How to Solve It

George Pólya's heuristics (1945) and twelve more problem-solving techniques, each set loose on a mathematical problem and a physics problem you can move, drag and test.

Four phases, twenty-four levers

“If you can't solve a problem, then there is an easier problem you can solve: find it.”

Pólya splits problem solving into four phases. The techniques below are the levers you pull in the second phase, when you are hunting for a plan: first Pólya's own twelve, then twelve more from the olympiad tradition and from physics. Each page states a problem, lets you play with it, shows how the technique cracks it, and looks back; then the same technique cracks a problem in physics.

Understand

What is the unknown? What are the data? What is the condition? Draw a figure, introduce notation.

Plan

Find the connection between data and unknown. Have you seen it before? Try the heuristics.

Carry out

Execute the plan, checking each step. Can you see clearly that each step is correct?

Look back

Check the result. Derive it differently. Can you use the result, or the method, for some other problem?

Pólya's heuristics

HeuristicPólya's questionWorked here onIn physics

More techniques

Collected in Engel's Problem-Solving Strategies, Zeitz's The Art and Craft of Problem Solving, and every physicist's back pocket.

TechniqueThe question to askWorked here onIn physics

Analogy

“Can you find a problem analogous to your problem and solve that?”

Problem. In a tetrahedron, join each vertex to the centroid of the opposite face. Do the four segments meet in one point, and in what ratio does that point divide them?

The plane analogue: drag A, B, C.

The space problem: drag to rotate.

The heuristic at work

  1. Space is hard to see. The analogous problem in the plane: in a triangle, join each vertex to the centroid of the opposite “face”, which is the midpoint of the opposite side. These are the medians.
  2. Solve the analogue in a form that can travel. Put G = (A+B+C)/3. With M = (B+C)/2, G = A + ⅔(M − A). The formula is symmetric in A, B, C, so all three medians pass through G, divided 2 : 1.
  3. Transplant the method, not the picture. Put G = (A+B+C+D)/4. With F = (B+C+D)/3, G = A + ¾(F − A).
  4. Symmetric in all four vertices, so the four segments concur at G, divided 3 : 1.
Looking back. Segment: midpoint, 1 : 1. Triangle: 2 : 1. Tetrahedron: 3 : 1. An n-simplex: n : 1. The analogy worked because the plane case was solved by vector averaging, a method with a three-dimensional counterpart, rather than by a trick that only lives in the plane.

In physics: the mass on a spring and the LC circuit

Problem. A charged capacitor C is connected across an inductor L and a resistor R. How does the charge on it behave?
  1. Kirchhoff's loop rule gives L q″ + R q′ + q/C = 0.
  2. That is term for term the damped spring m x″ + b x′ + k x = 0, with inductance as inertia, 1/C as stiffness and resistance as friction.
  3. So everything known about the spring transfers: ω₀ = 1/√(LC), the quality factor, the energy ½LI² + q²/2C. Both systems are driven by one simulation above.
  4. Analogue computers and Maxwell's mechanical models of the field were built on exactly this dictionary.

Auxiliary elements

“Can you add some new element to your problem to get closer to a solution?”

Problem. C is any point on a circle with diameter AB. Prove that the angle ACB is a right angle.
Drag C around the circle.

The heuristic at work

  1. The figure has a triangle and a circle, but nothing in it connects the circle's defining property, equal distance from the centre, to the angle we want.
  2. Add the element that brings that unused datum into play: the radius OC. Now OA = OB = OC.
  3. The triangle splits into two isosceles triangles: ∠OAC = ∠OCA = α and ∠OBC = ∠OCB = β.
  4. Angle sum in ABC: α + β + (α + β) = 180°, so ∠ACB = α + β = 90°.
Looking back. The auxiliary line was chosen because it touched both the unknown (the angle at C) and a datum not yet used (the centre O). The same radius proves the general inscribed-angle theorem: the central angle is twice the inscribed angle.

