W23 · Mastery gates26–30 hrs

Week 23 of 28 · Mastery · due 2026-10-31 · 26–30 hrs

Nightmare Ramp I

Nightmare Ramp I gate complete

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This week's lesson · 9 min read

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Nightmare ramp I: asymptotics, estimation, and the analysis endgame

The A5/B5 analysis problem has a house style: it asks not for an exact answer but for a RATE — how fast a sequence converges, the leading order of a sum, whether a product blows up. This week's ramp trains that style. The tools are ones you own (integral comparison, Taylor with remainder, Stolz-Cesàro, summation by parts); the new skill is estimation COURAGE — choosing what to throw away, and proving your trash bound.

Hard analysis is bookkeeping with conviction: the leading term is usually obvious — the exam grades whether you can bound everything you discarded.

Key ideas — the week on one card

  1. The asymptotics house method: find the driver, Taylor-expand WITH explicit remainders, sum main terms precisely and bound the trash crudely.
  2. Decaying recursions $x_{n+1} = x_n - c x_n^k$ linearize under $y_n = x_n^{-(k-1)}$: the differences converge, Cesàro gives linear growth.
  3. Expansions may only be iterated or summed when they carry explicit remainders — per-step errors compound.
  4. Split ranges out loud ('terms beyond $M$ contribute $o(1)$') and defend tails with worst-case times count or a geometric comparison.
  5. Reflexes on tap: $1 + t \le e^t$, the $\ln$ squeeze $\frac{t}{1+t} \le \ln(1+t) \le t$, $H_N = \ln N + \gamma + O(1/N)$.

1The asymptotic toolkit, assembled

The three-step house method for 'find the asymptotics of $x_n$' problems. IDENTIFY the driver: which operation dominates for large $n$ (the recursion's fixed point, the integral's peak, the sum's largest terms)? EXPAND around it: Taylor with EXPLICIT remainder — $\ln(1 + t) = t - \frac{t^2}{2} + O(t^3)$, $(1+t)^\alpha = 1 + \alpha t + O(t^2)$ — never bare '$\approx$'. SUM/ITERATE the expansion: the main terms accumulate into the answer; the $O$-terms accumulate into something you must SHOW is smaller, usually by comparison with a geometric or $p$-series.

The recurrence archetype: $x_{n+1} = x_n - c x_n^{k}$ (with $x_n \to 0^+$, $k > 1$) has $x_n \sim \left(\frac{1}{c(k-1)n}\right)^{1/(k-1)}$, found by the POWER TRANSFORM: set $y_n = x_n^{-(k-1)}$ and show $y_{n+1} - y_n \to c(k-1)$, so $y_n \sim c(k-1)n$ — differences that converge give linear growth (Cesàro/Stolz), and the transform is chosen precisely to make the differences converge. The famous instance: $x_{n+1} = \sin x_n$ gives $x_n \sim \sqrt{3/n}$ ($k = 3$, $c = \frac16$).

The sum archetype: for $\sum_{k=1}^n f(k)$ with $f$ smooth monotone, the ladder of precision is: integral comparison (main term), endpoint correction $\frac{f(1)+f(n)}{2}$ (next order), Euler-Maclaurin only if the problem visibly demands a third order (it almost never does). Matching the precision to the question is itself a graded skill — computing three orders when one suffices burns clock; computing one when two are needed loses the problem.

√(3/n)xₙ₊₁ = sin xₙ
Computed, not sketched: forty iterates of sine converging into the asymptotic √(3/n) envelope the power transform predicts.

Worked example

$x_1 = 1$, $x_{n+1} = \sin(x_n)$. Show $\lim_{n\to\infty} \sqrt{n}\, x_n = \sqrt{3}$.

  1. Nudge 1

    Establish the driver first: prove $x_n$ decreases to 0 before any expansion.

