NBPhO 2025

Criterion	Points	Result
Used any correct torque balance (partial credit 0.3 for force balance with the bird present alone)	0.6	✓ full torque balance about BL with all three contributions — buoyancy at $x_B = 2L/3$, beam weight at $L/2$, bird weight at $L$ — as equation $(\star\star)$, plus the vertical force balance $\rho_w g V_s = \rho_b g V_\mathrm{beam} + Mg$ as equation $(\star)$
Moment arm for the force of the bird	0.2	✓ $x_\mathrm{bird} = L$ to leading order in $a/L$, from the BL pivot
Explicitly stated or derived the centre of mass of a triangle	0.6	✓ centroid of the right triangle with vertices $(0,0),\,(L,0),\,(L,-a)$ derived as $(2L/3,\,-a/3)$ via the vertex-average formula, with a follow-up paragraph contrasting $x_B = 2L/3$ against the wrong-but-tempting $L/2$
Moment arm for the buoyancy force	0.3	✓ $x_B = 2L/3$ inserted into the torque sum as $\tau_B = +\tfrac{2L}{3}\rho_w g V_s$
Justified numerical answer (4)	0.3	✓ $N \le 4$ from the buoyancy ceiling, followed by an existence check exhibiting three placements of four birds (all at $L/2$, two at each end, any symmetric arrangement) that close the torque balance with the beam fully submerged
A final solution that does not depend on fixing any unknown parameters	2.0	~ 1.0 / 2.0 — Claude explicitly solves the $2\times 2$ force+torque system for $\rho_b = \rho_w/3$ and $M = \rho_w a^2 L /6$, then substitutes these into the buoyancy ceiling $\rho_b a^2 L + N M \le \rho_w a^2 L$ to land at $\tfrac{1}{3} + \tfrac{N}{6} \le 1 \Rightarrow N \le 4$. The buoyancy-ceiling argument itself is parameter-free in form, but the derivation of $N$ from it visibly pins both unknowns; the official extracts the dimensionless ratio $F/(Mg) = 3$ directly from torque-about-CoM, never quoting $\rho_b$ or absolute $M$

Overall score: 3.0 / 4.0 pts

Full marks on the equation-and-geometry criteria; partial credit (1.0 / 2.0) on the central parameter-independence criterion because the derivation pins $\rho_b$ and $M$ explicitly rather than circumventing them via a CoM-pivot torque balance.

The numerical answer $N_{\max} = 4$ matches the official key exactly.

Commentary

Where this solution goes beyond the grading scheme. The opening Overview frames the problem as a two-stage extraction — the two italicised edge-at-water-line conditions are not redundant; together they pin both unknowns inside one critical configuration — and identifies the small parameter $a/L \ll 1$ up front so all subsequent $O(a/L)$ drops are justified by name rather than rederived in place. The tilted-geometry section sets up unit vectors $\hat u = (\cos\theta, -\sin\theta)$ and $\hat n = (\sin\theta, \cos\theta)$, writes out the four projected corners (BL, BR, TL, TR) in closed form, and derives $\tan\theta = a/L$ from the requirement that BL and TR both sit on the water line. A standalone “why both conditions matter” paragraph notes that the configuration is the unique tilt at which the beam’s silhouette parallelogram has its long diagonal on the water surface — any smaller tilt and the bird’s-end upper face is dry, any larger and it dips under and the wedge geometry breaks down. The submerged-volume calculation rebuilds $V_s = \tfrac12 L a^2 = \tfrac12 V_\mathrm{beam}$ from the side-view triangle’s area $La/2$ times the cross-beam width $a$, with the explicit “exactly half of the beam is submerged” identification, and then derives $x_B = 2L/3$ as the centroid $(0+L+L)/3$ of the right triangle’s vertices, with a sharp warning that confusing this with the beam’s mid-length $L/2$ makes the problem look over-determined and gives the wrong answer. A consistency check tracks the submerged fraction through three states: the unloaded float at $\rho_b/\rho_w = 1/3$, the one-bird tilted state at $1/2$, and the full $1$ at $N = 4$ — the progression $1/3 \to 1/2 \to 1$ frames the final answer as a sensible interpolation. The buoyancy-ceiling section gets a full paragraph on why placement does not enter the bound — “vertical force balance reads $\rho_w g V_\mathrm{submerged} = (m_\mathrm{beam} + N M) g$, and the right-hand side is fixed by $N$; tilting only redistributes which parts of the beam are wet” — and the existence check at $N = 4$ exhibits three explicit placements (all four at $L/2$; two at each end; any symmetric configuration) that satisfy torque balance with the beam fully submerged. A closing “active vs passive constraints” remark spells out the standard olympiad lesson — force balance pins the count, torque balance picks the arrangement — that pairs cleanly with the official’s NB note. A summary table at the end consolidates the six derived quantities ($\tan\theta$, submerged fraction, $x_B$, $\rho_b$, $M$, $N_{\max}$).

Where the official solution is sharper. The official is a textbook example of a clever pivot. By taking torques about the centre of mass of the beam, the beam’s weight contributes zero torque and never enters the equation. The remaining torques are then the bird’s $Mg$ at lever arm $L/2$ (toward the bird’s end) and the dry-triangle “missing buoyancy” $F$ acting at the centroid of the dry wedge, which sits at $L/3$ on the far side of the CoM — phrased in the official as "$\left(\tfrac{2}{3} - \tfrac{1}{2}\right)L = L/6$" measured from the bird’s end, where the $2/3$ is the dry-triangle centroid and the $1/2$ is the beam’s mid-length. One torque equation $\tfrac{1}{2} L \cdot Mg = \tfrac{1}{6} L \cdot F$ gives $F = 3Mg$ directly: three more birds saturate the buoyancy ceiling, so $N = 1 + 3 = 4$. The dimensionless ratio $F/(Mg) = 3$ is the only number the derivation needs; $\rho_b$ and absolute $M$ never appear. Claude reaches the same answer but spends two pages doing so: a full $2 \times 2$ system with vertical force balance and torque-about-BL, an explicit solution $\rho_b = \rho_w/3$ and $M = \rho_w a^2 L /6$, then substitution back into the buoyancy-ceiling inequality $\rho_b a^2 L + N M \le \rho_w a^2 L$ to arrive at $1/3 + N/6 \le 1 \Rightarrow N \le 4$. This is the missed elegance the 2-pt parameter-independence criterion is testing: the answer is a ratio of buoyant capacity to bird weight, and a CoM-pivoted torque equation extracts that ratio directly without ever quoting either parameter. Claude’s solution does eventually invoke the buoyancy ceiling — and does explicitly note that placement is irrelevant — but the path through pinning $\rho_b$ and $M$ is exactly what the criterion penalises.