="text-align: center;"> 10</td>
<td style="text-align: center;"> 11</td>
</tr>
<tr>
<td style="text-align: center;">Frequency</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;">2</td>
<td style="text-align: center;">2</td>
<td style="text-align: center;">0</td>
<td style="text-align: center;">2</td>
<td style="text-align: center;">2</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;">1</td>
</tr>
</table>
<br></br>
The mean edge total is 7, and the edge totals are distributed
symmetrically about the mean.<br></br>
<br></br>
<span style="font-weight: bold;">What is the mean edge
total for your dodecahedral die?</span><br></br>
<span style="font-weight: bold;">Are the edge totals distributed
symmetrically?</span><br></br>
<br></br>
Ignoring rotations and reflections, there is only one way to number
a cube to create a six-sided die with the constraint that opposite
faces sum to 7, but there are multiple ways to create a
dodecahedral die with opposite faces that sum to 13.<br></br>
<br></br>
<span style="font-weight: bold;">Can you make any general
statements about which dodecahedral dice will have edge totals with
a symmetric distribution? Can you prove your
statements?</span><br></br>
<br></br>
For the six-sided die, the corner totals were also distributed
symmetrically. Will the same be true for the corner totals of a
dodecahedral die?<br></br>
<br></br>
Now use your insights to make and justify some statements about the
edge and corner totals of an icosahedral (20-sided) die with
opposite faces that sum to 21.<br></br>
<br></br></mdoxml>
<mdoxml version="1.0"><p>Here's an example of a net for a dodecahedral die with the opposite numbers summing to 13:</p>
<p style="text-align: center;"><mdo:image src="dodecahedralnet2.png" style="width: 300px; height: 406px;"></mdo:image></p>
<p>For such a dodecahedral die with the numbers 2,3,4,5,6 around 1, and opposite faces summing to 13, the edge totals are:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 500px;">
<tbody>
<tr>
<td style="text-align: center;"><strong>Edge 1</strong></td>
<td style="text-align: center;"><strong>Edge 2</strong></td>
<td style="text-align: center;"><strong>Sum</strong></td>
<td style="text-align: center;"> </td>
<td style="text-align: center;"><strong>Edge 1</strong></td>
<td style="text-align: center;"><strong>Edge 2</strong></td>
<td style="text-align: center;"><strong>Sum</strong></td>
</tr>
<tr>
<td>1</td>
<td>2</td>
<td>3</td>
<td> </td>
<td>1</td>
<td>3</td>
<td>4</td>
</tr>
<tr>
<td>1</td>
<td>4</td>
<td>5</td>
<td> </td>
<td>1</td>
<td>5</td>
<td>6</td>
</tr>
<tr>
<td>1</td>
<td>6</td>
<td>7</td>
<td> </td>
<td>2</td>
<td>3</td>
<td>
<p>5</p>
</td>
</tr>
<tr>
<td>2</td>
<td>6</td>
<td>8</td>
<td> </td>
<td>2</td>
<td>9</td>
<td>11</td>
</tr>
<tr>
<td>2</td>
<td>8</td>
<td>10</td>
<td> </td>
<td>3</td>
<td>4</td>
<td>7</td>
</tr>
<tr>
<td>3</td>
<td>4</td>
<td>7</td>
<td> </td>
<td>3</td>
<td>8</td>
<td>11</td>
</tr>
<tr>
<td>3</td>
<td>7</td>
<td>10</td>
<td> </td>
<td>4</td>
<td>5</td>
<td>9</td>
</tr>
<tr>
<td>4</td>
<td>7</td>
<td>11</td>
<td> </td>
<td>4</td>
<td>11</td>
<td>15</td>
</tr>
<tr>
<td>5</td>
<td>6</td>
<td>11</td>
<td> </td>
<td>5</td>
<td>11</td>
<td>16</td>
</tr>
<tr>
<td>5</td>
<td>10</td>
<td>15</td>
<td> </td>
<td>6</td>
<td>10</td>
<td>16</td>
</tr>
<tr>
<td>6</td>
<td>9</td>
<td>15</td>
<td> </td>
<td>9</td>
<td>8</td>
<td>17</td>
</tr>
<tr>
<td>9</td>
<td>10</td>
<td>19</td>
<td> </td>
<td>9</td>
<td>12</td>
<td>21</td>
</tr>
<tr>
<td>8</td>
