NBPhO 2026

The official solution gives two grading schemes — the primary one for the Thales-circle / power-of-a-point geometric route, and an alternative one for the pure algebraic / complex-impedance route. Claude takes the algebraic route end-to-end, so the assessment below is scored against the alternative grading scheme, as the official explicitly invites: “Students who reach the correct final form via either route receive full credit.”

Criterion	Points	Result
Correctly computing $V_B$ as a phasor	0.5	✓ $V_B = V/(1+ja)$ with $a = \omega R_1 C$, derived as the standard low-pass voltage divider
Correctly computing $V_D$ as a phasor	0.5	✓ $V_D = Vjb/(1+jb)$ with $b = \omega L/R_2$; algebraically equivalent to the official’s $V/(1 - iR_2/(\omega L))$ after multiplying numerator and denominator by $jb$
Identifying $V_P = 2V/3$ (or $V/3$)	0.3	✓ $V_P = 2V/3$ derived from $AP:PE = 1{:}2$ with $A$ on the left, with the explicit reasoning that the potentiometer voltage drops linearly with arc length
Expressing $V_{PB}/V$ and $V_{PD}/V$ in the form $X + iY$ with explicit real and imaginary parts	0.5	✓ Claude works directly with the ratio $V_y/V_x$ and rationalises it via $(2-jb)(1+ja) = (2+ab) + j(2a-b)$ and $(1+jb)(1-2ja) = (1+2ab) - j(2a-b)$, then reads off the imaginary part of the rationalised numerator. The $X + iY$ data is fully present, though packaged as ratio-numerator/ratio-denominator rather than separately for $V_{PB}$ and $V_{PD}$
Recognising that the Lissajous curve degenerates to a line iff $V_{PB}/V_{PD}$ is real (i.e., $X_B Y_D - Y_B X_D = 0$, or equivalent condition)	0.7	✓ stated up front in the Overview and used as the central organising principle: “It collapses to a line iff $V_x$ and $V_y$ are in phase — equivalently, iff the complex ratio $V_y/V_x$ is real”
Correctly deriving the bridge condition $R_1 R_2 C = L/2$ (equivalently $uw = 1/2$)	0.5	✓ $\delta = 2a - b = 0 \Rightarrow b = 2a \Rightarrow L = 2 R_1 R_2 C$, with the (small but useful) observation that the $(1+ab)$ factor never vanishes, so $\delta = 0$ is the unique line condition
Expressing $\tan\alpha$ in terms of $u$ and $w$ (or equivalent parameters)	0.5	✓ before substituting the bridge condition, Claude has $V_y/V_x = -(2+ab)/(1+2ab)$ in the form holding for all $a, b$ at the line-condition point
Substituting the bridge condition to write $\tan\alpha$ as a function of $u$ alone	0.5	✓ $\tan\alpha = -2(1+a^2)/(1+4a^2)$
Correctly computing $\lvert V_{PB}\rvert^2$ in terms of $u$ (before factoring)	0.7	✓ $\lvert V_{BP}\rvert^2 = (V/3)^2 \cdot (1+4a^2)/(1+a^2)$, computed as $\lvert N\rvert/\lvert D\rvert$ on the unrationalised fraction
Recognising the factoring $4u^4 + 5u^2 + 1 = (4u^2+1)(1+u^2)$ that simplifies $\lvert V_{PB}\rvert^2$	0.8	✓ Claude bypasses the polynomial factoring entirely by computing $\lvert 1 - 2ja\rvert/\lvert 1 + ja\rvert = \sqrt{1+4a^2}/\sqrt{1+a^2}$ on the original ratio of complex numbers, arriving at the simplified $(1+4a^2)/(1+a^2)$ form directly without ever needing the $X^2 + Y^2$ polynomial expansion
Eliminating $u$ using the $\tan\alpha$ expression	0.3	✓ $(1+4a^2)/(1+a^2) = -2/\tan\alpha$ inverted from the slope formula and substituted
Obtaining the final answer $\lvert V_{PB}\rvert = (V/3)\sqrt{2/\lvert\tan\alpha\rvert}$	0.5	✓ $\lvert V_{BP}\rvert = (V/3)\sqrt{-2\cot\alpha} = (V/3)\sqrt{2/\lvert\tan\alpha\rvert}$ — equivalent because $\tan\alpha < 0$ in this circuit, so $-\cot\alpha = 1/\lvert\tan\alpha\rvert$

Overall score: 6.0 / 6.0 pts — full marks

All three keyed numerical / symbolic results match the official answer key exactly: bridge condition $L = 2 R_1 R_2 C$, slope $\tan\alpha = -2(1+a^2)/(1+4a^2)$, and the boxed amplitude $\lvert V_{BP}\rvert = (V/3)\sqrt{2/\lvert\tan\alpha\rvert}$.