    Reveal step 1

    Driver: $x_n \downarrow 0$ (positive, decreasing since $\sin t < t$ on $(0, \pi]$, and the only fixed point of $\sin$ in $[0,1]$ is 0 — one line each). Expand the recursion near 0: $\sin t = t - \frac{t^3}{6} + O(t^5)$.

  2. Nudge 2

    Transform so the DIFFERENCES converge: try $y_n = x_n^{-2}$ and expand $\sin$ with its remainder.

    Reveal step 2

    Power transform to linearize: with $y_n = \frac{1}{x_n^2}$, compute $y_{n+1} - y_n = \frac{1}{\sin^2 x_n} - \frac{1}{x_n^2}$. Substituting the expansion: $\sin^2 t = t^2\left(1 - \frac{t^2}{3} + O(t^4)\right)$, so $\frac{1}{\sin^2 t} = \frac{1}{t^2}\left(1 + \frac{t^2}{3} + O(t^4)\right)$, giving $y_{n+1} - y_n = \frac{1}{3} + O(x_n^2) \to \frac{1}{3}$.

  3. Nudge 3

    Differences with a limit force linear growth — quote Stolz–Cesàro and translate back.

    Reveal step 3

    Cesàro: differences converging to $\frac13$ force $\frac{y_n}{n} \to \frac13$ (Stolz-Cesàro, or average the differences), i.e. $n x_n^2 \to 3$, i.e. $\sqrt{n}\,x_n \to \sqrt 3$. The whole proof is: one monotonicity paragraph, one Taylor expansion WITH remainder, one Cesàro citation — the A5 analysis template in its entirety.

  4. Answer

    $\sqrt n \, x_n \to \sqrt 3$: transform $y_n = x_n^{-2}$, show $\Delta y_n \to \frac13$, Cesàro. The transform exponent comes from the cubic term in $\sin$.

Pitfall. Writing $\sin t \approx t - \frac{t^3}{6}$ without the $O(t^5)$ and then summing the approximation over $n$ steps. Errors compound under iteration; only an expansion with EXPLICIT remainder survives being summed — and the grader checks exactly that line.

2Estimation courage: discard, then defend

The skill that separates attempted A5s from solved ones is deciding what NOT to compute. A hard sum splits: main range (where the mass lives) plus tails (which you bound crudely — by the largest term times the count, by a geometric comparison, by monotonicity). The courage is in making the split BEFORE knowing it works, computing the main range precisely, and then defending the tail bound honestly. Professional write-ups signal it openly: 'we show the terms with $k > n^{2/3}$ contribute $o(1)$' — announce the split, owe the proof.

Peak analysis for products and integrals: locate where the integrand/term is maximized (calculus on the log), expand to second order at the peak, check the remainder is uniformly small near the peak and the tails contribute negligibly, and the Gaussian integral does the rest — Laplace's method in contest form. Stirling is the canonical output; problems that contain $\binom{2n}{n}$-type factors or $n!$ ratios almost always want $\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$ used with its error acknowledged, not re-derived.

The inequality reflexes that make bounds work: $1 + t \le e^t$ everywhere — and for factors with $t_k \ge -1$ (so nothing negative flips the product), $\prod(1 + t_k) \le e^{\sum t_k}$; $\frac{t}{1+t} \le \ln(1+t) \le t$ for $t > -1$ (logs are squeezed by rationals); convexity/Jensen when averaging fights you. Ninety percent of contest tail-bounds are one of these three plus a geometric series — having them as reflexes converts 'estimate courage' from bravery into routine.

Worked example

Determine $\lim_{n \to \infty} \left(\prod_{k=1}^{n}\left(1 + \frac{k}{n^2}\right)\right)$.

  1. Nudge 1

    Logs turn the product into a sum — expand $\ln(1+t)$ with an explicit remainder.

    Reveal step 1

    Take logs and expand with the remainder: $\ln\prod = \sum_{k=1}^n \ln\left(1 + \frac{k}{n^2}\right)$, and $\ln(1+t) = t + O(t^2)$ uniformly for $t \in [0, 1]$, with $t = \frac{k}{n^2} \le \frac1n$.