In physics: the method of images

Problem. A point charge +q sits at height d above a large grounded conducting plane. What force pulls it toward the plane, and how is the induced charge spread?
Drag the charge.
  1. The conductor's charges rearrange in some unknown way: too many unknowns.
  2. Add an element that is not there: an image charge −q at depth d below the plane.
  3. The pair has zero potential everywhere on the plane, which is exactly the conductor's boundary condition. By the uniqueness theorem the field above the plane is the pair's field.
  4. So the force is kq²/(2d)², and the field at the surface gives σ = −qd / 2π(x² + d²)^(3/2), totalling −q.

Auxiliary problem

“Can you find a subproblem or side problem whose solution will help you solve your problem?”

Problem. Move a stack of n disks from peg A to peg C, one disk at a time, never placing a larger disk on a smaller one. What is the least number of moves?

The heuristic at work

  1. Focus on the largest disk. At the moment it moves from A to C, every other disk must be sitting on B.
  2. That isolates a side problem: move n − 1 disks from A to B. It is the original problem, one size smaller.
  3. Plan: solve the auxiliary problem A→B, move the big disk A→C, solve it again B→C. So T(n) = 2T(n−1) + 1, T(1) = 1.
  4. Unrolling, T(n) = 2ⁿ − 1. It is also a lower bound, since step 1 forces two complete (n−1)-transfers.
Looking back. The left panel shows the stack of auxiliary problems in force at each move. When the side problem is a smaller copy of the main one, the heuristic becomes recursion, the engine behind divide-and-conquer algorithms.

In physics: the two-body problem

Problem. Two stars orbit each other under their mutual gravity. What paths do they follow?
  1. Two coupled bodies, six unknown coordinates: hard.
  2. Auxiliary problem: the relative vector r = r₁ − r₂ obeys μ r″ = −Gm₁m₂ r̂/r², with reduced mass μ = m₁m₂/(m₁ + m₂).
  3. That is Kepler's one-body problem, already solved: an ellipse with a focus at the origin (right, dashed).
  4. Undo the substitution: each star moves on a scaled copy, r₁ = (m₂/M) r and r₂ = −(m₁/M) r, about the fixed centre of mass.

Decomposing and recombining

“Can you decompose the problem and recombine its elements in some new manner?”

Problem. Find the area of a disc of radius r, knowing only that its circumference is 2πr.

The heuristic at work

  1. Decompose the disc into N equal sectors.
  2. Recombine them in a new manner: alternately tip-up and tip-down, side by side.
  3. The strip's height is the radius; each scalloped long edge uses half the circumference, πr.
  4. Rearranging does not change area. As N grows the strip becomes a rectangle πr × r, so the area is πr².
Looking back. Archimedes made this rigorous by exhaustion. Decomposing the same way gives the sector area ½r²θ, and in space (cones from the centre to the surface) it gives the sphere's volume as ⅓ r times its surface area.

In physics: superposition of driven responses

Problem. A damped oscillator with natural frequency ω₀ is pushed by a square-wave force of frequency ω. What is its steady motion?
  1. Decompose the drive: a square wave is (4/π)Σ sin(nωt)/n over odd n.
  2. Each sinusoidal push has the textbook steady response, with amplitude F_n/√((ω₀² − n²ω²)² + (γnω)²) and a phase lag.
  3. The equation is linear, so recombine: the motion is the sum of the separate responses.
  4. Set ω₀ near 3 or 5 and the oscillator responds mainly to that overtone; the bars on the right show why.

Draw a figure

“Can you draw a picture of the problem?”

Problem. Two friends agree to meet between 12:00 and 1:00. Each arrives at an independent, uniformly random time and waits w minutes before leaving. With w = 15, what is the probability they meet?

The heuristic at work

  1. Two unknown arrival times become one point (x, y) in a 60 × 60 square. Uniform and independent means probability is proportional to area.
  2. They meet exactly when |x − y| ≤ w: the shaded band along the diagonal.
  3. The complement is two corner triangles with legs 60 − w; together they make a square of area (60 − w)².
  4. P = 1 − (1 − w/60)² = 1 − (3/4)² = 7/16 for w = 15.
Looking back. Without the figure this is a double integral with case splits; with it, the answer is read off as area. The same square settles many problems about two uniform variables, such as the distribution of x + y or the expected gap |x − y| = 20 minutes.

