IAS Math Olympiad

Solution

Correct: C

Given f(-1) = -10 and f(1) = 2, we have -1 + a - b + c = -10 and 1 + a + b + c = 2. Adding the equations, we get 2a + 2c = -8, or a + c = -4. Subtracting, we get 2 + 2b = 12, so b = 5. Thus, f(x) = x^3 + ax^2 + 5x + (-4 - a). Now, f'(x) = 3x^2 + 2ax + 5. Since f'(x) = 0 has real roots, the discriminant must be non-negative: (2a)^2 - 4(3)(5) >= 0, so 4a^2 >= 60, or a^2 >= 15. Thus, a >= sqrt(15) or a <= -sqrt(15). Since f(-1) < 0 and f(1) > 0, there must be a root in (-1, 1). Let's try to narrow down the root by considering the intermediate value theorem. Using a + c = -4, c = -4 - a. f(x) = x^3 + ax^2 + 5x - 4 - a. f(0) = -4 - a. If a > -4, f(0) < 0 and since f(1) > 0 there is a root in (0, 1). If a < -4, f(0) > 0 and since f(-1) < 0 there is a root in (-1, 0). We know a >= sqrt(15) or a <= -sqrt(15). sqrt(15) > 3, so a > 3 or a < -3. Because both (-1,0) and (0,1) are in range (-2, -1), (-1, 0), (0, 1), this is wrong. When we try x=2, f(2) = 8 + 4a + 10 - 4 - a = 14 + 3a. if f(2)>0 and f(1)>0, which it could be when a= -3, so it's not contained (1,2). When we try x=-2, f(-2) = -8 + 4a - 10 - 4 - a = -22 + 3a. if f(-2)<0, and f(-1) < 0, which it could be when a = 7, so it's not contained (-3, -2). so when a=-5, f(0) = -4 -a = 1. f(-1) = -10, so contained in (-1, 0).

Solution

Correct: B

The curves intersect when x^2 = sqrt(x), which implies x^4 = x, or x(x^3 - 1) = 0. The solutions are x = 0 and x = 1. The area is given by the integral of (sqrt(x) - x^2) from 0 to 1. Integral of sqrt(x) is (2/3)x^(3/2), and the integral of x^2 is (1/3)x^3. Evaluating at 1 and 0 gives (2/3)(1)^(3/2) - (1/3)(1)^3 - (0 - 0) = 2/3 - 1/3 = 1/3.

Solution

Correct: D

First, solve f(x) = 0: x^2 - 3x + 2 = (x-1)(x-2) = 0, so x = 1 or x = 2. Now, we need to solve f(x) = 1 and f(x) = 2. For f(x) = 1: x^2 - 3x + 2 = 1, so x^2 - 3x + 1 = 0. The solutions are x = (3 +/- sqrt(5))/2. For f(x) = 2: x^2 - 3x + 2 = 2, so x^2 - 3x = 0, which gives x = 0 or x = 3. The sum of all solutions is (3 + sqrt(5))/2 + (3 - sqrt(5))/2 + 0 + 3 = 3 + 3 = 6.

Solution

Correct: C

By Cauchy-Schwarz inequality, (a + b + c)(1/a + 1/b + 1/c) >= (1 + 1 + 1)^2 = 9. Since a + b + c = 1, we have 1 * (1/a + 1/b + 1/c) >= 9, so (1/a + 1/b + 1/c) >= 9. Equality holds when a = b = c = 1/3, which gives 1/a + 1/b + 1/c = 3 + 3 + 3 = 9.

Solution

Correct: D

Let the side of the square be s. The radius of the smaller circle is s/2, and its area is pi*(s/2)^2 = pi*s^2/4. The diagonal of the square is s*sqrt(2). The radius of the larger circle is (s*sqrt(2))/2, and its area is pi*((s*sqrt(2))/2)^2 = pi*s^2/2. The ratio of the smaller area to the larger area is (pi*s^2/4) / (pi*s^2/2) = (1/4) / (1/2) = 1/2.

Solution

Correct: A

Using the Taylor series expansion for sin(x) around x=0: sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... So, sin(x) - x + x^3/6 = sin(x) - x + x^3/3! = x^5/5! - x^7/7! + ... Then, (sin(x) - x + x^3/6) / x^5 = 1/5! - x^2/7! + ... As x approaches 0, the limit is 1/5! = 1/120.

Solution

Correct: A

There are two ways to draw two balls of different colors: red then blue (RB), or blue then red (BR). The probability of RB is (5/8) * (3/7) = 15/56. The probability of BR is (3/8) * (5/7) = 15/56. The total probability is 15/56 + 15/56 = 30/56 = 15/28.

