Solution　Let the sides be a, a+1, a+2, the angle oppose a be A, the angle opposite a+1 be B, and the angle opposite a+2 be C.
Using the cosine rule, we find cos A = (a+5)/(2a+4), cos B = (a+1)/2a, cos C = (a-3)/2a. Finally, using cos 2x = 2 cos²x - 1, we find solutions a = 4 for C = 2A, a = 1 for B = 2A, and no solutions for C = 2B.
a = 1 is a degenerate solution (the triangle has the three vertices collinear). The other solution is 4, 5, 6.

Solution　Suppose n has m > 1 digits. Let the first digit be d. Then the product of the digits is at most d.9^m-1 < d.10^m-1 <= n. But (n² - 11n - 22) - n = n(n - 11) - 22 >= 0 for n >= 13. So there are no solutions for n > = 13. But n² - 10n - 22 < 0 for n <= 11, so the only possible solution is n=12 and indeed that is a solution.

Solution　Let f(x) = ax² + bx + c - x. Then f(x)/a = (x + (b-1)/2a)² + (4ac - (b-1)²)/4a². Hence if 4ac - (b-1)² > 0, then f(x) has the same sign for all x. But f(x) > 0 means ax² + bx + c > x, so if {x_i} is a solution, then either x₁ < x₂ < ... < x_n < x₁, or x₁ > x₂ > ... > x_n > x₁. Either way we have a contradiction. So if 4ac - (b-1)² > 0 there cannot be any solutions.
If 4ac - (b-1)² = 0, then we can argue in the same way that either x₁ <= x₂ <= ... <= x_n <= x₁, or x₁ >= x₂ >= ... >= x_n >= x₁. So we must have all x_i = the single root of f(x) = 0 (which clearly is a solution).
If 4ac - (b-1)² < 0, then f(x) = 0 has two distinct real roots y and z and so we have at least two solutions to the equations: all x_i =y, and all x_i = z. We may, however, have additional solutions. For example, if a = 1, b = 0, c = -1 and n is even, then we have the additional solution x₁ = x₃ = x₅ = ... = 0, x₂ = x₄ = ... = -1.

Solution　The trick is to consider the longest side. That avoids getting into lots of different possible cases for which edge is longer than the sum of the other two.
So assume the result is false and let AB be the longest side. Then we have AB > AC + AD and BA > BC + BD. So 2AB > AC + AD + BC + BC. But by the triangle inequality, AB < AC + CB, AB < AD + DB, so 2AB < AC + CB + AD + DB. Contradiction.

Solution　Directly from the equality given: f(x+a) >= 1/2 for all x, and hence f(x) >= 1/2 for all x. So f(x+2a) = 1/2 + Ö(f(x+a) - f(x+a)²) = 1/2 + Ö(f(x+a) (1 - f(x+a)) = 1/2 + Ö(1/4 - f(x) + f(x)²) = 1/2 + (f(x) - 1/2) = f(x). So f is periodic with period 2a.
We may take f(x) to be arbitary in the interval [0,1). For example, let f(x) = 1 for 0 <= x < 1, f(x) = 1/2 for 1 <= x < 2. Then use f(x+2) = f(x) to define f(x) for all other values of x.

Solution For any real x we have [x] = [x/2] + [(x+1]/2]. For if x = 2n + 1 + k, where n is an integer and 0 <= k < 1, then lhs = 2n + 1, and rhs = n + n + 1. Similarly, if x = 2n + k.
Hence for any integer n, we have: [n/2^k] - [n/2^k+1] = [(n/2^k + 1)/2] = [(n + 2^k)/2^k+1]. Hence summing over k, and using the fact that n < 2^k for sufficiently large k, so that [n/2^k ] = 0, we have: n = [(n + 1)/2] + [(n + 2)/4] + [(n + 4)/8] + ... .