Berkeley Math Tournament style problems: algebra, geometry, discrete math, number theory, and calculus. Try each problem first, then open the strategy analysis and worked solution.
Real numbers x and y satisfy x + y = 7 and x² + y² = 29. Compute x³ + y³.
SYMMETRIC SUMS — NEVER SOLVE FOR x AND y: get xy from the sum and the sum of squares, then use the factored identity for x³ + y³.
From (x + y)² = x² + 2xy + y², we get 49 = 29 + 2xy, so xy = 10. Then x³ + y³ = (x + y)³ − 3xy(x + y) = 343 − 3 · 10 · 7 = 343 − 210 = 133.
Answer: 133
Find the real number x satisfying √(x + 3) + √(x − 1) = 4.
ISOLATE ONE RADICAL BEFORE SQUARING: moving one square root to the other side keeps the algebra linear after a single squaring, and the domain check at the end kills any extraneous root.
Isolate: √(x + 3) = 4 − √(x − 1). Squaring gives x + 3 = 16 − 8√(x − 1) + x − 1, so 8√(x − 1) = 12, hence √(x − 1) = 3/2 and x = 13/4. Checking: √(25/4) + √(9/4) = 5/2 + 3/2 = 4, so the root is valid.
Answer: 13/4
Compute the minimum value of x² − 6x + 11 + |x − 2| over all real x.
SPLIT AT THE ABSOLUTE-VALUE KINK: treat x ≥ 2 and x < 2 as separate quadratics, find each vertex, and only accept a vertex that lies in its own region.
For x ≥ 2, the expression is x² − 5x + 9, whose vertex is at x = 5/2 (inside the region), giving 25/4 − 25/2 + 9 = 11/4. For x < 2, it is x² − 7x + 13, which is decreasing on this region since its vertex is at x = 7/2, so its values exceed the limiting value 3 at x = 2. The smaller value is 11/4.
Answer: 11/4
A circle has radius 5. Two parallel chords of lengths 6 and 8 lie on opposite sides of the center. Compute the distance between the two chords.
DROP PERPENDICULARS FROM THE CENTER: each chord's distance from the center comes from a right triangle with the radius as hypotenuse and half the chord as a leg — and because the chords are on opposite sides, the two distances add.
Half the chord of length 6 is 3, so its distance from the center is √(5² − 3²) = √16 = 4. Half the chord of length 8 is 4, so its distance from the center is √(5² − 4²) = √9 = 3. Since the chords lie on opposite sides of the center, the distance between them is 4 + 3 = 7.
Answer: 7
Point P lies outside a circle. A tangent from P to the circle has length 12, and a secant through P and the center of the circle meets the circle at a nearer point 6 units from P. Compute the radius of the circle.
POWER OF A POINT: the square of the tangent length equals the product of the external segment and the whole secant, which turns the radius into a one-step linear equation.
Let the radius be r. The whole secant through the center has length 6 + 2r. Power of a point gives 12² = 6(6 + 2r), so 144 = 36 + 12r, giving 12r = 108 and r = 9.
Answer: 9
Point P lies inside rectangle ABCD with PA = 7, PB = 10, and PD = 11. Compute PC.
BRITISH FLAG THEOREM: for any point in a rectangle, the sums of the squares of distances to opposite vertices are equal, so PC comes out without ever finding where P is.
By the British Flag Theorem, PA² + PC² = PB² + PD². Thus 49 + PC² = 100 + 121 = 221, so PC² = 172 and PC = √172 = 2√43.
Answer: 2√43
Triangle ABC has AB = 7, AC = 8, and BC = 9. The internal bisector of angle A meets BC at D. Compute AD.
ANGLE BISECTOR THEOREM PLUS STEWART: split BC in the ratio of the adjacent sides, then use AD² = AB · AC − BD · DC to avoid solving the full Stewart equation.
By the Angle Bisector Theorem, BD / DC = AB / AC = 7/8, and BD + DC = 9, so BD = 21/5 and DC = 24/5. Then AD² = AB · AC − BD · DC = 56 − (21/5)(24/5) = 56 − 504/25 = 896/25. Hence AD = √896 / 5 = 8√14 / 5.
