A1/B1 Slot Mastery

Study the pattern catalog and a shape emerges: A1/B1 problems have SHORT statements, small or concrete objects (a specific year number, a small board, an explicit function), and exactly one hidden observation that collapses them — parity, a telescoping product, a clever pairing, a symmetry, an extreme case. They are engineered so that ANY systematic entry (try $n = 1, 2, 3$; check the extreme case; compute the first few terms) walks into the observation within minutes.

The professional's opening is always the same: MAKE THE PROBLEM CONCRETE. Plug in small values, draw the $n = 3$ case, compute the product for three terms. A1/B1 designers hide the pattern one or two examples deep — the solver who computes three data points and stares beats the solver who theorizes for ten minutes, essentially always.

Time discipline for these slots: target 20-25 minutes including write-up. If a supposed A1 resists for 30, you have either missed the intended observation (reread the EXACT statement — these problems are worded precisely, and 'positive integer', 'distinct', 'nonzero' are usually load-bearing) or you are overcomplicating; the intended solution almost never needs machinery beyond Week 8.

Pitfall. Skipping the construction half. 'Impossible when $3 \| n$' is half the problem; A1 graders want BOTH the obstruction and the explicit families that achieve everything else. Determine-all problems are always two proofs.

The dictionary: a point is a complex number; translation is addition; rotation by $\theta$ about the origin is multiplication by $e^{i\theta}$; rotation about $p$ is $z \mapsto p + e^{i\theta}(z - p)$. The instant payoffs: three points are collinear iff $\frac{a - b}{a - c} \in \mathbb{R}$; angles at a vertex become arguments of quotients; 'equilateral' becomes the crisp identity $a + \omega b + \omega^2 c = 0$ (for one orientation, with $\omega = e^{2\pi i/3}$) or equivalently $a^2+b^2+c^2 = ab+bc+ca$ covering both.

The method shines when the problem is built from rotations, midpoints, and centroids — squares erected on sides, equilateral triangles on edges (Napoleon configurations), spiral similarities. Each construction is one line of algebra ('the square on $AB$ outward has center $\frac{a+b}{2} + \frac{i}{2}(b-a)$'), and the conclusion is an identity you EXPAND rather than a diagram you stare at. What synthetic geometry calls insight, complex bashing calls collecting terms.

Know when NOT to use it: heavy circle-intersection problems want the unit-circle setup (put the circumcircle at $|z| = 1$, where reflections and feet of perpendiculars have clean formulas: the reflection of $z$ over chord $ab$ is $a + b - ab\bar{z}$) — and problems with many unrelated circles or lengths-only conclusions often stay synthetic or go to coordinates. The Week 14 drill builds the judgment: bash three, do one synthetically, compare costs.

Pitfall. Orientation errors. $x + \omega y + \omega^2 z = 0$ encodes ONE rotational orientation; the other is $x + \omega^2 y + \omega z = 0$. 'Outward' vs 'inward' triangles differ exactly here — fix the convention with one concrete check (an actual equilateral triangle with known vertices) before trusting the algebra.

Ceva's theorem: cevians $AD, BE, CF$ of a triangle concur iff $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$ (signed ratios). Menelaus: a line cuts the (extended) sides in points with the same product equal to $-1$. Mnemonic that keeps them straight: Ceva is about CONCURRENCE (three cevians through one point), Menelaus about COLLINEARITY (one line through three sides). Both convert incidence statements into ratio bookkeeping.

The trig form of Ceva ($\prod \frac{\sin \angle BAD}{\sin \angle DAC} = 1$) handles cevians specified by angles — bisectors, isogonals, symmedians — without ever finding the feet. Between the two forms, most 'show these three lines meet' problems on the exam become mechanical: name the ratios, multiply, cite the theorem.

Inversion earns a recognition-level entry (full drills come later): it is the map $z \mapsto \frac{r^2}{\bar{z}}$ fixing a circle, exchanging lines-through-center with circles-through-center. The cue is a configuration drowning in circles through a common point — inverting AT that point turns them all into lines. For now, learn to SPOT the cue; execution reps are scheduled with the geometry intensive.

Pitfall. Dropping signs when points leave the segments. Menelaus REQUIRES a negative product ($-1$) with signed ratios — an external division you treated as internal silently flips the verdict. Draw the configuration once, mark which divisions are external, then compute.