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 A Near Diophantine Octagon Problem (Posted on 2007-04-22)
The cyclic octagon ABCDEFGH has the sides a√2, a√2, a√2, a√2, b, b, b and b respectively in that order. Each of a, b and r are positive integers, where r is the radius of the circumcircle.

Analytically determine:

(i) The minimum value of a with a < b

(ii) The minimum value of b with a > b

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 non-analytical "solution" | Comment 2 of 10 |

If alpha  is  half the central angle of one side a*sqrt(2), and beta is half the central angle of one side b, then

a*sqrt(2) = 2 r sin alpha
b = 2r sin beta

and 4 * 2 * (alpha + beta) = 2 * pi

or  alpha + beta = pi/4

A table of values with increasing a is:

` a   b   r  2alpha  2 beta (degrees) 1   6   5  16.260  73.740 1  40  29   2.794  87.206 1 238 169   0.479  89.521 2  12  10  16.260  73.740 2  80  58   2.794  87.206 2 476 338   0.479  89.521 3  18  15  16.260  73.740 3 120  87   2.794  87.206 4  24  20  16.260  73.740 4 160 116   2.794  87.206 5  30  25  16.260  73.740 5 200 145   2.794  87.206 6  36  30  16.260  73.740 6 240 174   2.794  87.206 7  10  13  44.760  45.240 7  16  17  33.855  56.145 7  42  35  16.260  73.740 7  96  73   7.776  82.224 7 130  97   5.850  84.150 So a can be 1, making b = 6 and r = 5. (r was limited to 500 for the above table) For part 2:   For increasing b:    a   b   r  2alpha  2 beta (degrees)  17  14  25  57.480  32.520  31  18  41  64.639  25.361  49  22  61  69.222  20.778  71  26  85  72.405  17.595  34  28  50  57.480  32.520  97  30 113  74.744  15.256  47  32  65  61.500  28.500 127  34 145  76.534  13.466  62  36  82  64.639  25.361 161  38 181  77.949  12.051  79  40 101  67.158  22.842  51  42  75  57.480  32.520 199  42 221  79.095  10.905  98  44 122  69.222  20.778 241  46 265  80.042   9.958 119  48 145  70.945  19.055 287  50 313  80.838   9.162 142  52 170  72.405  17.595  93  54 123  64.639  25.361 `

So b can be as small as 14, requiring a=17 and r=25.
(r was allowed to go to 5000 to be reasonably sure of not missing any possibilities). Of course it's remotely possible that using values of r larger than 5000 might alow lower b with increased a.

DEFDBL A-Z
CLS
pi = ATN(1) * 4

FOR a = 1 TO 500
FOR r = 1 TO 500
salpha = a * SQR(2) / (2 * r)
IF ABS(salpha) <= 1 THEN
alpha = ATN(salpha / SQR(1 - salpha * salpha))
beta = pi / 4 - alpha
sbeta = SIN(beta)
b = 2 * r * sbeta
IF b > 0 AND ABS((INT(b + .5) - b) / b) < 1E-12 THEN
ct = ct + 1
IF ct < 20 THEN
PRINT USING "### ### ### ###.### ###.###"; a; b; r; alpha * 360 / pi; beta * 360 / pi
END IF
END IF
END IF
NEXT
NEXT

nextPart:
ct = 0
PRINT
FOR b = 1 TO 500
FOR r = 1 TO 5000
sbeta = b / (2 * r)
IF ABS(sbeta) <= 1 THEN
IF ABS(sbeta) = 1 THEN
beta = pi / 2 * SGN(sbeta)
ELSE
beta = ATN(sbeta / SQR(1 - sbeta * sbeta))
END IF
alpha = pi / 4 - beta
salpha = SIN(alpha)
a = 2 * r * salpha / SQR(2)
IF a > b AND ABS((INT(a + .5) - a) / a) < 1E-12 THEN
ct = ct + 1
IF ct < 20 THEN
PRINT USING "### ### ### ###.### ###.###"; a; INT(b + .5); r; alpha * 360 / pi; beta * 360 / pi
END IF
END IF
END IF
NEXT
NEXT

 Posted by Charlie on 2007-04-22 15:40:57

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