  2. Nudge 2

    Sum the linear terms exactly; they carry the whole answer.

    Reveal step 2

    Main term: $\sum_k \frac{k}{n^2} = \frac{n(n+1)}{2n^2} \to \frac12$. Discarded pile: $\sum_k O\left(\frac{k^2}{n^4}\right) = O\left(\frac{n^3}{n^4}\right) = O\left(\frac1n\right) \to 0$ — the tail defense is one line BECAUSE the expansion carried its remainder.

  3. Nudge 3

    Bound the quadratic pile by its worst case times its count and let it die.

    Reveal step 3

    Conclusion: $\ln \prod \to \frac12$, so the product $\to \sqrt{e}$. Anatomy: log, expand-with-remainder, main term by exact summation, trash bounded by its worst case times its count. Every product limit on the exam is this four-line skeleton with different numbers.

  4. Answer

    $\sqrt e$: logs turn the product into a sum, the linear terms give $\frac12$, the quadratic trash is $O(1/n)$.

Pitfall. Bounding the tail with the SAME precision as the main term — a page of needless Taylor orders. The tail needs any bound that vanishes; spend accuracy only where the answer lives.

Before you open the gates

  • House method: find the driver, expand with explicit remainders, sum main terms precisely and trash crudely.
  • Decaying recursions → power transform so differences converge, then Cesàro; the exponent comes from the recursion's leading nonlinear term.
  • Announce range splits ('terms beyond $M$ contribute $o(1)$') and defend each tail with worst-case × count or a geometric comparison.
  • Reflexes on tap: $1+t \le e^t$, the $\ln$ squeeze, Jensen, $\binom{2n}{n} \sim 4^n/\sqrt{\pi n}$, $H_n = \ln n + \gamma + O(1/n)$.
  • Precision budgeting: the main term deserves your accuracy; the tail deserves any bound that dies.

Check yourself

1. For $x_{n+1} = x_n - c x_n^k$ with $x_n \to 0^+$, the linearizing transform is:

2. Iterating or summing a Taylor approximation is only legal when:

3. The reflex inequality that converts products into exponential bounds is:

Practice ladder — three rungs, rising

Each rung: attempt cold, one hint if stuck, worked resolution only after a real try.

Rung 1 (product limit, direct). Evaluate $\displaystyle\lim_{n\to\infty} \prod_{k=1}^{n}\left(1 + \frac{k^2}{n^3}\right)$.

One hint

Take logs; expand $\ln(1+t) = t + O(t^2)$ with $t = k^2/n^3 \le 1/n$. Sum the linear part exactly using $\sum_{k=1}^n k^2 = \tfrac{n(n+1)(2n+1)}{6}$, and bound the quadratic pile by its worst case times its count.

Worked resolution

Let $P_n$ be the product. Then $\ln P_n = \sum_{k=1}^n \ln\!\left(1 + \tfrac{k^2}{n^3}\right)$, and since $0 \le \tfrac{k^2}{n^3} \le \tfrac1n$, the expansion $\ln(1+t) = t + O(t^2)$ gives $\ln P_n = \sum_{k=1}^n \tfrac{k^2}{n^3} + \sum_{k=1}^n O\!\left(\tfrac{k^4}{n^6}\right)$. The main sum is $\tfrac{1}{n^3}\cdot\tfrac{n(n+1)(2n+1)}{6} \to \tfrac13$; the remainder is $O\!\left(\tfrac{1}{n^6}\sum_{k=1}^n k^4\right) = O\!\left(\tfrac{n^5}{n^6}\right) = O\!\left(\tfrac1n\right) \to 0$. So $\ln P_n \to \tfrac13$ and the limit is $e^{1/3}$. [Source: this week's estimation-courage section — the log, expand-with-remainder, sum-the-mains-precisely, bound-the-trash skeleton for product limits.]

