answer:By the Cauchy-Schwarz inequality, begin{align*} &[(1^2 + 1^2 + 1^2 + dots + 1^2 + 1^2)][(1 - x_1)^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + dots + (x_9 - x_{10})^2 + x_{10}^2] &ge [(1 - x_1) + (x_1 - x_2) + (x_2 - x_3) + dots + (x_9 - x_{10}) + x_{10}]^2 = 1. end{align*}From the given condition, we have equality, so by the equality condition for Cauchy-Schwarz, [frac{1 - x_1}{1} = frac{x_1 - x_2}{1} = frac{x_2 - x_3}{1} = dots = frac{x_9 - x_{10}}{1} = frac{x_{10}}{1}.]Let [d = 1 - x_1 = x_1 - x_2 = x_2 - x_3 = dots = x_9 - x_{10} = x_{10}.]Then [(1 - x_1) + (x_1 - x_2) + dots + (x_9 - x_{10}) + x_{10} = 11d,]so 11d = 1. Then d = frac{1}{11}, so [(x_1, x_2, x_3, dots, x_{10}) = left( frac{10}{11}, frac{9}{11}, frac{8}{11}, dots, frac{1}{11} right).]In particular, there is only boxed{1} solution.

answer:The radical conjugate of this number is 10 + sqrt{2018}, so when we add them, the radical parts cancel, giving 10 + 10 = boxed{20}.

answer:Squaring x - frac{1}{x} = i sqrt{2}, we get [x^2 - 2 + frac{1}{x^2} = -2.]Hence, x^2 + frac{1}{x^2} = 0, so x^4 + 1 = 0, or x^4 = -1. Then [x^{2187} = (x^4)^{546} cdot x^3 = x^3,]so begin{align*} x^{2187} - frac{1}{x^{2187}} &= x^3 - frac{1}{x^3} &= left( x - frac{1}{x} right) left( x^2 + 1 + frac{1}{x^2} right) &= boxed{i sqrt{2}}. end{align*}

answer:Let P = left( frac{a^2}{2}, a right) be a point on the parabola. First, we find the equation of the tangent to the parabola at P. [asy] unitsize(0.5 cm); real y; pair P = (8,4); path parab = ((-5)^2/2,-5); for (y = -5; y <= 5; y = y + 0.01) { parab = parab--(y^2/2,y); } draw(parab,red); draw((P + (-4,-4/4))--(P + (4,4/4)),dashed); draw((-2,0)--(15,0)); draw((0,-5)--(0,5)); dot("P", P, S); [/asy] Since the tangent passes through left( frac{a^2}{2}, a right), the equation of the tangent is of the form [y - a = m left( x - frac{a^2}{2} right) = mx - frac{a^2 m}{2}.]Substituting x = frac{y^2}{2}, we get [y - a = frac{my^2}{2} - frac{a^2 m}{2}.]This simplifies to my^2 - 2y + 2a - a^2 m = 0. Since this is the equation of a tangent, the quadratic should have a double root of y = a, which means its discriminant is 0, which gives us [4 - 4m(2a - a^2 m) = 0.]Then 4a^2 m^2 - 8am + 4 = 4(am - 1)^2 = 0, so m = frac{1}{a}. Now, consider the point P that is closest to (6,12). [asy] unitsize(0.5 cm); real y; pair P = (8,4); path parab = ((-2)^2/2,-2); for (y = -2; y <= 5; y = y + 0.01) { parab = parab--(y^2/2,y); } draw(parab,red); draw((-2,0)--(15,0)); draw((0,-2)--(0,15)); draw(P--(6,12)); draw((P + (-4,-4/4))--(P + (4,4/4)),dashed); dot("(6,12)", (6,12), N); dot("P", P, S); [/asy] Geometrically, the line connecting P and (6,12) is perpendicular to the tangent. In terms of slopes, this gives us [frac{a - 12}{frac{a^2}{2} - 6} cdot frac{1}{a} = -1.]This simplifies to a^3 - 10a - 24 = 0, which factors as (a - 4)(a^2 + 4a + 6) = 0. The quadratic factor has no real roots, so a = 4. Therefore, P = (8,4), and the shortest distance is sqrt{(8 - 6)^2 + (4 - 12)^2} = boxed{2 sqrt{17}}.