<td>7</td>
<td>15</td>
<td> </td>
<td>8</td>
<td>12</td>
<td>20</td>
</tr>
<tr>
<td>7</td>
<td>11</td>
<td>18</td>
<td> </td>
<td>7</td>
<td>12</td>
<td>19</td>
</tr>
<tr>
<td>11</td>
<td>10</td>
<td>21</td>
<td> </td>
<td>11</td>
<td>12</td>
<td>23</td>
</tr>
</tbody>
</table>
<p><br></br>
If we plot the frequency distribution for the possible scores for this die we get:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 400px;">
<tbody>
<tr>
<td><strong>Score</strong></td>
<td><strong>Frequency</strong></td>
<td> </td>
<td><strong>Score </strong></td>
<td><strong>Frequency</strong></td>
</tr>
<tr>
<td>3</td>
<td>1</td>
<td> </td>
<td>13</td>
<td>0</td>
</tr>
<tr>
<td>4</td>
<td>1</td>
<td> </td>
<td>14</td>
<td>0</td>
</tr>
<tr>
<td>5</td>
<td>2</td>
<td> </td>
<td>15 </td>
<td>4</td>
</tr>
<tr>
<td>6</td>
<td>1</td>
<td> </td>
<td>16</td>
<td>2</td>
</tr>
<tr>
<td>7</td>
<td>2</td>
<td> </td>
<td>17</td>
<td>1</td>
</tr>
<tr>
<td>8</td>
<td>1</td>
<td> </td>
<td>18</td>
<td>1</td>
</tr>
<tr>
<td>9</td>
<td>1</td>
<td> </td>
<td>19</td>
<td>2</td>
</tr>
<tr>
<td>10</td>
<td>2</td>
<td> </td>
<td>20</td>
<td>1</td>
</tr>
<tr>
<td>11</td>
<td>4</td>
<td> </td>
<td>21</td>
<td>2</td>
</tr>
<tr>
<td>12</td>
<td>0</td>
<td> </td>
<td>22</td>
<td>1</td>
</tr>
<tr>
<td> </td>
<td> </td>
<td> </td>
<td>23</td>
<td>1</td>
</tr>
</tbody>
</table>
<p>This is a symmetric distribution with a mean score of 13. It turns out all possible dodecahedral dice with faces summing to 13 have this property. Consider swapping two of the faces, say 2 and 3, which will also swap 11 and 10 on the opposite side of the die. The sum of the edge between 2 and 3, and 11 and 10 will stay the same, however the other edges of these faces will either increase
or decrease by 1. The edge between faces A and B, say, will change in the opposite direction to the edge between the faces (13-A) and (13-B), as if A, say, increases by 1, (13-A) will decrease by 1. It follows from this that the distribution of the row sums will remain symmetric. </p>
<p>In general, you can prove that swapping two arbitrary faces can be done as the product of swapping pairs of adjacent faces, and as we've just shown the simpler process of just swapping one pair of faces leaves the distribution symmetric, we've therefore shown that all swaps will. <br></br>
<br></br>
For our example, we have the corner sums:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 500px;">
<tbody>
<tr>
<td><strong>Faces</strong></td>
<td><strong>Sum</strong></td>
<td> </td>
<td><strong>Faces </strong></td>
<td><strong>Sum</strong></td>
<td> </td>
<td><strong>Faces</strong></td>
<td><strong>Sum</strong></td>
<td> </td>
<td><strong>Faces</strong></td>
<td><strong>Sum</strong></td>
</tr>
<tr>
<td>1,2,3</td>
<td>6</td>
<td> </td>
<td>2,3,8</td>
<td>13</td>
<td> </td>
<td>4,5,11</td>
<td>20</td>
<td> </td>
<td>7,8,12</td>
<td>27</td>
</tr>
<tr>
<td>1,3,4</td>
<td>8</td>
<td> </td>
<td>2,6,9</td>
<td>17</td>
<td> </td>
<td>4,7,11</td>
<td>22</td>
<td> </td>
<td>7,11,12</td>
<td>30</td>
</tr>
<tr>
<td>1,4,5</td>
<td>10</td>
<td> </td>
<td>2,8,9</td>
<td>19</td>
<td> </td>
<td>5,6,10</td>
<td>21</td>
<td> </td>
<td>8,9,12</td>
<td>29</td>
</tr>
<tr>
<td>1,2,6</td>
<td>9</td>
<td> </td>
<td>3,4,7</td>
<td>14</td>
<td> </td>
<td>5,10,11</td>
<td>26</td>
<td> </td>
<td>9,10,12</td>
<td>31</td>
</tr>
<tr>
<td>1,5,6</td>
<td>12</td>
<td> </td>
<td>3,7,8</td>
<td>18</td>
<td> </td>
<td>6,9,10</td>
<td>25</td>
<td> </td>
<td>10,11,12</td>
<td>33</td>
</tr>