Solution

Correct: E

The series can be written as 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ... This is a telescoping series. 1/(n(n+1)) = 1/n - 1/(n+1). The sum is (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... The partial sum is 1 - 1/(n+1). As n approaches infinity, 1/(n+1) approaches 0, so the sum of the series is 1.

Solution

Correct: C

The determinant is 1(5*9 - 6*8) - 2(4*9 - 6*7) + 3(4*8 - 5*7) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 - 2(-6) + 3(-3) = -3 + 12 - 9 = 0.

Solution

Correct: B

Using the change of base formula, log4(x) = log2(x) / log2(4) = log2(x) / 2. So, log2(x) + log2(x)/2 = 3. (3/2)log2(x) = 3, which means log2(x) = 2. Thus, x = 2^2 = 4.

Solution

Correct: D

We are given f(x) + f(1/(1-x)) = x. Let x = 2, then f(2) + f(1/(1-2)) = f(2) + f(-1) = 2. Now, let x = -1, then f(-1) + f(1/(1-(-1))) = f(-1) + f(1/2) = -1. Next, let x = 1/2, then f(1/2) + f(1/(1-1/2)) = f(1/2) + f(2) = 1/2. We have the following system of equations: 1) f(2) + f(-1) = 2; 2) f(-1) + f(1/2) = -1; 3) f(1/2) + f(2) = 1/2. Subtract equation (2) from (1) to get: f(2) - f(1/2) = 3. Adding this to equation (3), we have: 2f(2) = 3 + 1/2 = 7/2. Therefore, f(2) = 7/4.

Solution

Correct: C

f(x) = x/(x-1). f(f(x)) = f(x/(x-1)) = (x/(x-1)) / ((x/(x-1)) - 1) = (x/(x-1)) / ((x - (x-1))/(x-1)) = (x/(x-1)) / (1/(x-1)) = x. Therefore, f(f(f(x))) = f(x) = x/(x-1).

Solution

Correct: C

First, find the prime factorization of 360: 360 = 2^3 * 3^2 * 5^1. The number of divisors is found by adding 1 to each exponent and multiplying the results: (3+1)(2+1)(1+1) = 4 * 3 * 2 = 24.

Solution

Correct: A

sqrt(x+5) = 3 - sqrt(x+2). Squaring both sides: x+5 = 9 - 6sqrt(x+2) + x + 2. Simplifying: 6sqrt(x+2) = 6. sqrt(x+2) = 1. Squaring both sides: x+2 = 1. So, x = -1. Checking the solution: sqrt(-1+2) + sqrt(-1+5) = sqrt(1) + sqrt(4) = 1 + 2 = 3. Therefore, x = -1.

Solution

Correct: B

By Vieta's formulas, a + b = -3 and ab = 1. We want to find a^2 + b^2. a^2 + b^2 = (a+b)^2 - 2ab = (-3)^2 - 2(1) = 9 - 2 = 7.

Solution

Correct: C

The equation of an ellipse is x^2/a^2 + y^2/b^2 = 1. The area of the ellipse is πab. In this case, a = 3 and b = 2. The area is π(3)(2) = 6π.

Solution

Correct: C

Since the circle is tangent to the x-axis at the origin and has a radius of 5, its center must be at (0, 5) or (0, -5). Since the radius is 5, the circle must be above x-axis. Therefore, the center is at (0, 5). The equation of a circle with center (h, k) and radius r is (x-h)^2 + (y-k)^2 = r^2. In this case, (x-0)^2 + (y-5)^2 = 5^2, which simplifies to x^2 + (y-5)^2 = 25.

Solution

Correct: C

Using the disk method, the volume V is given by the integral of π[f(x)]^2 dx from a to b. In this case, f(x) = x^2, a = 0, and b = 2. So, V = integral from 0 to 2 of π(x^2)^2 dx = π integral from 0 to 2 of x^4 dx = π[x^5/5] from 0 to 2 = π(2^5/5 - 0^5/5) = π(32/5) = 32π/5.

Solution

Correct: B

We want to find 2^100 mod 3. Notice that 2 mod 3 is -1. Thus, 2^100 mod 3 is (-1)^100 mod 3 = 1 mod 3 = 1.

Solution

Correct: B

∫0 to π/2 sin^2(x) cos^3(x) dx = ∫0 to π/2 sin^2(x) cos^2(x) cos(x) dx = ∫0 to π/2 sin^2(x) (1 - sin^2(x)) cos(x) dx. Let u = sin(x), then du = cos(x) dx. When x = 0, u = 0. When x = π/2, u = 1. So, the integral becomes ∫0 to 1 u^2 (1 - u^2) du = ∫0 to 1 (u^2 - u^4) du = [u^3/3 - u^5/5] from 0 to 1 = (1/3 - 1/5) - (0 - 0) = (5 - 3)/15 = 2/15.