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Decagon (Posted on 2005-01-30) Difficulty: 4 of 5
There is a decagon with eight 150 degree angles and two 120 degree angles. The lengths of its sides are a set of ten consecutive integers. Maximize the length of its longest side.

See The Solution Submitted by Tristan    
Rating: 4.0000 (1 votes)

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Solution The answer seems to be... | Comment 5 of 6 |

12, meaning the sides range in size from 3 to 12, maximum.

Based on my previous thoughts, The vertical displacement multipliers on the one side, top to bottom or bottom to top could be {.5,1,sqrt(3)/2,.5} or {.5,sqrt(3)/2,1,sqrt(3/2)}.  If the former, then the other side could be either  {sqrt(3)/2,1,sqrt(3/2),.5} or {.5,sqrt(3)/2,sqrt(3/2),.5}; if the latter, then only {.5,sqrt(3)/2,sqrt(3/2),.5} is possible on the other side.

If the vertical displacement multipliers on left and right are {.5,1,sqrt(3)/2,.5} and  {sqrt(3)/2,1,sqrt(3/2),.5}, then the horizontal displacement multipliers are {-sqrt(3)/2,0,.5,sqrt(3)/2} and {-.5,0,.5,sqrt(3)/2} respectively.  If the side lengths corresponding to these multipliers on the left are called a,b,c,d and on the right w,x,y,z, then, because of the necessity of the total left vertical displacement equalling that on the right and the irrational portions of the horizontal displacements cancelling each other, the following must be true:

(a+d)/2 + b = x + z/2
c = w + y
abs(a-d) = z
and the difference between the remaining two sides (the top and the bottom) must be +/- c/2  +/-  (w-y)/2.

The following program looks for such circumstances:

CLS
FOR bse = 3 TO 9
  REDIM had(bse TO bse + 9)
  FOR w = bse TO bse + 9
   had(w) = 1
  FOR x = bse TO bse + 9
   IF had(x) = 0 THEN
     had(x) = 1
  FOR y = bse TO bse + 9
   IF had(y) = 0 THEN
     had(y) = 1
     c = w + y
     IF c < bse + 10 THEN
     IF had(c) = 0 THEN
     had(c) = 1
  FOR a = bse TO bse + 9
   IF had(a) = 0 THEN
     had(a) = 1
  FOR d = bse TO bse + 9
   IF had(d) = 0 THEN
     had(d) = 1
     z = ABS(a - d)
     IF z < bse + 10 AND z >= bse THEN
     IF had(z) = 0 THEN
      had(z) = 1
      b = x + z / 2 - (a + d) / 2
      IF b = INT(b) AND b >= bse AND b < bse + 10 THEN
       IF had(b) = 0 THEN
         had(b) = 1
         ct = 0
         FOR i = bse TO bse + 9
          IF had(i) = 0 THEN ct = ct + 1: tb(ct) = i
         NEXT i
         diff = ABS(tb(1) - tb(2))
         IF diff = c / 2 + (w - y) / 2 OR diff = -c / 2 + (w - y) / 2 OR diff = -c / 2 - (w - y) / 2 OR diff = c / 2 - (w - y) / 2 THEN
           PRINT bse, a; b; c; d; TAB(30); w; x; y; z; TAB(60); tb(1); tb(2)
           sCt = sCt + 1
         END IF
         had(b) = 0
       END IF
      END IF ' b =int(b) etc.
      had(z) = 0
     END IF  'had(z)=0
     END IF ' z<bse+10
     had(d) = 0
   END IF
  NEXT d
     had(a) = 0
   END IF
  NEXT a
     had(c) = 0
     END IF  ' had(c)=0
     END IF ' c<bse+10
     had(y) = 0
   END IF
  NEXT y
     had(x) = 0
   END IF
  NEXT x
   had(w) = 0
  NEXT w
NEXT bse
PRINT sCt