In physics: the twin paradox

Problem. One twin flies to a star D light-years away at speed v and comes straight back. Who is younger at the reunion, by how much, and why isn't the situation symmetric?
  1. Draw the spacetime diagram, with light rays at 45°. Each twin is a worldline between the same two events.
  2. The Earth twin's worldline is straight; the traveller's is bent at the turnaround. That bend breaks the symmetry.
  3. Ages are proper times, τ = ∫√(dt² − dx²). In Minkowski geometry the straight worldline is the longest, so the traveller is younger by a factor γ.
  4. The dashed lines show the traveller's “now” on Earth jumping forward at the turnaround: the years that seem to go missing.

Generalization

“Can you find a problem more general than your problem?”

Problem. Evaluate ∫₀^∞ (sin x)/x dx. The integrand has no elementary antiderivative.

The integrand e−tx sin x / x for the chosen t.

The whole family I(t): dots computed numerically, curve from the formula.

The heuristic at work

  1. Embed the problem in a family: I(t) = ∫₀^∞ e^(−tx) (sin x)/x dx, t ≥ 0. The original problem is I(0).
  2. The general problem is easier. Differentiating in t removes the troublesome 1/x: I′(t) = −∫₀^∞ e^(−tx) sin x dx = −1/(1 + t²).
  3. So I(t) = C − arctan t. As t → ∞ the damping kills the integral, so C = π/2.
  4. Set t = 0: ∫₀^∞ (sin x)/x dx = π/2.
Looking back. Pólya called this the inventor's paradox: the more ambitious problem may have the better chance of success. Feynman made the parameter trick famous; the same move gives ∫₀^∞ xⁿe^(−x) dx = n! by differentiating ∫ e^(−tx) dx = 1/t n times.

In physics: from two coupled masses to waves

Problem. Masses joined by identical springs between two walls: what are the natural frequencies? For two masses it is a 2 × 2 determinant. What about N?
  1. Solve the general problem instead: N masses. Guess u_j = sin(jkπ/(N+1)) cos ωt, a standing wave.
  2. It satisfies every equation at once, with ω_k = 2ω₀ sin(kπ/2(N+1)).
  3. N = 2 drops out as a special case: ω₀ and √3 ω₀.
  4. The general solution shows what the small one hides: a dispersion relation, and in the limit N → ∞ the wave equation for a continuous string.

Induction and mathematical induction

“Can you solve your problem by deriving a generalization from some examples?”

Problem. What is 1 + 3 + 5 + … + (2n − 1)?

Observe. 1, 4, 9, 16, 25: squares. Guess the sum is n².

See why. Each odd number 2k − 1 is an L-shaped gnomon that grows a (k−1) × (k−1) square into a k × k square.

Prove it by mathematical induction. Base: 1 = 1². Step: if the first n odd numbers sum to n², adding the next one gives n² + (2n + 1) = (n + 1)². So the guess holds for every n.

A guess must be tested

Problem. Put n points on a circle and draw every chord between them. Into how many regions is the disc cut?

Counting the first cases gives 1, 2, 4, 8, 16, which practically shouts 2ⁿ⁻¹. Six points give 31. The real rule comes from understanding, not pattern-matching: each new chord adds one region plus one for every chord it crosses, and every crossing is fixed by a choice of 4 points, so the count is 1 + C(n,2) + C(n,4).

Looking back. Induction, inferring a law from examples, proposes; mathematical induction disposes. In the first problem the figure explained the pattern and the proof confirmed it. In the second, an equally convincing pattern was wrong.

In physics: Galileo's odd-number rule

Problem. How does the distance covered by a falling or rolling body grow with time? Galileo slowed the fall with a gentle incline and timed it by ear against frets on the ramp.
  1. Examples first: in equal beats the ball covers distances in the ratio 1 : 3 : 5 : 7 …, whatever the incline.
  2. The running totals are 1, 4, 9, 16 …: conjecture that distance grows as t².
  3. Mathematical induction finishes it, exactly as for the odd numbers: constant acceleration makes the k-th beat cover a distance proportional to 2k − 1, and the first n odd numbers sum to n².
  4. Galileo stated it in Two New Sciences (1638); the odd-number rule is still how strobe photographs of falling balls are read.

