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SMT Algebra Mastery Guide
Stanford Math Tournament · 2019–
A Complete Topic Reference & 1.5-Month Prep Plan
Built from Analysis of 80+ Problems Across 6 Years
Target Profile: Strong AIME performer (10–11 on AIME), aiming for USAMO-level mastery Time Budget: 1.5 hours per day × 45 days = 67.5 hours total Focus: SMT Algebra specifically — topics, difficulty tiers, patterns, and drills
Covers: Vieta’s Formulas · Polynomial Theory · Symmetric Functions · Functional Equations · Inequalities · Complex Numbers · Sequences & Series · Logarithms · Optimization · and more
Contents
- Introduction: Understanding SMT Algebra
1.1. What SMT Algebra Actually Tests
The Stanford Math Tournament Algebra round is a 10-problem, 50-minute exam. After an- alyzing all available years (2019–2023, 2025) — over 80 main round and tiebreaker problems — a clear picture emerges of what the test rewards. SMT Algebra is not a standard AIME algebra exam. It is significantly more diverse, faster-paced per problem, and rewards three skills above all else:
- Pattern recognition speed. Problems 1–3 are designed to fall quickly if you see the right manipulation. You should solve them in under 5 minutes each.
- Algebraic fluency. Mid-range problems (4–7) require seamless command of Vieta’s formulas, symmetric functions, polynomial manipulation, and inequalities. There is rarely a long computation; the solution almost always hinges on one key identity or substitution.
- Olympiad-style cleverness. Problems 8–10 are full olympiad problems compressed into short-answer format. They demand functional equations, Cauchy-Schwarz in clever forms, group theory, and deep polynomial theory. These are solvable but require genuine mathematical creativity.
1.2. Problem Distribution by Topic (2019–2025)
Based on full analysis of all years, here is the approximate frequency of each topic across main and tiebreaker rounds:
Topic Approx. Frequency Typical Difficulty Vieta’s Formulas & Polynomial Theory Very High Medium–Hard Symmetric Functions & Newton’s Identities High Hard Functional Equations High Hard–Olympiad Sequences & Recurrences High Medium–Olympiad Inequalities (AM-GM, Cauchy-Schwarz) Medium-High Hard Complex Numbers Medium-High Hard Logarithms & Exponentials Medium Medium–Hard Optimization (min/max) Medium Hard Modular Arithmetic / Number Theory crossover Low-Medium Hard Generating Functions / Power Series Low Olympiad
1.3. Difficulty Tier Framework
Throughout this guide, every topic and technique is tagged with a difficulty tier based on how problems using it appear in SMT:
- Tier 1 (Foundation): Required for problems 1–3. You must be able to execute these instantly.
- Tier 2 (Core): Required for problems 4–6. Your main score driver.
- Tier 3 (Advanced): Problems 7–8. Needs deliberate practice.
- Tier 4 (Olympiad): Problems 9–10. Think of these as USAMO warm-ups.
2.3. Radical & Algebraic Identity Recognition
Identities to Internalize
a^3 + b^3 = (a + b)(a^2 − ab + b^2 ) a^3 − b^3 = (a − b)(a^2 + ab + b^2 ) a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 − 2 ab) (Sophie Germain) (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
SMT Example 2.4 (SMT 2022, Problem 1). Compute
P 31
n=
n+ √ √ n^2 −^1 n+√n− 1. Solution: Rationalize: n+
√ √ n^2 −^1 n+√n− 1 =^
(√n+ √ n−1)(n+
√ (n)(n−1)) n+√n− 1.^ After rationalization, the numerators become differences of cubes, yielding a telescoping sum equal to 32
SMT Example 2.5 (SMT 2020, Problem 1). Evaluate
Solution: Note 2223^2 − 8888 = (2222 + 1)^2 − 4(2222) = (2222 − 1)^2 = 2221^2. Answer:
2.4. Logarithm and Exponential Basics
Key Logarithm Facts for SMT
loga b =
logc b logc a
(change of base)
loga b · logb c = loga c (chain rule) aloga^ b^ = b loga(xy) = loga x + loga y
When the equation is of the form ax^ = f (x), look for bijectivity: if both sides are strictly monotone in opposite directions, the solution is unique and can often be found by inspection or substitution.
SMT Example 2.6 (SMT 2020, Problem 2). log 2020 a = 2020^20 − a has a unique solution a, and 2020b^ = 2020^20 − b has a unique solution b. Find a + b. Solution: Let x = 2020^20 − a. Then 2020x^ = 2020^20 − x, which means x = b. So a + b = a + x = 2020^20.
- Tier 2: Core Topics
3.1. Vieta’s Formulas — The SMT Workhorse
Vieta’s formulas appear in virtually every SMT Algebra round. At the AIME→USAMO level, the key is using Vieta’s indirectly — not to find individual roots, but to evaluate symmetric expressions of the roots without ever knowing them.
