Proximity

326 /www/nrich/html/content/00/09/15plus3/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <table border="0"> <tbody> <tr> <td> <div>We are given a regular icosahedron having three red vertices. Show that it has a vertex that has at least two red neighbours.</div> <div>If you're stuck, try this <a href="/content/00/09/15plus3/icosahedron.doc" style="font-style: italic;">proof sorting activity</a>.</div> </td> <td><mdo:image alt="icosahedron" height="346" src="icosahedron.jpg" width="418"></mdo:image></td> </tr> </tbody> </table> <div class="framework">This problem was taken from the Hungarian magazine KoMaL. There are many other challenging problems in English on <a href="http://www.cs.elte.hu/komal/index.e.html">the KoMal website</a>.</div></mdoxml> <mdoxml version="1.0"> Solution to Proximity by Tony and John, State College Area High School, PA, USA. <table border="0"> <tbody> <tr> <td><mdo:image height="346" width="418" alt="icosahedron" src="icosahedron.jpg"></mdo:image></td> <td> <div>We give a proof by contradiction.</div> <div>Suppose no vertex has more than one red neighbour.</div> <div>Without loss of generality say the vertex at the top is red. Then none of the five vertices around the upper horizontal pentagon, which are neighbours of the top vertex, are red. This is because if any of them were red then there would be two vertices each having two red neighbours which is not allowed.</div> </td> </tr> </tbody> </table> Now consider the five vertices around the lower horizontal pentagon. Only one of these can be red because if two were red then there would be a vertex with two red neighbours which is not allowed. The argument shows that of the eleven vertices discussed only two can be red. We know that there is a third red vertex so it must be the vertex at the bottom. However if the bottom vertex is red then there will be two vertices on the lower horizontal pentagon having two red neighbours which is not allowed. We have reached a contradiction. So the assumption that no vertex has more than one red neighbour is false. We have proved that there is at least one vertex with two red neighbours. </mdoxml> <mdoxml version="1.0"> Why do this problem? <a href="http://nrich.maths.org/public/viewer.php?obj_id=326&part=index"> The result</a> can be proved by an argument using proof by contradiction and it is a useful example of this type of reasoning. It also calls for visualisation and to clearly explain the reasoning provides learners with another challenge. Possible approach You might discuss arguments by contradiction with the class first. We often use arguments by contradiction in ordinary conversations that are nothing to do with mathematics. You might want the class to realise that a statement and its contrapositive are always logically equivalent. If we consider a statement and its contrapositive, and we can prove one of them, then we have also proved the other statement. One approach is to ask the learners to make up 'If ...then... " statements of their own and write down their contrapositives. (See Possible support below). Invite the class to try to prove the Proximity result using an argument by contradiction. A good strategy in cases like this is to ask the learners to work individually for a short time, then to work in pairs and explain their arguments to their partner and agree on the best argument, then to work in fours so that each pair explains their argument to the other pair and they try to get the best argument possible between them. Then invite groups to come to the board and try to convince the whole class that their argument works. An alternative approach to consider is to work with your learners on the NRICH resource<a href="http://nrich.maths.org/public/viewer.php?obj_id=6332&part=index"> Contrary Logic</a> first and then to tackle the Proximity problem. Key Questions If you turn the statement you are trying to prove round to use a proof by contradiction what would you start by assuming? If the icosahedron has 3 red vertices is there any loss of generality in taking the top vertex to be red? If the top vertex is red, what can you say about the other vertices around it? Possible extension Learners might try <a href="http://nrich.maths.org/public/viewer.php?obj_id=1404&part=index"> Proof Sorter</a> to get the proof that the square root of 2 is irrational in order/ A natural follow up, and re-inforcement for confidence in using this sort of argument, would be to work on the resource <a href="http://nrich.maths.org/public/viewer.php?obj_id=6332&part=index"> Contrary Logic.</a> Read the article on <a href="http://nrich.maths.org/public/viewer.php?obj_id=4717&part=index"> Proof by Contradiction</a> . Possible support Even small children will understand the logical equivalence of the following statements, supposing the only money they have to spend is the pocket money they get on Saturdays: (1) If you spend all your money on Saturday you will have none to spend for the rest of the week. (2) If you have money to spend in the rest of the week you did not spend it all on Saturday. These two statements are contrapoitives of each other. For another example, if we are talking about polygons, then the two statements (3) and (4) below are logically equivalent. (3) If this figure is a triangle then it has three sides (4) If this figure does not have three sides then it is not a triangle. </mdoxml> <mdoxml version="1.0"> We have to prove the statement: If a regular icosahedron has three red vertices then it has a vertex that has at least two red neighbours. We can use an argument by contradiction. We suppose that NOvertex has more than one red neighbour and reach a contradiction thus showing that this statement must be false Without loss of generality you can think about the top vertex being red and decide what that means for the other vertices around it. </mdoxml> <mdoxml version="1.0"> Jenny writes: An argument might go something like this: Let's assume I can make an icosahedrons with 3 red vertices but no vertex has 2 red neighbours. First, I mark any vertex red and call it A (see the diagram), I can now think of the icosahedrons as made up of 4 layers of vertices.<mdo:image height="373" width="441" src="proximity_sol_canonical.jpg" alt=""></mdo:image> <ul> <li>The one I marked at the top, which is A</li> </ul> <ul> <li>The ring of 5 vertices adjacent to A, call it B</li> </ul> <ul> <li>The ring of 5 vertices below ring B, call it C</li> </ul> <ul> <li>The vertex that is adjacent to ring C, call it D</li> </ul> Since A is red, I cannot now mark any of the vertices on B red because there will be a third vertex adjacent to this vertex and A so that it has 2 red neighbours. So let me put a second red vertex on the ring C, now there will always be a vertex on the ring B with two adjacent red vertices. Thus, I cannot mark any of the vertices on C red. The only place left for the second red vertex is the vertex D. Now I have to place a third vertex on the ring B or C but I have shown that by putting one there I can find a vertex with 2 red neighbours. Therefore it is not possible to have 3 red vertices without at least one vertex having 2 adjacent red vertices. Steve writes: Suppose that three vertices are coloured red and that no vertex has more than one red neighbour. Because no vertex can be adjacent to two red vertices we can see that no two red vertices can be on the same triangular face or on two adjacent triangular faces. Furthermore, all of the 10 vertices on the two middle layers (B and C) of the dodecahedron are on a face adjacent to a face containing the point A. Therefore if A is coloured red then the only other vertex which can be coloured red is D, which contradicts the assumption that three vertices were coloured red. </mdoxml> 2 4 0 0 0 1 0 Proximity We are given a regular icosahedron having three red vertices. Show that it has a vertex that has at least two red neighbours. Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof Using, Applying and Reasoning about Mathematics Visualising Using, Applying and Reasoning about Mathematics Proof by contradiction 3D Geometry, Shape and Space Icosahedra Using, Applying and Reasoning about Mathematics Proof Sorting