It finds the following 32 sets

  3             4  6  8  11    3  10  5  7                   9  12
  3             11  6  8  4    3  10  5  7                   9  12
  3             4  6  12  11   3  10  9  7                   5  8
  3             11  6  12  4   3  10  9  7                   5  8
  3             4  7  8  10    3  11  5  6                   9  12
  3             10  7  8  4    3  11  5  6                   9  12
  3             4  7  12  10   3  11  9  6                   5  8
  3             10  7  12  4   3  11  9  6                   5  8
  3             3  5  10  12   4  8  6  9                    7  11
  3             12  5  10  3   4  8  6  9                    7  11
  3             3  5  11  12   4  8  7  9                    6  10
  3             12  5  11  3   4  8  7  9                    6  10
  3             3  9  10  8    4  12  6  5                   7  11
  3             8  9  10  3    4  12  6  5                   7  11
  3             3  9  11  8    4  12  7  5                   6  10
  3             8  9  11  3    4  12  7  5                   6  10
  3             4  6  8  11    5  10  3  7                   9  12
  3             11  6  8  4    5  10  3  7                   9  12
  3             4  7  8  10    5  11  3  6                   9  12
  3             10  7  8  4    5  11  3  6                   9  12
  3             3  5  10  12   6  8  4  9                    7  11
  3             12  5  10  3   6  8  4  9                    7  11
  3             3  9  10  8    6  12  4  5                   7  11
  3             8  9  10  3    6  12  4  5                   7  11
  3             3  5  11  12   7  8  4  9                    6  10
  3             12  5  11  3   7  8  4  9                    6  10
  3             3  9  11  8    7  12  4  5                   6  10
  3             8  9  11  3    7  12  4  5                   6  10
  3             4  6  12  11   9  10  3  7                   5  8
  3             11  6  12  4   9  10  3  7                   5  8
  3             4  7  12  10   9  11  3  6                   5  8
  3             10  7  12  4   9  11  3  6                   5  8


 
where the left hand column indicates the lowest number in the sequence of ten (originally the program started with 1, but since results were found with 3, the starting value was changed to reduce the size of the output set); the next group of 4 numbers lists the side lengths on one side (left side for sake of argument) that correspond to the vertical displacement multipliers of .5, 1, sqrt(3)/2, .5; the next group is the sequence on the other side corresponding to multipliers of sqrt(3)/2,1,sqrt(3)/2,.5.  The top and bottom are at the extreme right. In the last example above, for example, the total vertical displacement on the one side is therefore 10*.5 + 7*1 + 12*sqrt(3)/2 + 4*.5 = 24.3923048454133, while the right side equals 9*sqrt(3)/2+11*1+3*sqrt(3)/2+6*.5, which totals the same.  The horizontal net movement on the left is -10*sqrt(3)/2+7*0+12*.5+4*sqrt(3)/2=.80384757729336. On the right it is -9*.5-11*0+3*.5+6*sqrt(3)/2=2.19615242270665.  When added, these make the 3 unit difference. The actual sequence on the left is therefore 10, 7, 12, 4, from bottom to top, rather than top to bottom, and the bottom is 8 while the top is 5. So, counterclockwise from just to the left of the top, we have 4,12,7,(here's where one 120-degree angle is), 10; the bottom is 8, then proceding counterclockwise, 6,3,11,9, (here's the other 120-degree angle), and the top, 5.  The reason the 10,7,12,4 sequence was inverted (bottom-to-top rather than top-to-bottom) was to allow the decagon to the convex, while still causing the difference from top to bottom add to the bottom sides length.

Similar considerations apply to the situation where the vertical displacement multipliers are {.5,1,sqrt(3)/2,.5} and {.5,sqrt(3)/2,sqrt(3/2),.5}.  In this instance the equations are

(a+d)/2 + b = (w+z)/2
c = x + y
abs(a-d) = abs(w-z)
and the difference in top vs. bottom sides is +/- c/2 +/- (x-y)/2

and a similar program also finds the maximum range for the sequence of side lengths is 3 to 12.

A similar set for {.5,sqrt(3)/2,1,sqrt(3/2)} and {.5,sqrt(3)/2,sqrt(3/2),.5} using equations

d/2 + b = (w+z)/2
a + c = x + y
d = abs(w-z)
and a difference between the top and bottom sides of +/- (c-a)/2 +/- (x-y)/2

yields no solutions.


  Posted by Charlie on 2005-01-31 04:46:57
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