Vieta’s Formulas
For p(x) = anxn^ + an− 1 xn−^1 + · · · + a 0 with roots r 1 ,... , rn: X ri = −
an− 1 an
X
i
3.4. Geometric Sequences and AM-GM
AM-GM at the Competition Level
Beyond the basic inequality a+ 2 b≥
ab, SMT uses:
- Weighted AM-GM: λa + (1 − λ)b ≥ aλb^1 −λ.
- AM-GM with split terms: To bound A+B +C ≥ k, split terms so exponents match.
- AM-GM for ratios: uv + vu ≥ 2; more generally uv + k vu ≥ 2
k.
SMT Example 3.7 (SMT 2019, Problem 7). Minimize 5 x
(^2) − 2 xy+y 2 x^2 −y^2 for^ x > y >^ 0. Solution: Via partial fractions: 5 x^2 − 2 xy + y^2 x^2 − y^2
x + y x − y
2(x − y) x + y
by AM-GM on the two ratio terms. Equality when (x + y)^2 = 2(x − y)^2.
SMT Example 3.8 (SMT 2022, Problem 5). With xyz = 10, maximize x^3 y^3 z^3 − 3 x^4 − 12 y^2 − 12 z^4. Solution: Note x^3 y^3 z^3 = (xyz)^3 = 1000. Minimize 3x^4 + 12y^2 + 12z^4. Split 12y^2 = 6 y^2 +6y^2. Apply AM-GM to 3x^4 +6y^2 +6y^2 +12z^4 ≥ 4 4
p 3 · 6 · 6 · 12 · x^4 y^4 z^4 = 4· 4
1296 ·xyz = 24 · 10 = 240. Maximum is 1000 − 240 = 760.
3.5. Sequences and Recurrences
Recurrence Types in SMT
- Linear: an+1 = pan + qan− 1. Characteristic equation r^2 = pr + q. General solution an = c 1 rn 1 + c 2 rn 2.
- Chebyshev-type: an+1 = kan − an− 1 with trig substitution an = 2 cos(nθ) or similar.
- Quadratic-type: an+1 = a^2 n − c. Grows doubly-exponentially; use an ≈ r^2 n .
- Nonlinear: Convert to linear via clever algebraic manipulation (SMT 2025 #9).
SMT Example 3.9 (SMT 2021, Tiebreaker 3). x^2 − 3 x + 1 = 0. Then x^16 − kx^8 + 1 = 0. Find k. Solution: Rearrange: x + 1/x = 3. Let Pn = xn^ + x−n. Then P 2 n = P (^) n^2 − 2. So: P 1 = 3, P 2 = 7, P 4 = 47, P 8 = 2207. Since x^16 − kx^8 + 1 = x^8 (x^8 + x−^8 − k) = 0, we get k = P 8 = 2207.
SMT Example 3.10 (SMT 2020, Problem 7). Sequence with a 0 =
3 , a 1 =
2 , a 3 = −1, an = an− 1 an− 2 − an− 3.
Solution: Recognize an = 2 cos(15F (^) n◦+3) where Fn is the Fibonacci sequence. Since Fibonacci is periodic mod 24 (period 24), an is periodic with period 24. So a 2020 = a 4 =
2 cos(195) = −
√ 6+ √ 2
3.6. Nested Radicals and Denesting
Cube Root Denesting
When you see (A + B
C)^1 /^3 + (A − B
C)^1 /^3 , let the sum equal S and cube both sides: S^3 = 2A + 3(A^2 − B^2 C)^1 /^3 · S. This gives a depressed cubic in S.
SMT Example 3.11 (SMT 2019, Problem 4). Evaluate (350+
15)^1 /^3 +(350− 90
15)^1 /^3.
Solution: Let S be the sum. Cube: S^3 = 700 + 3(350^2 − 902 · 15)^1 /^3 · S = 700 + 30S. So S^3 − 30 S − 700 = (S − 10)(S^2 + 10S + 70) = 0. The real root is S = 10.
4.2. Power Series and Generating Functions
Differentiation of Power Series
To compute
P∞
n=1 n
(^2) rn, use:
X^ ∞
n=
rn^ =
1 − r X^ ∞
n=
nrn^ =
r (1 − r)^2
(differentiate and multiply by r)
X^ ∞
n=
n(n − 1)rn^ =
2 r^2 (1 − r)^3
(differentiate again)
X^ ∞
n=
n^2 rn^ =
r(1 + r) (1 − r)^3
SMT Example 4.4 (SMT 2019, Tiebreaker 3). Compute
P∞
n=
n 2
4
n . Solution: Let f (x) =
P∞
n=2 n(n^ −^ 1)x
n (^) = 2 x^2 (1−x)^3. At^ x^ = 3/4:^ f^ (3/4) =^
2(9/16) (1/4)^3 =^
9 / 8 1 / 64 =
- The sum equals 12 f (3/4) = 36.
