The length of each of the sides of a triangle ABC is a positive integer with: ∠ BAC = 2* ∠ ABC and, ∠ ACB is obtuse.
Find the minimum length of the perimeter.
(In reply to
re: this program finds... (spoiler) by Jer)
I should have put a check in the program to generate only fundamental solutions. Such a list is:
sides peri angles (deg.)
16 28 33 77 28.9550244 57.9100487 93.1349269
25 45 56 126 25.8419328 51.6838655 102.4742017
36 66 85 187 23.5564643 47.1129286 109.3306071
49 91 120 260 21.7867893 43.5735786 114.6396321
64 120 161 345 20.3641348 40.7282696 118.9075956
81 144 175 400 27.2660445 54.5320889 98.2018666
81 153 208 442 19.1881365 38.3762729 122.4355906
100 190 261 551 18.1948723 36.3897447 125.4153830
121 220 279 620 24.6199773 49.2399547 106.1400680
121 231 320 672 17.3414428 34.6828856 127.9756716
144 276 385 805 16.5978421 33.1956843 130.2064736
169 299 360 828 27.7957725 55.5915450 96.6126825
169 312 407 888 22.6198649 45.2397299 112.1404052
169 325 456 950 15.9423686 31.8847372 132.1728942
196 350 429 975 26.7655006 53.5310012 99.7034983
225 390 451 1066 29.9264349 59.8528697 90.2206954
196 378 533 1107 15.3588856 30.7177712 133.9233433
225 420 559 1204 21.0394698 42.0789396 116.8815907
225 435 616 1276 14.8351116 29.6702232 135.4946653
256 464 585 1305 25.0078332 50.0156665 104.9765003
289 510 611 1410 28.0724869 56.1449739 95.7825392
256 496 705 1457 14.3615116 28.7230231 136.9154653
289 527 672 1488 24.2496286 48.4992572 107.2511142
289 544 735 1568 19.7499228 39.4998456 120.7502316
289 561 800 1650 13.9305546 27.8611091 138.2083363
361 627 728 1716 29.7243291 59.4486583 90.8270126
361 646 795 1802 26.5253520 53.0507040 100.4239440
324 630 901 1855 13.5362027 27.0724055 139.3913918
361 665 864 1890 22.9195422 45.8390843 111.2413735
361 684 935 1980 18.6717181 37.3434361 123.9848458
361 703 1008 2072 13.1735511 26.3471022 140.4793467
400 740 969 2109 22.3316450 44.6632900 113.0050650
441 777 928 2146 28.2425372 56.4850744 95.2723884
441 798 1003 2242 25.2087653 50.4175306 104.3737041
400 780 1121 2301 12.8385681 25.6771363 141.4842956
484 858 1037 2379 27.5801992 55.1603984 97.2594024
441 840 1159 2440 17.7527902 35.5055803 126.7416295
529 920 1071 2520 29.5918458 59.1836916 91.2244626
441 861 1240 2542 12.5279054 25.0558109 142.4162837
484 902 1197 2583 21.2799665 42.5599329 116.1601006
529 943 1152 2624 26.9623890 53.9247780 99.1128330
529 966 1235 2730 24.0706146 48.1412292 107.7881561
484 946 1365 2795 12.2387558 24.4775115 143.2837327
529 989 1320 2838 20.8069478 41.6138956 117.5791567
576 1032 1273 2881 26.3843297 52.7686595 100.8470108
529 1012 1407 2948 16.9574263 33.9148526 129.1277211
625 1100 1311 3036 28.3576366 56.7152732 94.9270903
529 1035 1496 3060 11.9687456 23.9374911 144.0937633
DECLARE FUNCTION gcd# (x#, y#)
DEFDBL A-Z
CLS
pi = 4 * ATN(1#)
FOR p = 1 TO 9999
FOR s1 = 1 TO p / 3
FOR s2 = s1 + 1 TO (p - s1) / 2
s3 = p - s1 - s2
g = gcd(s1, gcd(s2, s3))
IF g = 1 AND s3 < s1 + s2 THEN
c2a = (s1 * s1 + s3 * s3 - s2 * s2) / (2 * s1 * s3)
ca = (s2 * s2 + s3 * s3 - s1 * s1) / (2 * s2 * s3)
s2a = SQR(1 - c2a * c2a): sa = SQR(1 - ca * ca)
IF c2a > 0 AND ca > 0 AND sa > 0 THEN
twoA = ATN(s2a / c2a)
a = ATN(sa / ca)
ratio = twoA / a
obt = pi - twoA - a
IF obt + .000000001# > pi / 2 AND ABS(ratio - 2) < .000000001# THEN
PRINT USING "#### #### #### #### ###.####### ###.####### ###.#######"; s1; s2; s3; p; a * 180 / pi; twoA * 180 / pi; (pi - twoA - a) * 180 / pi
END IF
END IF
END IF
NEXT
NEXT
NEXT p
FUNCTION gcd (x, y)
dnd = x: dvr = y
IF dnd < dvr THEN SWAP dnd, dvr
DO
q = INT(dnd / dvr)
r = dnd - q * dvr
dnd = dvr: dvr = r
LOOP UNTIL r = 0
gcd = dnd
END FUNCTION
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Posted by Charlie
on 2010-12-10 20:08:23 |
re(2): this program finds... (spoiler)
