AMC 12-style practice problems

Hint 1: Express both logarithms using base 2.

Hint 2: If t = log_2(x), then log_4(x) = t/2.

Answer: C. 16

Let t = log_2(x). Since 4 = 2^2, log_4(x) = t/2. Thus 3t/2 = 6, so t = 4 and x = 2^4 = 16. The positive value 16 satisfies the original equation.

Hint 1: Square the given equation.

Hint 2: Use sin^2(theta) + cos^2(theta) = 1 and 2 sin(theta) cos(theta) = sin(2 theta).

Answer: A. -8/9

Squaring gives 1 + 2 sin(theta) cos(theta) = 1/9. Therefore sin(2 theta) = 1/9 - 1 = -8/9. Such angles exist because sin(theta) + cos(theta) ranges from -sqrt(2) to sqrt(2).

Hint 1: Square 1 + i first.

Hint 2: Its square is 2i; square twice more.

Answer: D. 16

We have (1 + i)^2 = 1 + 2i + i^2 = 2i. Thus (1 + i)^4 = (2i)^2 = -4 and (1 + i)^8 = (-4)^2 = 16.

Hint 1: Factor the equation as a quadratic in t^2.

Hint 2: The factors are (t^2 - 1)(t^2 - 4).

Answer: C. 5/2

The four roots are 1, -1, 2, and -2, all nonzero. The reciprocal squares sum to 1 + 1 + 1/4 + 1/4 = 5/2.

Hint 1: An integer logarithm means b = a^k for an integer k.

Hint 2: Here k must be positive. Count the permitted bases separately for each exponent.

Answer: D. 24

Since a and b are at least 2, k = log_a(b) must be a positive integer. For k = 1, all 19 bases work. For k = 2, the bases are 2, 3, and 4. For k = 3 only base 2 works, and for k = 4 only base 2 works.

For k >= 5, even 2^k exceeds 20. Each (a,b) has a unique exponent, so the total is 19 + 3 + 1 + 1 = 24.

Hint 1: Find the third side using the law of cosines.

Hint 2: The extended sine rule gives 2R = c/sin(120 degrees).

Answer: B. sqrt(327)/3

If c is opposite the 120-degree angle, then c^2 = 5^2 + 7^2 - 2 x 5 x 7 x cos(120 degrees) = 25 + 49 + 35 = 109.

By the extended sine rule, R = c/[2 sin(120 degrees)] = sqrt(109)/sqrt(3) = sqrt(327)/3.

Hint 1: Multiply by z to find a quadratic equation.

Hint 2: The equation implies z^3 = -1, so powers repeat with period 6.

Answer: B. -1

Multiplying by z gives z^2 - z + 1 = 0. Multiplying this by z + 1 yields z^3 + 1 = 0, so z^3 = -1 and z^6 = 1. As 2026 has remainder 4 on division by 6, the required sum is z^4 + z^(-4) = z^4 + z^2.

Using z^3 = -1 gives z^4 = -z, while z^2 = z - 1. Hence the sum is -z + z - 1 = -1 for either root.

Hint 1: Try transforming x by T(x) = (1 + x)/(1 - x).

Hint 2: For 0 <= x < 1, T(f(x)) = 3T(x).

Answer: D. 121/122

If 0 <= x < 1, then 0 < f(x) < 1, so all required denominators are nonzero. Direct simplification gives (1 + f(x))/(1 - f(x)) = 3(1 + x)/(1 - x).

Because T(x_0) = 1, we have T(x_5) = 3^5 = 243. Solving (1 + x_5)/(1 - x_5) = 243 gives 244x_5 = 242, so x_5 = 121/122.

Hint 1: Remove square factors from each integer.

Hint 2: A product is a square exactly when the two remaining squarefree parts are equal.

Answer: C. 5/66

Write each number as a square times a squarefree integer. The only groups with at least two members are {1,4,9}, with squarefree part 1; {2,8}, with part 2; and {3,12}, with part 3. All other groups are singletons.

These give 3 + 1 + 1 = 5 successful pairs. There are 12 x 11 / 2 = 66 equally likely distinct unordered pairs, so the probability is 5/66.

Hint 1: Set u = x/3 and v = y/2, so u^2 + v^2 = 1.

Hint 2: Then 2x + 3y = 6(u + v); use (u - v)^2 >= 0.

Answer: C. 6 sqrt(2)

Let u = x/3 and v = y/2. Then u^2 + v^2 = 1, and 2x + 3y = 6(u + v). Since (u - v)^2 >= 0, (u + v)^2 <= 2(u^2 + v^2) = 2. Therefore u + v <= sqrt(2).

Equality is attainable at u = v = 1/sqrt(2), corresponding to x = 3/sqrt(2) and y = 2/sqrt(2). Thus the greatest value is 6 sqrt(2).

Hint 1: If k of the nine factors contribute x^2, the others contribute 1/x.

Hint 2: The resulting exponent is 2k - (9 - k). Set it equal to zero.

Answer: C. 84

A term with k choices of x^2 has exponent 2k - (9 - k) = 3k - 9. This is zero exactly when k = 3. Choosing those three factors can be done in 9 choose 3 = 9 x 8 x 7 / 6 = 84 ways, each with coefficient 1.

Hint 1: Replace cos(2 theta) by sin(pi/2 - 2 theta).

Hint 2: For sin(u) = sin(v), use u = v + 2k pi or u = pi - v + 2k pi, and check for overlap.

Answer: B. 5

The equation becomes sin(3 theta) = sin(pi/2 - 2 theta). Thus either 3 theta = pi/2 - 2 theta + 2k pi, or 3 theta = pi - (pi/2 - 2 theta) + 2k pi, for an integer k.

The first family is theta = pi/10 + 2k pi/5. Exactly k = 0,1,2,3,4 lie in the given interval, giving five angles. The second family is theta = pi/2 + 2k pi, whose only angle in the interval is pi/2. It is already the first family with k = 1. Thus there are 5 distinct solutions.