4.3. Polynomial Divisibility and Modular Polynomial Arithmetic
Key Technique: Evaluate at Roots
If p(x) | q(x), then every root r of p satisfies q(r) = 0. For x^2 − x + 1, its roots are primitive 6th roots of unity satisfying ω^6 = 1, ω^3 = −1. Always reduce ωk^ mod the minimal polynomial.
Remainder Theorem Trick
To find the remainder when f (x) is divided by (x − a)(x − b): Write remainder as Rx + S. Then f (a) = Ra + S and f (b) = Rb + S. Two equations, two unknowns.
SMT Example 4.5 (SMT 2021, Problem 1). Find the remainder when x^6 is divided by x^2 − 3 x + 2. Solution: Roots of x^2 − 3 x + 2 are 1 and 2. If remainder is ax + b: a + b = 1 and 2 a + b = 64. So a = 63, b = −62. Remainder: 63x − 62.
4.4. Cauchy-Schwarz and Its Variants
Forms of Cauchy-Schwarz
- Standard: (
P
aibi)^2 ≤ (
P
a^2 i )(
P
b^2 i ).
- Engel/Sedrakyan (titu’s lemma):
P (^) a (^2) i bi ≥^
(P P^ a i)^2 bi.
- Quadratic form: (λ 1 a^21 + λ 2 a^22 )(b^21 /λ 1 + b^22 /λ 2 ) ≥ (a 1 b 1 + a 2 b 2 )^2.
- RMS-AM-GM chain: RMS ≥ AM ≥ GM ≥ HM.
SMT Example 4.6 (SMT 2022, Tiebreaker 1). Maximize x+2y +3z given x^2 +2y^2 +3z^2 =
Solution: By Cauchy-Schwarz: (1 + 2 + 3)(x^2 + 2y^2 + 3z^2 ) ≥ (x + 2y + 3z)^2. So (x + 2y + 3 z)^2 ≤ 576, giving max 24.
SMT Example 4.7 (SMT 2021, Problem 9). Let f (m, n) = m^4 (8 − m^4 ) + 2m^2 n^2 (12 − m^2 n^2 ) + n^4 (18 − n^4 ) − 100. Find smallest a with f (m, n) ≤ a.
p^ Solution:^ Rewrite: 2(2m^2 + 3n^2 )^2 −^ (m^4 +^ n^4 )^2 ≤^ c. By Cauchy-Schwarz, 2m^2 + 3n^2 ≤ (4 + 9)(m^4 + n^4 ) =
p 13(m^4 + n^4 ). So minimum k =
13, giving c = 169 and a = 69.
4.5. Telescoping: Advanced Forms
Telescoping Variants
- Cotangent telescope: cot−^1 (n(n + 1) + 1) = cot−^1 (n) − cot−^1 (n + 1)
- Log telescope: log(f (n + 1)/f (n)) sums to log(f (N + 1)/f (1))
- Partial fraction telescope: (^) n(n^1 +1) = (^) n^1 − (^) n^1 +1 , extended to higher degree
SMT Example 4.8 (SMT 2022, Problem 6). Compute cot
P 23
n=1 cot
− (^1) (1 + Pn k=1 2 k)
Solution: Note
Pn k=1 2 k^ =^ n(n+1). So we sum cot − (^1) (1+n(n+1)) = cot− (^1) (n)−cot− (^1) (n+
1). This telescopes to cot−^1 (1) − cot−^1 (24) = cot−^1 (25/23), giving cot = 25/23.
SMT Example 4.9 (SMT 2023, Problem 4). For k = 2,... , 2023, the equation 2^2 x^ − 2 x+1^ + 1 − 1 /k^2 = 0 has real roots summing to N. Find 2N^.
Solution: Roots are x = log 2 (1 ± 1 /k). Sum for each k: log 2
k^2 − 1 k^2
. Total: N =
log 2
Q 2023
k=
(k−1)(k+1) k^2 = log^2
1 · 2024 2 · 2023 = log^2
1012
- So 2
N = 1012
5.3. Strictly Increasing Functions and Olympiad Sequences
SMT Example 5.4 (SMT 2019, Problem 9). f : Z+^ → Z+^ strictly increasing. Both f (f (1)), f (f (2)),... and f (f (1) + 1), f (f (2) + 1),... are arithmetic sequences. f (1) = 1, f (2) = 3. Find max
P 100
j=1 f^ (j). Solution: The key claim is that f itself must be arithmetic. Proof: Let dn = f (n + 1) − f (n). Since both sequences above are arithmetic, careful bounding shows dn is constant. With d 1 = f (2) − f (1) = 2, we get f (j) = 2j − 1 and
P 100
j=1(2j^ −^ 1) = 100
5.4. Group Theory and Algebraic Structures
SMT Example 5.5 (SMT 2020, Tiebreaker 3). S is closed under composition, contains
r(x) = x
√3+ /(
3 − x) and s(x) = 1/x. What is the minimum size of S? Solution: Note r^6 = id and s^2 = id. The group generated is isomorphic to the dihedral group D 6 , with |D 6 | = 12. The minimum set S has size 12.