Rung 2 (decaying recursion, power transform). Let $x_1 = \tfrac12$ and $x_{n+1} = x_n - x_n^2$. Prove that $x_n \to 0$ and that $\displaystyle\lim_{n\to\infty} n\, x_n = 1$.

One hint

First show $x_n \downarrow 0$ on $(0,1)$. Then set $y_n = 1/x_n$ (the $k=2$ case of the transform $y_n = x_n^{-(k-1)}$) and show $y_{n+1} - y_n \to 1$; differences with a limit force linear growth.

Worked resolution

If $x_n \in (0,1)$ then $x_{n+1} = x_n(1 - x_n) \in (0, \tfrac14] \subset (0,1)$, so by induction $x_n \in (0,1)$ for all $n$; and $x_{n+1} = x_n - x_n^2 < x_n$, so $(x_n)$ strictly decreases to a limit $L$ with $L = L - L^2$, forcing $L = 0$. Set $y_n = 1/x_n$. Then $y_{n+1} - y_n = \dfrac{1}{x_n(1-x_n)} - \dfrac{1}{x_n} = \dfrac{1}{x_n}\cdot\dfrac{x_n}{1-x_n} = \dfrac{1}{1-x_n} \to 1$ as $x_n \to 0$. By Stolz-Cesaro, $y_n/n \to 1$, i.e. $n\, x_n \to 1$. [Source: this week's asymptotic-toolkit section — the power-transform archetype $x_{n+1} = x_n - c x_n^k$ with $y_n = x_n^{-(k-1)}$ and Cesaro; here $c = 1,\ k = 2$, giving the predicted rate $x_n \sim 1/n$.]

Rung 3 (sum asymptotics, integral comparison). Prove that the sequence $a_n = \displaystyle\sum_{k=1}^{n} \frac{1}{\sqrt{k}} - 2\sqrt{n}$ converges to a finite limit.

One hint

Show $(a_n)$ is decreasing by estimating $a_n - a_{n-1} = \tfrac{1}{\sqrt n} - 2(\sqrt n - \sqrt{n-1})$, and bounded below by comparing $\sum \tfrac{1}{\sqrt k}$ with $\int \tfrac{dx}{\sqrt x}$. Then invoke monotone convergence.

Worked resolution

Monotone: $a_n - a_{n-1} = \dfrac{1}{\sqrt n} - 2(\sqrt n - \sqrt{n-1}) = \dfrac{1}{\sqrt n} - \dfrac{2}{\sqrt n + \sqrt{n-1}}$. Since $\sqrt n + \sqrt{n-1} < 2\sqrt n$, the subtracted term exceeds $\dfrac{2}{2\sqrt n} = \dfrac{1}{\sqrt n}$, so $a_n - a_{n-1} < 0$: $(a_n)$ is decreasing. Bounded below: as $\tfrac{1}{\sqrt x}$ is decreasing, $\sum_{k=1}^n \tfrac{1}{\sqrt k} \ge \int_1^{n+1}\tfrac{dx}{\sqrt x} = 2\sqrt{n+1} - 2 > 2\sqrt n - 2$, so $a_n > -2$. A decreasing sequence bounded below converges (monotone convergence theorem). [Source: this week's asymptotic-toolkit section — the sum archetype's integral-comparison rung ($\sum f(k)$ against $\int f$, the ladder of precision).]

Prove it — constructed response

Define $x_1 = 1$ and $x_{n+1} = \sin(x_n)$. Prove that $\lim_{n\to\infty} \sqrt{n}\, x_n = \sqrt{3}$. Establish that $x_n$ decreases to $0$ BEFORE expanding; use a Taylor expansion WITH explicit remainder to linearize; and name the theorem that turns convergent differences into linear growth.

