Magical Tetrahedron
The edges and vertices of the magical tetrahedron PQRS are labelled with one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 11. Each number is used exactly once. For any two vertices, the sum of the numbers on the vertices is equal to the number on the edge connecting those two vertices. If the edge PQ is labelled with the number 9, then what number is labelling the edge RS?
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2 solutions
We can also show that there is, to within symmetry, exactly one way of labelling this magic tetrahedron. There are exactly five ways of choosing four numbers out of 1,2,3,4,5,6,7,8,9,11 that sum to 14, namely 1 , 2 , 3 , 8 1 , 2 , 4 , 7 1 , 2 , 5 , 6 1 , 3 , 4 , 6 2 , 3 , 4 , 5 The first is not possible as a vertex-labelling, since it would require an edge as well as a vertex to be labelled 3 . The third is not possible, since it would require a vertex and an edge to be labelled 6 . The fourth and fifth are not possible, since they do not have either an edge or a vertex labelled with 1 1 . It is easy to check that the second vertex-labelling works.
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Another way of looking at this is that certain numbers are definitely on vertices while others are definitely on edges.
Each vertex is connected to three others. If 8 were on a vertex, we'd need three larger numbers in the list for its edges; but there are only two. So straight away, 8 , 9 , 1 1 must be edges.
1 and 2 are too small to be edges, so they are vertices. All pairs of vertices are connected, so 3 must be on the edge joining 1 and 2 .
Now 4 can't be an edge, since we know where 1 , 2 , 3 are; so it's a vertex. Finally, the only way to get 1 1 as an edge is 4 + 7 (since we know 9 is not a vertex), so the last vertex is 7 .
(This is the way I first solved it before spotting the quicker route I gave in my solution)
Do other magic Platonic polyhedra exist with consecutive (or - as in this case - almost consecutive) labels?
When you were adding the p's, q's, etc did you put up a picture or was it just text? How did you make your text or picture?
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Do you mean the horizontal braces (with the labels "vertices" and "edges")? The LaTeX for this is
\underbrace{p+q+r+s}_{\text{vertices}}
Brute force: r = ParallelTable [ If [ p [ [ 1 ] ] + p [ [ 2 ] ] = p [ [ 5 ] ] , If [ p [ [ 2 ] ] + p [ [ 3 ] ] = p [ [ 6 ] ] , If [ p [ [ 1 ] ] + p [ [ 3 ] ] = p [ [ 7 ] ] , If [ p [ [ 1 ] ] + p [ [ 4 ] ] = p [ [ 8 ] ] , If [ p [ [ 2 ] ] + p [ [ 4 ] ] = p [ [ 9 ] ] , If [ p [ [ 3 ] ] + p [ [ 4 ] ] = p [ [ 1 0 ] ] , p , Nothing ] , Nothing ] , Nothing ] , Nothing ] , Nothing ] , Nothing ] , { p , l } ] ⇒
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Vertices to the left and edges to the right. 24 permutations of the vertex labelings. Then it is a matter of labeling the vertices, computing the edges and finding the edge opposite the edge labeled 9, which is 5.
The fun way to do this is to draw a tetrahedral graph and have at it - lots of nice deductions to be made.
The quick way is this: the sum of all the vertex and edge labels is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 1 = 5 6 . Alternatively, we can write everything in terms of the vertex labels p , q , r , s ; the label on the edge P Q has value p + q , and so on. So the sum is
vertices p + q + r + s + edges ( p + q ) + ( p + r ) + ( p + s ) + ( q + r ) + ( q + s ) + ( r + s ) = 5 6
Each vertex label appears four times in the sum, so p + q + r + s = 1 4 .
Since p + q = 9 , the label on the edge R S is r + s = 5 .