SMT Example 5.6 (SMT 2022, Problem 7). f : M × M → M (where M = { 0 ,... , 2022 }) satisfying f (a, f (b, a)) = b and f (x, x) ̸= x. How many such f exist? Solution: Each triple (x, y, z) with f (x, y) = z generates a cyclic triple {(x, y, z), (y, z, x), (z, x, y)}. So 3 | |M |^2 = 2023^2. But 2023 ≡ 1 (mod 3), so 3 ∤ 20232. Contradiction. Answer: 0.
5.5. Advanced Series and Convergence
SMT Example 5.7 (SMT 2025, Problem 9). Define a 1 = 5, an+1 = 5 an+
21 a^2 n+ P∞^2.^ Find n=
1 anan+. Solution: From the recurrence, a^2 n − 5 anan+1 + a^2 n+1 = 1, and an− 1 + an+1 = 5an. This
gives (^) ana^1 n+1 = (^) aann+1 − an a−n 1 for n ≥ 2, a telescoping sum. As N → ∞, aN /aN +1 → 5 −
√ 21
Final answer: 23 −^5
√ 21
The 1.5-Month SMT Prep Plan
7.1. Overview and Philosophy
Core Principles
- Prioritize coverage over perfection. At AIME 10–11 level, you already know most foundations. The gaps are in olympiad-level topics. Don’t over-drill what you already do well.
- Practice under time pressure from Week 2 onward. SMT is 50 minutes / 10 problems. Speed matters.
- Daily review of one SMT solution. Even on days not focused on a specific topic, reading one elegant SMT solution builds pattern vocabulary.
- Track your weak topics. After each session, note which technique caused difficulty. Return to it within 3 days.
7.2. Phase 1: Foundation and Core (Days 1–15, 1.5 hrs/day)
The goal of this phase is to close any gaps in Tier 1 and Tier 2 material and build fluency in the tools that appear in 70% of SMT problems.
Day Focus Session Plan (1.5 hrs)
1 Vieta’s Formulas 45 min: Study Section 3.1 of this guide. 45 min: Solve SMT 2019 #5, 2022 #4, 2025 #3 fully from scratch. 2 Newton’s Identities 40 min: Derive p 1 , p 2 , p 3 , p 4 from scratch. 50 min: Solve 2019 #3, 2025 #5, 2020 Tie #2. 3 Symmetric Functions 30 min: Review e 1 , e 2 , e 3 and Vieta pairing. 60 min: Solve 2021 #4, 2021 #7. Write up both solutions cleanly. 4 AM-GM (advanced) 40 min: Study splitting and ratio forms. 50 min: Solve 2019 #7, 2022 #5, 2021 #9. 5 Cauchy-Schwarz 30 min: All four forms. 60 min: Solve 2022 Tie #1, 2021 #9, 2023 #8. 6 Sequences (linear) 40 min: Characteristic equation method. 50 min: Solve 2021 Tie #3, 2023 #7. 7 Review Day 1 30 min: Re-do any problem from Days 1–6 that took >10 min. 60 min: Full timed attempt on SMT 2020 problems 1–5 (25 min), then grade and review. 8 Logarithms 40 min: Key identities + bijectivity. 50 min: Solve 2020 #2, 2021 Tie #1, 2023 #4. 9 Polynomial divisibil- ity
40 min: Roots of unity, remainder the- orem. 50 min: Solve 2021 #1, 2020 #4. 10 Complex Numbers (basic)
45 min: Review modulus, conjugation, unit circle. 45 min: Solve 2019 Tie #2, 2023 Tie #1. 11 Complex Numbers (advanced)
40 min: M¨obius transforms, inversion. 50 min: Solve 2022 Tie #2, 2023 #6. 12 Nested radicals 30 min: Denesting technique. 60 min: Solve 2019 #4, 2020 Tie #1, 2022 #2. 13 Telescoping 40 min: All telescope forms. 50 min: Solve 2022 #1, 2022 #6, 2023 #4. 14 Power series 40 min: Differentiating power series. 50 min: Solve 2019 Tie #3, study 2025 #7. 15 Review Day 2 Full 90-min timed mock: SMT 2021 Problems 1–10. Target: 7 correct. Grade, review every missed problem.
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