Answer: 8√14/5
How many integers between 1 and 200, inclusive, are divisible by 3 or by 5 but not by 7?
INCLUSION–EXCLUSION TWICE: first count multiples of 3 or 5, then subtract the ones that are also multiples of 7, handling the 3-and-5 overlap with least common multiples in both passes.
Multiples of 3 or 5 up to 200: ⌊200/3⌋ + ⌊200/5⌋ − ⌊200/15⌋ = 66 + 40 − 13 = 93. Among these, the ones also divisible by 7 are multiples of 21 or 35: ⌊200/21⌋ + ⌊200/35⌋ − ⌊200/105⌋ = 9 + 5 − 1 = 13. So the desired count is 93 − 13 = 80.
Answer: 80
How many binary strings of length 10 contain no two consecutive 1s?
RECURSION ON THE FIRST DIGIT: a valid string either starts with 0 (leaving any valid string of length n − 1) or starts with 10 (leaving any valid string of length n − 2), giving a Fibonacci-type recurrence instead of a brute-force count.
Let aₙ be the count for length n. Then aₙ = aₙ₋₁ + aₙ₋₂, with a₁ = 2 and a₂ = 3. Continuing: 5, 8, 13, 21, 34, 55, 89, 144. So a₁₀ = 144.
Answer: 144
How many 3-element subsets of {1, 2, …, 12} have a sum divisible by 3?
SORT BY RESIDUES MOD 3: the residues 0, 1, 2 each occur exactly 4 times, so a subset sum is divisible by 3 only for residue patterns (0, 0, 0), (1, 1, 1), (2, 2, 2), or (0, 1, 2).
Each residue class has 4 elements. The counts are C(4, 3) = 4 for each of the three monochromatic patterns, plus 4 · 4 · 4 = 64 for the mixed pattern (0, 1, 2). Total: 4 + 4 + 4 + 64 = 76.
Answer: 76
Find the smallest positive integer n such that 2ⁿ − 1 is divisible by 21.
COMBINE THE ORDERS WITH CRT: the order of 2 modulo 21 is the least common multiple of its orders modulo 3 and modulo 7, so check the two small moduli instead of testing powers blindly.
We need 2ⁿ ≡ 1 (mod 3) and 2ⁿ ≡ 1 (mod 7). Modulo 3, the order of 2 is 2; modulo 7, the powers are 2, 4, 1, so the order is 3. Hence n must be a multiple of lcm(2, 3) = 6, and 2⁶ = 64 ≡ 1 (mod 21), so the smallest such n is 6.
Answer: 6
How many pairs (x, y) of positive integers satisfy 1/x + 1/y = 1/6?
SIMON'S FAVORITE FACTORING TRICK: clear denominators and rewrite the equation as (x − 6)(y − 6) = 36, turning a reciprocal equation into a divisor count.
Clearing denominators gives 6x + 6y = xy, which rearranges to (x − 6)(y − 6) = 36. Since x, y are positive and 1/x < 1/6, both factors are positive divisors of 36. The number 36 has 9 positive divisors, each giving one pair, so there are 9 pairs.
Answer: 9
Let f(x) = x²e^(−x). Find the positive x-coordinate at which f attains its maximum value.
PRODUCT RULE PLUS A SIGN CHART: differentiate, factor out e^(−x), and read the sign change of the remaining quadratic factor to locate the maximum without graphing.
f′(x) = 2xe^(−x) − x²e^(−x) = xe^(−x)(2 − x). The critical points are x = 0 and x = 2. For 0 < x < 2, f′ > 0, and for x > 2, f′ < 0, so f increases then decreases and attains its maximum at x = 2.
Answer: 2
Compute ∫₀¹ x√(1 + x²) dx.
U-SUBSTITUTION SPOTTED FROM THE DERIVATIVE: the factor x outside the square root is (up to a constant) the derivative of 1 + x² inside it, so set u = 1 + x² and the integral becomes a power rule.
Let u = 1 + x², so du = 2x dx. When x goes from 0 to 1, u goes from 1 to 2. The integral becomes (1/2)∫₁² √u du = (1/2) · (2/3)[u^(3/2)]₁² = (1/3)(2√2 − 1).
Answer: (2√2 − 1)/3