Advanced high-school contest problems: trigonometry, complex numbers, polynomials, and functions at real AMC 12 difficulty. Try each problem first, then open the strategy analysis and worked solution.
Problem 1AMC 12 · Trigonometry
How many solutions does the equation sin 2x + cos x = 0 have for 0 ≤ x < 2π?
A. 2 · B. 3 · C. 4 · D. 5 · E. 6
Strategy analysis
FACTOR FIRST: use sin 2x = 2 sin x cos x so a common factor of cos x appears, turning one trig equation into two simple ones instead of a messy single equation.
Worked solution and answer
sin 2x + cos x = 2 sin x cos x + cos x = cos x (2 sin x + 1) = 0. So cos x = 0 or sin x = −1/2. On [0, 2π), cos x = 0 gives x = π/2 and x = 3π/2, while sin x = −1/2 gives x = 7π/6 and x = 11π/6. That is 4 solutions in total.
Answer: C
Problem 2AMC 12 · Complex Numbers
Let z be a complex number satisfying z + 1/z = 1. What is the value of z⁶ + 1/z⁶?
A. 0 · B. 1 · C. −1 · D. 2 · E. −2
Strategy analysis
ROOTS OF UNITY IN DISGUISE: clearing denominators gives a quadratic whose roots are 6th roots of unity, so the powers of z cycle with period 6 and the 6th power collapses to 1.
Worked solution and answer
Multiplying by z gives z² − z + 1 = 0, so z = e^(±iπ/3). Then z⁶ = e^(±2πi) = 1, and likewise 1/z⁶ = 1. Hence z⁶ + 1/z⁶ = 1 + 1 = 2.
Answer: D
Problem 3AMC 12 · Logarithms
What is the sum of all real numbers x satisfying log₂ x + log₄ x + log₈ x = 11?
A. 32 · B. 64 · C. 128 · D. 16 · E. 256
Strategy analysis
CHANGE OF BASE TO A SINGLE LOG: rewrite every term as a multiple of log₂ x, so the left side becomes one logarithm with a fractional coefficient that is easy to clear.
Worked solution and answer
log₄ x = (1/2) log₂ x and log₈ x = (1/3) log₂ x, so the equation becomes (1 + 1/2 + 1/3) log₂ x = (11/6) log₂ x = 11. Thus log₂ x = 6, giving x = 64, which is valid since x > 0. The sum of all such x is 64.
Answer: B
Problem 4AMC 12 · Polynomials
Let a, b, and c be the roots of x³ − 7x + 6 = 0. What is a³ + b³ + c³?
A. −18 · B. −6 · C. 0 · D. 6 · E. 18
Strategy analysis
NEWTON'S IDENTITIES — NEVER SOLVE FOR THE ROOTS: use Vieta's formulas plus the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), and notice the root sum is 0, which kills the whole second factor.
Worked solution and answer
By Vieta, a + b + c = 0 and abc = −6. The identity gives a³ + b³ + c³ − 3abc = (a + b + c)(…) = 0, so a³ + b³ + c³ = 3abc = 3(−6) = −18.
Answer: A
Problem 5AMC 12 · Analytic Geometry
The line y = mx + 5 is tangent to the circle x² + y² = 16. What is the value of m²?
A. 1/4 · B. 3/4 · C. 1 · D. 25/16 · E. 9/16
Strategy analysis
DISTANCE FROM CENTER TO TANGENT LINE EQUALS THE RADIUS: skip solving the system and instead set the point-to-line distance from the origin equal to 4.
Worked solution and answer
Write the line as mx − y + 5 = 0. Its distance from the origin is |5| / √(m² + 1). Tangency means this equals the radius 4, so 25 / (m² + 1) = 16, giving m² + 1 = 25/16 and m² = 9/16.
Answer: E
Problem 6AMC 12 · Sequences and Series
A sequence is defined by a₁ = 2 and aₙ₊₁ = 3aₙ − 2 for n ≥ 1. What is a₁₀?
A. 19682 · B. 19683 · C. 19684 · D. 19685 · E. 59050
Strategy analysis
SHIFT TO KILL THE CONSTANT TERM: subtract the fixed point of the recurrence (here 1) so the shifted sequence becomes purely geometric and can be written in closed form.
A function f defined on all real numbers satisfies f(x + y) = f(x) + f(y) + 2xy for all real x and y, and f(1) = 3. What is f(5)?
A. 27 · B. 35 · C. 30 · D. 40 · E. 45
Strategy analysis
SUBTRACT THE CROSS TERM: set g(x) = f(x) − x², which turns the given relation into the clean additive equation g(x + y) = g(x) + g(y), solvable at integers by repeated addition.
How many four-digit positive integers have digits whose sum is 12?
A. 320 · B. 335 · C. 355 · D. 342 · E. 364
Strategy analysis
STARS AND BARS, THEN SUBTRACT THE DIGIT-CAP VIOLATIONS: count unrestricted digit sums first, then use inclusion–exclusion to remove the cases where a digit reaches 10 (the first digit also cannot be 0, so shift it first).
Worked solution and answer
Write the digits as a, b, c, d with a ≥ 1 and b, c, d ≥ 0. Setting a′ = a − 1 gives a′ + b + c + d = 11 with all variables nonnegative, which has C(14, 3) = 364 solutions. Subtract violations of the digit cap 9: a′ ≥ 9 leaves a sum of 2, giving C(5, 3) = 10 cases, and each of b, c, d ≥ 10 leaves a sum of 1, giving C(4, 3) = 4 cases each, for 12 more. No two violations can occur at once since 9 + 10 > 11. Hence the count is 364 − 10 − 12 = 342.
Answer: D
Problem 9AMC 12 · Trigonometric Identities
What is cos² 10° + cos² 50° + cos² 70°?
A. 3/2 · B. 1 · C. 2 · D. 5/2 · E. 3
Strategy analysis
THE 120°-SPACED SQUARES IDENTITY: angles 10°, 70°, and 130° are spaced 60° apart, and the sum of squared cosines over three angles spaced 120° apart in the double-angle sense is always 3/2 — and cos² 130° = cos² 50° closes the match.
Worked solution and answer
Use cos² θ = (1 + cos 2θ)/2. The sum equals 3/2 + (1/2)(cos 20° + cos 100° + cos 140°). But cos 100° + cos 140° = 2 cos 120° cos 20° = −cos 20°, so the cosine terms cancel, leaving 3/2.
Answer: A
Problem 10AMC 12 · Trigonometry
What is the maximum value of sin x + cos x + sin x cos x over all real numbers x?
A. √2 · B. 1 + √2 · C. 2 · D. (3 + √2)/2 · E. (1 + 2√2)/2
Strategy analysis
SUBSTITUTE t = sin x + cos x: then sin x cos x = (t² − 1)/2, turning a trig expression into a single quadratic in t whose range is the interval [−√2, √2], so the maximum sits at an endpoint.
Worked solution and answer
Let t = sin x + cos x. Then t² = 1 + 2 sin x cos x, so sin x cos x = (t² − 1)/2. The expression becomes t + (t² − 1)/2 = (t² + 2t − 1)/2, a parabola opening upward with vertex at t = −1. On t ∈ [−√2, √2] the maximum is at the right endpoint t = √2, giving (2 + 2√2 − 1)/2 = (1 + 2√2)/2.