<?xml version="1.0" encoding="ISO-8859-1" ?>
  <resource>
  <id>1929</id>
  <path>/www/nrich/html/content/97/11/15plus1/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;p&gt;&lt;mdo:image alt=&quot;Dodecahedron&quot; float=&quot;left&quot; src=&quot;5d.gif&quot; style=&quot;border-width: 0px; border-style: solid; height: 133px; float: left; width: 131px; margin-left: 10px; margin-right: 10px;&quot;&gt;&lt;/mdo:image&gt;A and B are the centres of two opposite faces of a regular solid dodecahedron.&lt;/p&gt;
&lt;p&gt;What is the shortest distance between A and B on the surface?&lt;/p&gt;
&lt;p&gt;&lt;br&gt;&lt;/br&gt;
 &lt;/p&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;p class=&quot;editorial&quot;&gt;There are two interpretations of this problem,
the shortest route between opposite faces of the dodecahedron going
across the outside of the solid and the shortest route boring a
hole through the middle.&lt;/p&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;While few tackled this challenging
problem, Luke, of Madras College found the shortest distance
through the centre. In his solution he used the centres of three
spheres: the circumsphere that passes through all the vertices of
the dodecahedron, the midsphere touching the midpoints of the
edges, and the insphere which touches the opposite faces of the
dodecahedron at their centres. Luke's answer is 2.23 approximately.
Before you look at Luke's solution here you might like to try to
try it for yourself.&lt;/span&gt;&lt;/p&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p&gt;On the surface this must be the distance between A and B on the
net of the solid.&lt;/p&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;581&quot; width=&quot;421&quot; src=&quot;dod2.gif&quot; alt=&quot;&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
Let the lengths of the edges be 2 units.&lt;br&gt;&lt;/br&gt;
AB = 2 x AO&lt;br&gt;&lt;/br&gt;
AO = ON + AN&lt;br&gt;&lt;/br&gt;
and AN = AP + PN.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Working to 4 decimal places,&lt;br&gt;&lt;/br&gt;
AP = 1/cos 54° = 1.7015&lt;br&gt;&lt;/br&gt;
PN = 2 sin72° = 1.9021&lt;br&gt;&lt;/br&gt;
NM = 2 cos72° = 0.6180&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
AN = 1/cos54° + 2sin72° = 3.6034&lt;br&gt;&lt;/br&gt;
NO = 2cos72° + 1 = 1.6180&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
AO² = (1.6180)² + (3.6034)²&lt;br&gt;&lt;/br&gt;
AO = 3.9499 AB = 7.8999&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;p&gt;How do you find the shortest distance between two points on the net of the dodecahedron?&lt;/p&gt;
&lt;p&gt; &lt;/p&gt;
&lt;p&gt;Does the distance on the surface change when the flat net is folded into the shape of the dodecahedron?&lt;/p&gt;&lt;/mdoxml&gt;</clueXML>
  <canonXML/>
  <end_user_role>2</end_user_role>
  <difficulty>5</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>0</keystage3>
  <keystage4>0</keystage4>
  <keystage4plus>1</keystage4plus>
  <title>The Dodecahedron</title>
  <description>What are the shortest distances between the centres of opposite
faces of a regular solid dodecahedron on the surface and through
the middle of the dodecahedron?</description>
  <spec_group>Trigonometry
    <specifier>Sine</specifier>
  </spec_group>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Pythagoras' theorem</specifier>
  </spec_group>
  <spec_group>Trigonometry
    <specifier>Cosine</specifier>
  </spec_group>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Regular polygons</specifier>
  </spec_group>
  <spec_group>3D Geometry, Shape and Space
    <specifier>Dodecahedra</specifier>
  </spec_group>
  <spec_group>Admin
    <specifier>Long problems</specifier>
  </spec_group>
</resource>