An equiangular hexagon has side lengths 6, 7, 8, 9, 10, 11 (not necessarily in this
order). If the area of the hexagon is k√3 , find the sum of all possible values of k.
DECLARE SUB permute (a$)
CLS
DEFDBL A-Z
pi = ATN(1) * 4
a$ = "6789ab": h$ = a$
DO
x = 0: y = 0: a = 0
psn = 0
FOR angle = 0 TO 300 STEP 60
psn = psn + 1
l = INSTR("123456789ab", MID$(a$, psn, 1))
nx = x + l * COS(angle * pi / 180)
ny = y + l * SIN(angle * pi / 180)
a = a + (x - nx) * (y + ny) / 2
x = nx
y = ny
NEXT angle
IF ABS(x) <= .000001 AND ABS(y) <= .000001 THEN
PRINT a$, a, a * 4 / SQR(3)
END IF
permute a$
LOOP UNTIL a$ = h$
finds the orders for the sides so as to form a closed hexagon, and the area of each:
order of
lengths
of sides area k*4
(a=10
b=11)
69a78b 184.0303983041932 424.9999999999999
6a897b 184.8964237079776 427
6b798a 184.8964237079777 427.0000000000001
6b87a9 184.0303983041932 425.0000000000001
78b69a 184.0303983041932 425.0000000000001
798a6b 184.8964237079777 427.0000000000001
7a96b8 184.0303983041932 425.0000000000001
7b6a89 184.8964237079776 427
87a96b 184.0303983041933 425.0000000000001
897b6a 184.8964237079777 427.0000000000001
8a6b79 184.8964237079777 427.0000000000001
8b69a7 184.0303983041933 425.0000000000001
96b87a 184.0303983041932 425.0000000000001
97b6a8 184.8964237079776 427
98a6b7 184.8964237079777 427.0000000000001
9a78b6 184.0303983041933 425.0000000000001
a6b798 184.8964237079777 427.0000000000001
a78b69 184.0303983041933 425.0000000000001
a897b6 184.8964237079777 427.0000000000001
a96b87 184.0303983041933 425.0000000000001
b69a78 184.0303983041932 425.0000000000001
b6a897 184.8964237079777 427.0000000000001
b798a6 184.8964237079777 427.0000000000001
b87a96 184.0303983041932 425.0000000000001
so 427/4 and 425/4 are the two possible values of k.
The sum of these two values is 213.
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Posted by Charlie
on 2014-02-17 17:32:03 |
computer solution