The gates

Technique Resources — nightmare ramp I — mixed hard Putnam (study before the reinforcing problems)

Learning support for this week's reinforcing IMO/Putnam problems. Read/watch these before (and after) attempting the nightmare ramp I — mixed hard Putnam problems so each attempt has a source to learn the method and to check your write-up against.

sources & assignments (5)

Nightmare Archive

Nightmare Archive accelerated only

- Archive: 6 Nightmare · USAMO/IMO/hard Putnam A4–A5 · 40 min each - After each: write certified sub-lemma; classify miss; route to repair source

sources & assignments (5)
  • Problems Archive BrowserPutnam 2023 A4, 2022 A4 · mixed · 40 min eachNightmare

archive pull: 2 problems · Analysis/Algebra/Combinatorics/Number Theory/Geometry · A4/A5 · Nightmare · 40 min

A1/A2/B1/B2 Maintenance

A1/A2/B1/B2 Maintenance

- Archive: 6 mixed A1/A2/B1/B2 · all topics · 20 min each

sources & assignments (5)
  • Problems Archive BrowserPutnam 1992 B1 · mixed · 20 min eachPutnam

archive pull: 1 problems · Analysis/Algebra/Combinatorics/Number Theory/Geometry · A1/A2/B1/B2 · Challenge · 20 min

Repair Block

Repair Block

- Do: 2 repairs from routing table · 30 min each

sources & assignments (5)
  • Problems 2 repairs from routing table · 30 min each2 repairs from routing table · 30 min eachPutnam

Spiral Reinforcement

Spiral Reinforcement

- Archive: 2 Putnam A2 · your 2 weakest topics from Wk 13 audit · any slot ---

sources & assignments (5)
  • Problems Archive Browser2 Putnam A2 · your 2 weakest topics from Wk 13 audit · any slotPutnam

archive pull: 2 problems · Mixed Putnam · A2 · Challenge · 20 min

Deep Study — Asymptotics + estimation

Deep Study — Asymptotics + estimation (the A5/B5 toolkit)

Mastery-phase learning gate: estimation discipline is the most common missing skill on A4-A6 — and it is teachable.

Ritual: Output: growth-rate ladder card + dominant-term checklist.

why this gate: the lesson's §1 The asymptotic toolkit, assembled is what it trains

sources & assignments (8)

USAMO/IMO Bridge — USAMO 1998/3

USAMO/IMO Bridge — USAMO 1998/3 (stretch) accelerated only

Ritual: Self-grade 0–7 against the AoPS/official solution: 7 = complete and rigorous · 5–6 = right idea, rigor gaps · 3–4 = key lemma proven · 1–2 = nontrivial progress · 0 = none. Log the score and the single biggest missing idea in the verify box.

sources & assignments (5)
  • Source USAMO 1998/3 — AoPS wiki (statement + solutions + discussion link) — tan(a₀−π/4)+⋯+tan(a_n−π/4) ≥ n−1 ⇒ tan a₀⋯tan a_n ≥ n^(n+1). Inequality + clever substitution + AM-GM at scale — A5/B5-band analysis. ALSO: cold re-solve IMO 1988/6 from last week's study (30 min, from memory). Time cap 75 min — do not scroll to the Solutions section until the attempt is over.
  • Source MIT 18.A34 — Putnam Seminar (OCW) — Strategy companion for “USAMO/IMO Bridge — USAMO 1998/3 (stretch)” — a live problem-solving session: watch how the solver chooses a first move under uncertainty, then apply the same selection discipline to this gate.
  • Source The Bright Side of Mathematics — Real Analysis (playlist) — Pacing companion for “USAMO/IMO Bridge — USAMO 1998/3 (stretch)” — note when the presenter abandons a line; compare with your own bail rule before starting this gate's timed work.
  • Source MIT 18.100B Lecture 5 — Monotone Convergence Theorem (OCW) — Writeup companion for “USAMO/IMO Bridge — USAMO 1998/3 (stretch)” — attend to how each claim is justified aloud; steal one justification phrase for this gate's written artifact.
  • Problems USAMO 1998/3tan(a₀−π/4)+⋯+tan(a_n−π/4) ≥ n−1 ⇒ tan a₀⋯tan a_n ≥ n^(n+1) — full writeup to the Evan Chen standard, 75 min cap. Self-grade 0–7 against the AoPS/official solution: 7 = complete and rigorous · 5–6 = right idea, rigor gaps · 3–4 = key lemma proven · 1–2 = nontrivial progress · 0 = none. Log the score and the single biggest missing idea in the verify box.USAMO bridge

