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Integer Side Lengths (Posted on 2009-03-05) Difficulty: 2 of 5
The side lengths of a triangle are consecutive integers.
The largest interior angle is twice the smallest.

What are the side lengths of the triangle?

See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

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Solution computer solution | Comment 2 of 8 |

DECLARE FUNCTION asin# (x#)
DECLARE FUNCTION acos# (x#)
DEFDBL A-Z
DIM SHARED pi
pi = 4 * ATN(1)

FOR s1 = 2 TO 45
  s2 = s1 + 1: s3 = s2 + 1
 
  ' a^ = b^2 + c^2 - 2bc cosA
  ' cos A = (b^2 + c^2 - a^2)/(2bc)
 
  cosSmall = (s2 ^ 2 + s3 ^ 2 - s1 ^ 2) / (2 * s2 * s3)
  cosLarge = (s2 ^ 2 + s1 ^ 2 - s3 ^ 2) / (2 * s2 * s1)

  small = acos(cosSmall)
  large = acos(cosLarge)

  PRINT USING "## ## ## "; s1; s2; s3;
  PRINT USING " ###.#####"; small; large; large / small
NEXT s1

FUNCTION acos (x)
 acos = 90 - asin(x)
END FUNCTION

FUNCTION asin (x)
 IF ABS(x) = 1 THEN
  asin = 90 * SGN(x)
 ELSE
  asin = ATN(x / SQR(1 - x * x)) * 180 / pi
 END IF
END FUNCTION

finds

s1 s2 s3   angle 1   angle 3     ratio
 2  3  4   28.95502 104.47751   3.60827
 3  4  5   36.86990  90.00000   2.44102
 4  5  6   41.40962  82.81924   2.00000
 5  6  7   44.41531  78.46304   1.76658
 6  7  8   46.56746  75.52249   1.62179
 7  8  9   48.18969  73.39845   1.52312
 8  9 10   49.45840  71.79004   1.45152
 9 10 11   50.47880  70.52878   1.39720
10 11 12   51.31781  69.51268   1.35455
11 12 13   52.02013  68.67631   1.32019
12 13 14   52.61680  67.97569   1.29190
13 14 15   53.13010  67.38014   1.26821
14 15 16   53.57643  66.86760   1.24808
15 16 17   53.96812  66.42182   1.23076
16 17 18   54.31467  66.03052   1.21570
17 18 19   54.62346  65.68426   1.20249
18 19 20   54.90037  65.37568   1.19081
19 20 21   55.15010  65.09894   1.18040
20 21 22   55.37646  64.84934   1.17106
21 22 23   55.58261  64.62307   1.16265
22 23 24   55.77113  64.41700   1.15502
23 24 25   55.94420  64.22854   1.14808
24 25 26   56.10364  64.05552   1.14174
25 26 27   56.25101  63.89612   1.13591
26 27 28   56.38763  63.74879   1.13055
27 28 29   56.51462  63.61220   1.12559
28 29 30   56.63299  63.48523   1.12099
29 30 31   56.74357  63.36688   1.11672
30 31 32   56.84711  63.25632   1.11274
31 32 33   56.94427  63.15279   1.10903
32 33 34   57.03561  63.05564   1.10555
33 34 35   57.12165  62.96431   1.10228
34 35 36   57.20283  62.87828   1.09922
35 36 37   57.27956  62.79711   1.09633
36 37 38   57.35218  62.72039   1.09360
37 38 39   57.42103  62.64777   1.09102
38 39 40   57.48639  62.57893   1.08859
39 40 41   57.54851  62.51357   1.08628
40 41 42   57.60763  62.45145   1.08408
41 42 43   57.66397  62.39233   1.08200
42 43 44   57.71772  62.33600   1.08001
43 44 45   57.76905  62.28225   1.07812
44 45 46   57.81812  62.23093   1.07632
45 46 47   57.86507  62.18186   1.07460

Left out at the beginning is the degenerate triangle with sides 1, 2 and 3, where the smallest angle is 0 degrees and the largest 180, for an infinite ratio.

There's one obtuse triangle, then the familiar 3:4:5 right triangle.

Then comes the solution to the problem: 4:5:6 gives a 2:1 ratio between its largest and smallest angles.

The ratio is monotonically decreasing, approaching 1 as the sides approach the ratio of 1:1:1 of an equilateral triangle.


  Posted by Charlie on 2009-03-05 12:48:10
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