Reductio ad absurdum and indirect proof

“Can you show the falsity of an assumption by deriving from it a manifest absurdity?”

Problem. Prove that there are infinitely many primes.

Assume the opposite: the primes you select below are all the primes there are.

The heuristic at work

  1. Assume the claim is false: the list p₁, …, pₖ contains every prime.
  2. The assumption hands you something concrete to build with: N = p₁p₂⋯pₖ + 1.
  3. N > 1, so N has some prime factor p. Every pᵢ leaves remainder 1 when dividing N, so p is not in the list.
  4. That contradicts the list being complete. The assumption is false. (N itself need not be prime: try 2, 3, 5, 7, 11, 13.)

The same move, again

√2 is irrational. Suppose √2 = p/q in lowest terms. Then p² = 2q², so p² is even, so p is even: p = 2m. Substituting, 4m² = 2q², so q² = 2m² and q is even too. Both even contradicts lowest terms.
Looking back. An indirect proof gives you an extra hypothesis to manipulate: a finite list, a fraction in lowest terms. A direct proof of “infinitely many” or “not a fraction” has nothing so tangible to start from.

In physics: Galileo's tied stones

Problem. Aristotle held that heavier bodies fall faster, at speeds proportional to weight. Is that consistent?
  1. Assume the premise: speed grows with weight.
  2. Tie a light stone to a heavy one. The pair is heavier than the heavy stone alone, so it should fall faster.
  3. But the slow stone should hold back the fast one, so the pair should fall slower than the heavy stone alone.
  4. The premise predicts both, which is absurd. Only one assignment is free of contradiction: all bodies fall at the same rate (in the absence of air).

Specialization

“Can you find a problem more specialized? What if the given value were zero?”

Problem. Two unit squares. A corner of the second is pinned at the centre of the first, and the second square turns freely about that point. What is the area of the overlap?

The heuristic at work

  1. Specialize to the easiest position, θ = 0. The overlap is a quarter of the square: area ¼.
  2. Another special case, θ = 45°. The overlap is a triangle with base 1 and height ½: area ¼ again.
  3. Conjecture: the area is ¼ for every θ. Drag the square: the computed overlap never changes.
  4. Why do the special cases agree? The overlap is the square cut by a 90° wedge at its centre. The quarter-turns of that wedge partition the plane, and the square is unchanged by quarter-turns about its centre, so four congruent copies of the overlap tile the square.
Looking back. Specialization delivered the answer before the proof, and asking why the special cases agree delivered the proof idea: symmetry under quarter-turns.

In physics: the Atwood machine

Problem. Two masses hang over a light, frictionless pulley. You derive the acceleration a = g(m₁ − m₂)/(m₁ + m₂). Should you trust it?
  1. Specialize to cases whose answers you already know.
  2. Equal masses: nothing should move, and the formula gives 0.
  3. Remove m₂: m₁ is in free fall, and the formula gives g, with tension 0.
  4. Swap the masses: the motion reverses, and the formula changes sign. A guess that fails any of these is wrong before you look for the algebra slip.

Variation of the problem

“Keep only a part of the condition, drop the other part; how far is the unknown then determined, how can it vary?”

Problem. Inscribe a square in a given triangle ABC, with two vertices on the base AB and one vertex on each of the other two sides. This is Pólya's own example.

Drag B and C to reshape the triangle.

The heuristic at work

  1. Four conditions at once are too many. Vary the problem: drop the condition that the fourth vertex lies on BC.
  2. The new problem has infinitely many solutions: any square with two vertices on AB and one on AC. Slide the size to see them.
  3. All of them are scaled copies of one another from centre A, so the free fourth vertex moves along a straight line through A.
  4. Now restore the dropped condition: the fourth vertex must also lie on BC. Intersect the line with BC. With base a and height h, the side is s = ah / (a + h).
Looking back. Varying the problem turned a construction with no obvious starting point into a family whose behaviour could be watched. The locus method solves a whole class of construction problems the same way.