Continuation — Algebra

Continuation — Algebra: interpolation + finite differences reps (≈2 hrs)

Ritual: Thread: Algebra. This lane keeps a second course moving during mastery weeks — new material, not review. Required artifact is written; videos are reinforcement only.

sources & assignments (5)
  • Source AoPS — Lagrange Interpolation Formula — Continuation-lane companion for: Algebra: interpolation + finite differences reps (≈2 hrs). Reused from the verified pool — watch/read with this week's lens.
  • Source Michael Penn — Newton's Sums — Continuation-lane companion for: Algebra: interpolation + finite differences reps (≈2 hrs). Reused from the verified pool — watch/read with this week's lens.
  • Source MIT 18.A34 — Putnam Seminar (OCW) — Pacing companion for “Continuation — Algebra: interpolation + finite differences reps (≈2 hrs)” — note when the presenter abandons a line; compare with your own bail rule before starting this gate's timed work.
  • Source The Bright Side of Mathematics — Real Analysis (playlist) — Writeup companion for “Continuation — Algebra: interpolation + finite differences reps (≈2 hrs)” — attend to how each claim is justified aloud; steal one justification phrase for this gate's written artifact.
  • Problems Continuation laneThree reps: (1) re-solve USAMO 1975/3 cold (you met it in W19's bridge); (2) compute the finite-difference table proof that deg-n polynomials have constant Δ^n; (3) one archive polynomial problem where plugging consecutive integers wins.Standard

Keep-Warm Rotation — probability tail bounds · asymptotics · abstract algebra

Keep-Warm Rotation — probability tail bounds · asymptotics · abstract algebra (30 min)

Ritual: If any micro-rep took over 10 minutes or failed, that family goes on this week's repair list — it was rusting.

why this gate: the lesson's §1 The asymptotic toolkit, assembled is what it trains

sources & assignments (2)
  • Problems Keep-warm rotationThree 10-minute micro-reps, no notes: (1) probability — one tail-bound rep: state Markov's inequality, derive Chebyshev from it (apply Markov to (X−μ)²), then use Chebyshev to bound P(|X − n/2| ≥ n/4) for X ~ Binomial(n, 1/2) — you should get 4/n (capped at 1); say which bound is tighter here and why; (2) asymptotics — one bound via weak Stirling, or re-derive log n! ≈ n log n − n by integral comparison; (3) abstract algebra — Gaussian integers / UFD recognition: state the norm N(a+bi) = a² + b², identify the units (±1, ±i), and name when ℤ[i] cracks a sum-of-two-squares problem. The dormancy rule: no tool family sleeps longer than 2 weeks — these are the three most overdue right now.Standard

archive pull: 1 problems · Probability/Analysis/Abstract Algebra · A1/B1 · Putnam · 10 min

Track A∗ — Analysis

Track A∗ — Analysis (parallel mastery track): Asymptotics & growth rates (the A5/B5 toolkit)

Runs EVERY mastery week so Analysis never goes cold — the Toolkit parallel-track model carried into Mastery. One depth problem + one A3–A6 reach problem each week.

why this gate: the lesson's §1 The asymptotic toolkit, assembled is what it trains

sources & assignments (4)
  • Problems Analysis (depth)Putnam 1996 B2 — work it with the reading open; one clean write-up.Putnam
  • Problems Analysis (reach)Putnam 1988 B4 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track D∗ — Linear Algebra

Track D∗ — Linear Algebra (parallel mastery track): Linear algebra over 𝔽_p / in combinatorics (oddtown, Fisher)