Commentary

Where this solution goes beyond the grading scheme. The Overview front-loads the two structural ideas — “Lissajous degenerates iff $V_y/V_x \in \mathbb{R}$” and “the slider’s role is to be real” — and frames the slider as the AC analogue of a Wheatstone null arm, a perspective that motivates the algebra rather than just performing it. The “Why this is a Wheatstone-bridge balance” subsection generalises the result to arbitrary slider position $V_P = kV$, deriving $b/a = k/(1-k)$ — so the experimentalist’s reading of $k = 2/3$ is shown to be the key that unlocks $L = 2 R_1 R_2 C$, and a different slider reading would calibrate a different component combination. The “Educational remark — what the slider really does” gives a coordinate-free interpretation: sliding $P$ adds the same real translation to $V_B$ and $V_D$ in the complex plane, and there is in general exactly one such translation that makes $V_B$ and $V_D$ collinear with the origin — which sets the slider position uniquely. The slope discussion extracts the limits $\tan\alpha \to -2$ at low frequency and $\tan\alpha \to -\tfrac{1}{2}$ at high frequency, and gives a physical narrative for the negative sign (capacitor open-circuit-like at low $\omega$ pushes $V_x$ positive, inductor short-circuit-like pushes $V_y$ negative). The amplitude section closes with four sanity checks: the range $\lvert V_{BP}\rvert/V \in (\tfrac{1}{3}, \tfrac{2}{3})$ verified at both endpoints, a direct check at $a = 1/\sqrt{2}$ where both the boxed formula and the unrationalised expression give $V\sqrt{2}/3$, an explicit dimensional check, and a $V_P = 0$ degeneracy comment that flags exactly when the boxed formula would not apply. The “Geometric reading” exposes that $\lvert V_{BP}\rvert$ is the on-screen $x$-extent of the line and that the full half-length is $\lvert V_x\rvert/\lvert\cos\alpha\rvert$ — a useful warning against measuring the wrong quantity when reading the trace. Finally, the “practical use” framing (balance gives a frequency-independent component check, while the residual slope encodes the frequency the bridge is being driven at) places the device in its experimental context.

Where the official solution is sharper. The official’s primary solution — which Claude does not take and which the alternative grading scheme implicitly cedes — is dramatically slicker. The argument is purely geometric: $Z_C/Z_{R_1} = -i/(\omega R_1 C)$ is a $90^\circ$ clockwise rotation, so $V_{AB} \perp V_{BE}$, which by Thales’ theorem puts $V_B$ on the circle of diameter $\overline{AE}$ in the lower half-plane; the same reasoning puts $V_D$ on the same circle in the upper half-plane. The line condition reduces to ”$P$, $B$, $D$ collinear in the phasor plane”, and because $V_P$ is real and lies between $V_B$ and $V_D$ on that chord, the power of the point $P$ gives $\lvert PB\rvert \cdot \lvert PD\rvert = V_P(V - V_P) = 2V^2/9$ — and crucially this value is the same for $V_P = V/3$ and $V_P = 2V/3$ by the symmetry of the quadratic about $V/2$, which is the official’s clean answer to the figure-orientation ambiguity. Combined with $\lvert\tan\alpha\rvert = \lvert PD\rvert/\lvert PB\rvert$, the boxed result $\lvert V_{PB}\rvert = (V/3)\sqrt{2/\lvert\tan\alpha\rvert}$ falls out in three lines. Claude’s algebraic route reaches the same answer but via roughly thirty lines of complex algebra, a polynomial simplification (or the polar-form bypass), a substitution of the bridge condition, and an inversion of the slope formula — perfectly correct, and pedagogically rich on the way through, but multiplicatively longer than the geometric path. The Thales-circle insight also makes the bridge condition itself almost obvious from the picture: balance occurs when $V_B$, $V_P$, and $V_D$ are colinear, which is a one-parameter family of lines through the chord — the slider just chooses which line. None of this is visible in the algebraic route, where the cancellation $\delta(u + w) = 0$ feels like a happy accident even though it has the same geometric origin.