Asian Pacific Mathematics Olympiad (APMO)

The sum of the first n positive integers is given by the formula n(n+1)/2. Substituting n = 100, we get 100(101)/2 = 5050.

The area of a circle is given by the formula πr^2. Substituting r = 10 cm, we get π(10)^2 = 100π cm^2.

The volume of a cube is given by the formula s^3. Substituting s = 5 cm, we get (5)^3 = 125 cm^3.

The equation of a line passing through two points (x1, y1) and (x2, y2) is given by the formula (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). Substituting the given points, we get (y - 3) / (x - 2) = (7 - 3) / (5 - 2) = 2. Solving for y, we get y = 2x + 1.

We can solve this equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Substituting a = 1, b = -4, and c = 3, we get x = (-(-4) ± √((-4)^2 - 4(1)(3))) / 2(1) = (4 ± √(16 - 12)) / 2 = (4 ± √4) / 2 = (4 ± 2) / 2. Therefore, x = 1 or x = 3.

When flipping a coin, there are two possible outcomes: head or tail. Since both outcomes are equally likely, the probability of getting a head is 1/2.

The derivative of a function f(x) is given by the formula f'(x) = lim_(h->0) [f(x+h) - f(x)] / h. Substituting f(x) = x^3 - 2x^2 + 3x - 4, we get f'(x) = lim_(h->0) [(x+h)^3 - 2(x+h)^2 + 3(x+h) - 4 - (x^3 - 2x^2 + 3x - 4)] / h = lim_(h->0) [x^3 + 3x^2h + 3xh^2 + h^3 - 2x^2 - 4xh - 2h^2 + 3x + 3h - 4 - x^3 + 2x^2 - 3x + 4] / h = lim_(h->0) [3x^2h + 3xh^2 + h^3 - 4xh - 2h^2 + 3h] / h = lim_(h->0) [3x^2 + 3xh + h^2 - 4x - 2h + 3] = 3x^2 - 4x + 3.

The integral of a function f(x) from x = a to x = b is given by the formula ∫f(x) dx = F(b) - F(a), where F(x) is an antiderivative of f(x). An antiderivative of f(x) = 2x + 3 is F(x) = x^2 + 3x + C, where C is a constant. Therefore, ∫f(x) dx = F(2) - F(0) = (2^2 + 3(2) + C) - (0^2 + 3(0) + C) = 4 + 6 + C - C = 10.

Using the distributive property, we can expand the expression as follows: (2 + 3i)(4 - 5i) = 2(4 - 5i) + 3i(4 - 5i) = 8 - 10i + 12i - 15i^2 = 8 + 2i - 15(-1) = 22 + 11i.

The equation of a plane that passes through a point (x0, y0, z0) and has normal vector n = (a, b, c) is given by the formula a(x - x0) + b(y - y0) + c(z - z0) = 0. Substituting the given values, we get 2(x - 1) - 1(y - 2) + 3(z - 3) = 0, which simplifies to 2x - y + 3z = 8.

The volume of the solid generated by revolving the region bounded by the curves y = f(x) and y = g(x) about the x-axis is given by the formula V = π∫[f(x)]^2 - [g(x)]^2 dx. Substituting f(x) = x^2 and g(x) = 4 - x^2, we get V = π∫(x^2)^2 - (4 - x^2)^2 dx = π∫x^4 - 16 + 8x^2 dx = π[(1/5)x^5 - 16x + (8/3)x^3] + C, where C is a constant. Evaluating the integral from x = -2 to x = 2, we get V = π[(1/5)(2)^5 - 16(2) + (8/3)(2)^3 - (1/5)(-2)^5 + 16(-2) - (8/3)(-2)^3] = π[(32/5) - 32 + (64/3) - (32/5) + 32 - (64/3)] = π(64/3).

We can solve this differential equation using the method of separation of variables. Rewriting the equation as (x - y)dy = (x + y)dx, we can integrate both sides to get ∫(x - y)dy = ∫(x + y)dx. This gives us (1/2)x^2 - xy + C1 = (1/2)x^2 + xy + C2, where C1 and C2 are constants. Simplifying this equation, we get 2xy = C, where C = C2 - C1. Therefore, the general solution of the differential equation is y = x + C.

The determinant of a matrix A = [[a11, a12, a13], [a21, a22, a23], [a31, a32, a33]] is given by the formula det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31). Substituting the values from matrix A, we get det(A) = 1[(5)(9) - (6)(8)] - 2[(4)(9) - (6)(7)] + 3[(4)(8) - (5)(7)] = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0.

The equation of a circle that passes through three points (x1, y1), (x2, y2), and (x3, y3) is given by the formula (x - x1)(x - x2) + (y - y1)(y - y2) + (x - x1)(x - x3) + (y - y1)(y - y3) + (x - x2)(x - x3) + (y - y2)(y - y3) = 0. Substituting the given points, we get (x - 1)(x - 3) + (y - 2)(y - 4) + (x - 1)(x - 5) + (y - 2)(y - 6) + (x - 3)(x - 5) + (y - 4)(y - 6) = 0. Expanding and simplifying this equation, we get x^2 + y^2 - 4x - 6y + 14 = 0.