</tbody>
</table>
<p>These have an average of 19.5, and are also symmetrically distributed about that point. The same argument given for the edges remaining symmetric under the exchange of faces can be adapted to prove the claim that the corner sums remain symmetrically distributed when pairs of faces are swapped. </p>
<p> </p>
<p>The important fact throughout these arguments is that the opposite faces always add up to 13. If this wasn't the case, then the distribution would not remain symmetric under the swapping of all pairs of faces. </p>
<p> </p></mdoxml>
<mdoxml version="1.0"><br></br>
<h3>Why do this problem?</h3>
This problem offers students the opportunity to consider the
symmetry of dodecahedra, and to develop insights from reasoning
about dodecahedra to help them to analyse symmetry in
icosahedra.<br></br>
<h3>Possible approach</h3>
<div>This problem follows on from <a href="http://nrich.maths.org/7346&part=">Dicey
Decisions</a>.</div>
<div>The problem could be given to students to work on
independently (perhaps as a homework task) but could also be used
as a whole class activity as follows:</div>
<div>Start by making sure students are happy with the idea that
opposite faces of a six-sided die sum to 7, and then challenge them
to come up with a dodecahedral die where the opposite faces sum to
13. Some students may wish to visualise the die, some may be happy
working with a two-dimensional representation, and some may prefer
to create a net and actually make the die in three
dimensions.</div>
<div>There is opportunity for rich discussion about the number of
distinct dodecahedral dice it is possible to make (with the
constraint that opposite faces sum to 13), and different groups of
students could be encouraged to work with distinct dice.</div>
<div> </div>
<div>Once students have created (or visualised) their die, ask them
to create a frequency table showing the frequency with which the
different edge scores occur.</div>
<div>"Does everyone get a symmetric distribution?"</div>
<div>"Are you surprised that everyone got a symmetric distribution
even if they started with different dice?"</div>
<div>"Can you explain why it was symmetric?"</div>
<div>This could lead into rich discussion about the symmetry of the
distribution and how it relates to the constraint that opposite
faces sum to 13.</div>
<div> </div>
<div>Finally, once students have explanations of where the symmetry
in the distribution comes from, they could be challenged to do a
similar analysis for corner totals of the dodecahedral die, or for
edges or corner totals on an icosahedral die.</div>
<h3>Key questions</h3>
<div>If a face of the die is numbered $n$, what would the opposite
face be numbered?</div>
<div>If adjacent faces were numbered $n$ and $m$ (with an edge
total between them of $n+m$, what would the opposite edge total
be?</div>
<br></br>
<h3>Possible extension</h3>
<div>Write up a really clear proof of the result concerning the
symmetric distribution of the edge totals.</div>
<div> </div>
<div>I have a 100-sided die. To the eye, it looks pretty regular,
but it can't be exactly regular because there is no such thing as a
100 sided Platonic solid. Consider the issues which might arise in
an analysis of the various totals.</div>
<br></br>
<h3>Possible support</h3>
Spend time considering the edge totals of the six-sided die to make
sense of the symmetry there before working on the
dodecahedron.<br></br>
<br></br>
<br></br></mdoxml>
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