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In any triangle with sides of lengths a, b, and c, and perimeter, semiperimeter, and area equal to p, s, and A respectively,
A^2 = s(s-a)(s-b)(s-c) --> (16A^2)/p = (p-2a)(p-2b)(p-2c)
Also, since the arithmetic mean is greater than or equal to the geometric mean,
((p-2a)+(p-2b)+(p-2c))/3 >= the cube root of ((p-2a)(p-2b)(p-2c)) --> (p^3)/27 >= (p-2a)(p-2b)(p-2c) --> (p^3)/27 >= (16A^2)/p --> (p^2)/A >= 12sqrt(3)
Since the given rectangle has A=6 and p=10, (p^2)/A = 16+2/3 which is less than 12sqrt(3) so there are no triangles with area 6 and perimeter 10. |