Runs EVERY mastery week so Linear Algebra never goes cold — the Toolkit parallel-track model carried into Mastery. One depth problem + one A3–A6 reach problem each week.

sources & assignments (4)
  • Problems Linear Algebra (depth)Putnam 1993 B4 — work it with the reading open; one clean write-up.Putnam
  • Problems Linear Algebra (reach)Putnam 1994 A4 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track E∗ — Number Theory

Track E∗ — Number Theory (parallel mastery track): Estimation / density flavor (Chebyshev-type bounds)

Runs EVERY mastery week so Number Theory never goes cold — the Toolkit parallel-track model carried into Mastery. One depth problem + one A3–A6 reach problem each week.

why this gate: the lesson's §2 Estimation courage: discard, then defend is what it trains

sources & assignments (4)
  • Problems Number Theory (depth)Putnam 2023 B2 — work it with the reading open; one clean write-up.Putnam
  • Problems Number Theory (reach)Putnam 1992 A5 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track C∗ — Combinatorics

Track C∗ — Combinatorics (parallel mastery track): Algebraic & polynomial method — Combinatorial Nullstellensatz

Continuous Combinatorics mastery spine (Track C∗). Combinatorics is the #1 Putnam genre (~20%); A1–B2 combinatorics is already drilled hard in the slots, so this spine adds the continuous theory ladder plus a weekly A3–A6 reach problem (the real gap). One depth re-solve + one fresh reach each week.

sources & assignments (4)
  • Source Combinatorics spine — Combinatorics text — This week: Algebraic & polynomial method — Combinatorial Nullstellensatz. Polynomial method, Combinatorial Nullstellensatz, dimension/counting bounds. Read the matching section, then do the two problems below.
  • Source Combinatorics continuous spine — Runs every Mastery week (parallel-track model) so Combinatorics — the single most frequent Putnam genre (~20%) — never goes cold. 1 depth re-solve + 1 fresh reach problem.
  • Problems Combinatorics (depth)Putnam 2005 A2 — cold re-solve with the reading open; one clean write-up.Putnam
  • Problems Combinatorics (reach)Putnam 1997 A5 — 45 min; extract one rigorous lemma / reach-point (2–4 of 10), do not force a full solve.Nightmare

Track F∗ — Geometry · Combinatorics · Probability

Track F∗ — Geometry · Combinatorics · Probability (parallel maintenance): Geometry — 3D, areas & geometric inequalities

Bundled parallel maintenance track so Geometry, Combinatorics, and Probability each get a fixed weekly cadence in Mastery (probability previously had multi-week gaps). Rotates focus; ~1.5 hrs.

sources & assignments (3)
  • Source Engel — Problem-Solving Strategies, Geometry — This week’s focus: 3D, areas & geometric inequalities. Volumes/cross-sections, projections, isoperimetric-type and triangle inequalities.
  • Source Geometry keep-warm — Lower-cadence topic on a guaranteed weekly rotation (Toolkit Track F model). 1 problem this week; Geometry comes round again every 3rd week.
  • Problems Geometry (F∗)Putnam 1989 A5 — 3d, areas & geometric inequalities; full attempt + one clean write-up.Putnam

Problem-Solving Reps — weekly homework

Problem-Solving Reps — weekly homework

Standing weekly homework — the problem-solving book stack, every week. Solve ALL listed; write ONE full clean solution (the rest may stay scratch). Up to 2 due re-solves from your review queue surface first. Slot reinforcement supplement: A2/B2 are deliberately overtrained so they become bankable, not occasional exposure. Slot reinforcement supplement: A3/B3 now get repeated lemma-hunt reps so slot 3 is trained as a strong reach tier. Slot reinforcement supplement: A4-B6 remain a good reach tier without displacing A1-B3 scoring practice.

