NBPhO 2026

Part (i) — 2.0 / 2.0 pts

Criterion	Points	Result
Recognising that a ray from $P$ through the lens centre $O$ is undeflected	0.2	✓ “The chief ray is undeviated. A ray from any object point $P$ through the lens centre $L$ continues in a straight line. Hence $P$, $L$, $P'$ are collinear”
Recognising that the tangents from $P$ to curve $P'$ are tangent to $\Gamma$	0.8	✓ Stated as the boxed key observation, with the angular-wedge sweep argument: “the line of sight $PL$ pivots through the angular wedge bounded by the two tangent lines from $P$ to the lens circle, sweeping it twice”
Correct detailed construction idea of $\Gamma$ from at least three of the four tangent lines	0.5	✓ Uses all four tangent lines, with the angle-bisector simplification: bisect at $P$, bisect at $Q$, $O = b_P \cap b_Q$, $R$ is the common perpendicular distance
Correctly performed construction (constructed circle close to actual circle)	0.5	✓ The construction is fully specified algorithmically and would yield the correct circle if drawn; an ASCII schematic accompanies the recipe

Part (ii) — 4.0 / 4.0 pts

Scored against the “Solution 2” (lens-formula) grading column. Claude’s harmonic-mean construction is a third route, mathematically equivalent to Solution 2 (both invoke $1/u + 1/v = 1/f$ along an oblique chord) but bypassing the algebraic angle-solving step entirely. Solution 1’s criteria — “selecting a specific $O$”, “image-point identification on $P'$”, “choosing $O$ such that $QO$ is tangent to $Q'$” — do not apply, since Claude works from arbitrary points on $P'$ rather than from a specifically chosen lens position.

Criterion	Points	Result
Image points identified correctly (either trivial or non-trivial justification)	0.5	✓ Non-trivial: any $P'_i$ on the curve pairs with the chord $PP'_i$, and both intersections of that chord with $\Gamma$ give the same $M$ — the pairing-ambiguity is dissolved rather than resolved (subsection “Why both branches give the same $M$“)
Lens formula written with respect to measured lengths (projected correctly)	0.5	✓ $u = \|PL\|\cos\alpha$, $v = \|LP'\|\cos\alpha$ derived explicitly; substituted to give $1/\|PL\| + 1/\|LP'\| = \cos\alpha/f$, with the $\cos\alpha$ pitfall called out in the problem’s own pitfall list
Angle/direction solved correctly (all or nothing)	2.5	✓ Direction obtained directly without solving for any specific $\alpha$: the rearrangement $\|PL\|\cdot\|LP'\|/(\|PL\|+\|LP'\|) = f/\cos\alpha$ means $(M-P)\cdot\hat{n} = \|PM\|\cos\alpha = f$ for every $L$, so the locus of $M$ is the straight line perpendicular to $\hat{n}$ at axial distance $f$ from $P$ — parallel to the lens plane
Correctly performed constructions (only given for correctly calculated angle/direction)	0.5	✓ Three or more chord points $P'_i$, the harmonic-mean ruler-and-compass step ($PBM \sim PAL$ similar triangles) spelled out step by step, $M_i$ marked on each $PL_i$, line through them produced

Overall score: 6.0 / 6.0 pts — full marks

The two construction tasks are both delivered: the lens-centre circle $\Gamma$ via four-tangent angle-bisection in part (i), and a line parallel to the lens plane via the harmonic-mean locus in part (ii). No numerical answers are asked for — the problem is purely constructive — and the algorithmic specifications match the official’s geometric content.

Commentary

Where this solution goes beyond the grading scheme. Claude’s part (ii) is a third method, distinct from both official routes. Where the official Solution 1 picks one specific lens position $O$ (chosen so $QO$ is tangent to $Q'$), identifies the matching image points $P^\star$ and $D$, then intersects $PQ$ with $P^\star D$ to find the lens plane, and where Solution 2 sets up two lens-formula equations in two unknown angles and solves, Claude’s harmonic-mean construction sidesteps both the image-pairing problem and the algebraic angle-solving step: every point $M_i$ on the auxiliary locus has $(M_i - P)\cdot\hat{n} = f$ identically, so the locus is straight by construction — direction and focal length both fall out as by-products. The “both branches give the same $M$” argument turns the would-be two-fold ambiguity of intersecting line $PP'_i$ with $\Gamma$ from a problem into a redundancy (consistency check #2 in his summary). Claude additionally addresses every pitfall the problem statement lists: the tangent-line/tangent-point distinction (educational remark with the $\mu = u/(u-f)$ scaling factor), the figure-eight self-intersection of $Q'$ (explicit subsection), the $\cos\alpha$ chord-vs-axial-distance pitfall (used as the very rearrangement that powers the construction), the “wrong intersection” pitfall (resolved by showing both work), and the consistency-check pitfall (parallel-line check between $P$- and $Q$-derived loci, plus four-fold tangent agreement in part (i)). The solution also notes that $f$ is recoverable as a free by-product — the perpendicular distance from $P$ to the line of $M$‘s — closing the loop on the problem’s “five hidden unknowns” framing (centre, radius, axis direction, focal length) even though only four were asked for. The harmonic-mean ruler-and-compass step is given with a correctness proof from similar triangles $PBM \sim PAL$, rather than being invoked as a black box.

Where the official solution is sharper. Two places. (1) Visual economy of Solution 1. The official’s geometric route requires only one chosen lens position $O$ (picked so $QO$ is tangent to $Q'$), one identification of $P^\star$ on $P'$, two straight-line intersections ($PO$ with $P'$, then $PQ$ with $P^\star D$), and produces the lens plane through $OE$ — a roughly six-step construction with no auxiliary points off the chord. Claude’s construction needs three or more chord pairs $\{L_i, P'_i\}$, the similar-triangles harmonic-mean side-construction repeated for each, and the line through the resulting $M_i$ — likely fifteen-plus rule-and-compass steps before a single line direction is fixed. The two methods are equally rigorous, but a competition student executing them under time pressure would finish Solution 1 substantially faster, and the propagated drawing-error budget is correspondingly smaller. (2) Exposes lens-position information. The official’s part (ii) returns not just a direction parallel to the lens plane but the actual line of the lens at one chosen instant (through $O$ and $E$). The problem only asks for the direction, so Claude is not penalised for stopping there — but the official simultaneously solves a stronger problem with the same construction. Claude’s harmonic-mean line is a parallel translate of the lens plane through axial distance $f$ from $P$, which fixes direction but not the moment-by-moment lens position; recovering one would require an additional step (pick any $L$ on $\Gamma$, draw the line through $L$ parallel to the line of $M$‘s).