S2T2

Jim Randell 9:42 am on 13 May 2021 Permalink Reply
Tags: by: Graham Smithers ( 61 )   

• 

  Jim Randell 9:42 am on 13 May 2021 Permalink | Reply

  If the face sum is F, and the vertex sum is V, and the total sum of the values of the edges is T.

  Then adding up the sums for all 6 faces counts each edge twice:

  6F = 2T

  Also adding up the sums for all 5 vertices counts each edge twice:

  5V = 2T

  Hence:

  6F = 5V

  So: F < V and F must be divisible by 5. And we know F and V are 3 digit numbers where one is the reverse of the other.

  There is only one possible solution for this scenario, so we can calculate T (the required answer).

  This Python program finds the values of F, V, T, and then also solves the equations to find the values on the edges. It runs in 53ms.

  Run: [ @replit ]

  from enigma import irange, div, matrix, catch, nreverse, join, printf
  
  # consider possible 3-digit values for F (must be divisible by 5)
  for F in irange(105, 999, step=10):
    # calculate V
    V = div(6 * F, 5)
    if V > 999: break
    # check V is the reverse of F
    if nreverse(F) != V: continue
    T = div(5 * V, 2)
    printf("F={F} V={V}; T={T}")
  
    # solve the equations to find edge values
    eqs = [
      # face sums = F
      #  a  b  c  d  e  f  g  h  i    k
      (( 1, 1, 0, 0, 1, 0, 0, 0, 0 ), F), # (a + b + e)
      (( 1, 0, 1, 1, 0, 0, 0, 0, 0 ), F), # (a + c + d)
      (( 0, 1, 1, 0, 0, 1, 0, 0, 0 ), F), # (b + c + f)
      (( 0, 0, 0, 1, 0, 0, 1, 0, 1 ), F), # (d + g + i)
      (( 0, 0, 0, 0, 1, 0, 1, 1, 0 ), F), # (e + g + h)
      (( 0, 0, 0, 0, 0, 1, 0, 1, 1 ), F), # (f + h + i)
      # vertex sums = V
      (( 1, 1, 1, 0, 0, 0, 0, 0, 0 ), V), # (a + b + c)
      (( 0, 0, 0, 0, 0, 0, 1, 1, 1 ), V), # (g + h + i)
      (( 1, 0, 0, 1, 1, 0, 1, 0, 0 ), V), # (a + d + e + g)
      (( 0, 1, 0, 0, 1, 1, 0, 1, 0 ), V), # (b + e + f + h)
      (( 0, 0, 1, 1, 0, 1, 0, 0, 1 ), V), # (c + d + f + i)
    ]
  
    s = catch(matrix.linear, eqs)
    if s:
      printf("-> {s}", s=join(s, sep=", ", enc="()"))
  

  Solution: The sum of the numbers on the edges is 1485.

  The 3 edges that form the “equator” have values of 99. The other 6 edges have values of 198 (= 2×99). And 3×99 + 6×198 = 1485.

  Each face has a sum of 495 (= 5×99).

  Each vertex has a sum of 594 (= 6×99).

  ---

  It makes sense that there are only 2 different edge values, as there is nothing to distinguish the “north pole” from the “south pole”, nor the rotation of the bipyramid. The only edges we can distinguish are the “equator” edges (q) from the “polar” edges (e).

  Which gives a much simpler set of equations to solve:

  F = 2e + q
  V = 3e = 2e + 2q
  ⇒ q = V − F; e = 2q

  So for: F = 495, V = 594, we get: q = 99, e = 198.

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• 

  GeoffR 3:52 pm on 14 October 2021 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  % Face, vertex and edge descriptions of tetrahedra.
  % Top tetrahedron has faces F2 and F3 visible, and back face F1 not visible.
  % Bottom tetrahedron has faces F5 and F6 visible, and back face F4 not visible.
  % Top and bottom vertices are V1 and V5 respectively.
  
  % Two tetrahedra join at vertices V2, V3 and V4.
  % Face F2 has vertices V1, V2 and V4, Face F3 has vertices V1, V3 and V4.
  % Face F1 has vertices V1, V2 and V3.
  % e1 = V1-V2, e2 = V1-V4, e3 = V1-V3
  % e4 = V2-V3, e5 = V2-V4, e6 = V3-V4
  % e7 = V2-V5, e8 = V4-V5, e9 = V3-V5
  
  % Edges
  var 34..333:e1; var 34..333:e2; var 34..333:e3;
  var 34..333:e4; var 34..333:e5; var 34..333:e6;
  var 34..333:e7; var 34..333:e8; var 34..333:e9;
  
  % Similar edges for two identical regular tetrahedra
  constraint e1 == e7 /\ e3 == e9 /\ e2 == e8; % sloping edges
  constraint e4 == e5 /\ e5 == e6;  % junction of two  tetrahedra
  
  % Sum of the nine numbers
  var 306..2997: Edge_Sum = e1 + e2 + e3 + e4 + e5 + e6 + e7 + e8 + e9;
  
  % Vertices
  var 102..999:V1; var 102..999:V2; var 102..999:V3;
  var 102..999:V4; var 102..999:V5; 
  
  % Faces
  var 102..999:F1; var 102..999:F2; var 102..999:F3;
  var 102..999:F4; var 102..999:F5; var 102..999:F6;
  
  % Face totals all the same
  constraint F1 == e1 + e2 + e4 /\ F2 == e1 + e2 + e5;
  constraint F3 == e2 + e3 + e6;
  constraint F4 == e4 + e7 + e9;
  constraint F5 == e5 + e7 + e8 /\ F6 == e6 + e8 + e9;
  constraint F1 == F2 /\ F1 == F3 /\ F1 == F4 /\ F1 == F5 /\ F1 == F6;
  
  % Vertex totals all the same
  constraint V1 == e1 + e2 + e3 /\ V2 == e1 + e4 + e5 + e7;
  constraint V3 == e3 + e4 + e6 + e9;
  constraint V4 == e5 + e6 + e8 + e2 /\ V5 == e7 + e8 + e9;
  constraint V1 == V2 /\ V1 == V3 /\ V1 == V4 /\ V1 == V5;
  
  % Three digits of  F1 = reverse three digits of V1
  constraint F1 div 100 == V1 mod 10;
  constraint F1 div 10 mod 10 == V1 div 10 mod 10;
  constraint F1 mod 10 == V1 div 100;
  
  solve satisfy;
  
  output [ "Total edge sum = " ++ show(Edge_Sum) 
  ++ "\n" ++ "Face sum = " ++ show(F1) 
  ++ "\n" ++ "Vertex sum = " ++ show(V1)];
  
  % Total edge sum = 1485
  % Face sum = 495
  % Vertex sum = 594
  % ----------
  % ==========
  
  
  
  

  Interesting to note that Euler’s polyhedron formula (V – E + F = 2) also applies to this triangular bipyramid, as V = 5, E = 9 and F = 6, so 5 – 9 + 6 = 2

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