sources & assignments (14)
  • Problems PnBPnB Ch. 6 probs 13–14 (maintenance)Putnam
  • Problems Problems from the BookAlways Cauchy-Schwarz chapter, probs 1–2Nightmare
  • Problems Alexanderson + GleasonPutnam 1978 A5,A6 · 1958 B5,B6Nightmare
  • Problems Putnam ArchivePutnam 1985 A2 - 18 min clean-solve; write the final proof and fatal-slip check.Putnam
  • Problems Putnam ArchivePutnam 2019 A3 - 35 min lemma hunt; record one rigorous lemma even if the full proof does not close.Nightmare
  • Problems Putnam ArchivePutnam 2013 B3 - 35 min lemma hunt; compare to official solution after the attempt.Nightmare
  • Problems Putnam ArchivePutnam 2018 A6 - 60 min reach; goal is a certified lemma, not forced completion.Nightmare
  • Problems Putnam ArchivePutnam 2018 B6 - 60 min reach; goal is a certified lemma, not forced completion.Nightmare
  • Problems Putnam ArchivePutnam 2002 A4 - 45 min reach; extract the first invariant/structure and stop only after a written lemma.Nightmare
  • Problems Putnam ArchivePutnam 2003 B4 - 45 min reach; extract the first invariant/structure and stop only after a written lemma.Nightmare
  • Problems Putnam Slot 2 EqualizerSlot-2 Equalizer Set 10: Putnam 2020 B2, Putnam 2021 B2, Putnam 2010 B2 - clean-solve focus; write final proofs and fatal-slip checks.Putnam
  • Problems AMC/AIME Fluency OpenerOptional 5-minute opener: solve the three linked AMC/AIME reps before Putnam work; stop at 15 minutes total.AMC/AIME · archive: AIME II2022 AIME IAIME standard
  • Problems Putnam Full Past ExamFull Past Exam 23: Putnam 2017 A1, Putnam 2017 A2, Putnam 2017 A3, Putnam 2017 A4, Putnam 2017 A5, Putnam 2017 A6, Putnam 2017 B1, Putnam 2017 B2, Putnam 2017 B3, Putnam 2017 B4, Putnam 2017 B5, Putnam 2017 B6 - Primary paper protocol: A-session A1-A6 in one 180-minute block; B-session B1-B6 in one 180-minute block within 48 hours; next day score all 12 slots, tag every miss, and choose three repair pulls.Putnam
  • Problems Putnam Full Past ExamFull Past Exam 24: Putnam 2018 A1, Putnam 2018 A2, Putnam 2018 A3, Putnam 2018 A4, Putnam 2018 A5, Putnam 2018 A6, Putnam 2018 B1, Putnam 2018 B2, Putnam 2018 B3, Putnam 2018 B4, Putnam 2018 B5, Putnam 2018 B6 - Secondary paper protocol: Day 1 bankable sweep A1-A2-B1-B2 in 100 minutes; Day 2 medium/hard sweep A3-A4-B3-B4 in 100 minutes with certified-lemma goal; Day 3 reach scan A5-A6-B5-B6 in 50 minutes; Day 4 score all 12 slots and re-solve one same-family miss.Putnam

Reflect

Reflect: reconstruct and log the pattern

Mandatory Reflect gate from AGENTS.md for every week 3–28.

Ritual: Pick the nightmare-tier problem where you found the key idea but ran out of road. Close all notes and the solution. Rewrite the full solution from memory, filling the gap you missed live. Then write: When I see [this configuration], the unlocking move is [the lemma/idea].

sources & assignments (4)

Verify: fatal-slip audit before closing

unlocks after: Reflect: reconstruct and log the pattern

Mandatory Verify gate from AGENTS.md for every week 3–28.

Ritual: Pick one solution. Re-check for: (1) false claim, (2) missing case, (3) wrong bound. Write claim, hypotheses, edge cases, score estimate, and remaining risk.

sources & assignments (4)

Exit contract — Week 23

Verification remaining

  • reading progress…

Carry-forward repairs

  • reading queue…

Next week opens with

W24 · Nightmare Ramp II
the exam follows the final taper week

Train Week 23 in the trainer →