In physics: the cannon and the safety parabola

Problem. A cannon fires with a fixed muzzle speed. At what angle must it aim to hit a target at (x, y)?
Drag the target.
  1. Drop part of the condition: forget the target, keep only the muzzle speed. Now every angle gives a trajectory (grey).
  2. The family has an envelope, the safety parabola y = v²/2g − gx²/2v²: every reachable point lies on or under it.
  3. Restore the condition: a target inside the envelope is hit by exactly two trajectories (lob and direct), one on the envelope by exactly one, and one outside by none.
  4. The same envelope gives the maximum range on any slope at once.

Working backwards

“Can you start with the goal and work backwards to something you already know?”

Problem. How can you bring up from the river exactly 6 quarts of water, when you have only a 4-quart pail and a 9-quart pail? (Pólya's example; change the numbers below.)

Every possible state: quarts in B across, quarts in A up. Numbers are moves remaining to the goal, found by searching backwards from it.

    The heuristic at work

    1. Start from the goal: 6 quarts in the big pail.
    2. What state could come just before it? A full 9-quart pail from which exactly 3 quarts are poured off, into the small pail when it already holds 1.
    3. How do we get 1 quart? Fill the big pail and pour off 4 twice: 9 − 4 − 4 = 1. That is something we know how to do.
    4. Reverse the chain and you have the solution. The grid does exactly this for every state at once.
    Looking back. A target is reachable exactly when it fits in the larger pail and is a multiple of gcd(a, b), by Bézout's identity. Working backwards is also how chess endgame tables and dynamic programming work: values propagate from the goal toward the start.

    In physics: the ballistic pendulum

    Problem. A bullet is fired into a wooden block hanging on strings; the block swings up to a height h. How fast was the bullet?
    1. Start from what you can measure at the end: the rise h.
    2. Before that, the swing: energy is conserved while the block rises, so just after impact the speed was V = √(2gh).
    3. Before that, the impact: momentum is conserved in the brief collision, so v = (m + M)V/m.
    4. Working forwards from v would require modelling the collision itself, which destroys almost all the kinetic energy as heat. Working backwards never needs it.

    Invariants

    “Find a quantity that no allowed move can change. If the start and the goal disagree on it, no sequence of moves connects them.”

    Problem. Sam Loyd's 15 puzzle starts in order except that tiles 14 and 15 are swapped. Loyd offered $1,000 for a solution. Can it be solved?

    Click a tile next to the gap, or use the arrow keys.

    The heuristic at work

    1. Every move slides a tile into the gap. As a permutation of the 16 squares (the gap counted as tile 16) that is one transposition, so the permutation's parity flips.
    2. The same move shifts the gap one square, so the parity of its taxicab distance from the bottom-right corner flips too.
    3. So parity + distance (mod 2) never changes. That is the invariant.
    4. Solved: even permutation, distance 0, invariant 0. Loyd's position: one transposition, gap at home, invariant 1. Unreachable; the $1,000 was safe.
    Looking back. The converse also holds (Johnson and Story, 1879): every position with invariant 0 can be solved, so the invariant splits the 16! arrangements exactly in half. Sliding never changes it; lifting out two tiles always does. Conservation laws in physics are the same idea.

    In physics: Kepler's second law

    Problem. A planet speeds up near the Sun and slows down far away. Is anything about its motion constant?
    1. Gravity points straight at the Sun, so it exerts no torque about the Sun.
    2. So the angular momentum L = m r × v is invariant, while speed and distance both change.
    3. The area swept per unit time is dA/dt = L/2m, also invariant.
    4. Equal times, equal areas: the shaded sectors, each one twelfth of an orbit, all have the same area.

    Colouring

    “Colour the objects so that every allowed move treats the colours the same way, then count colours.”

    Problem. Remove two diagonally opposite corners from a chessboard. Can the remaining 62 squares be covered by 31 dominoes?

    The heuristic at work

    1. Trying tilings by hand keeps failing, but failing is not a proof: there are far too many arrangements to exhaust.
    2. Colour the board in the usual checkerboard way. Switch the colouring on.
    3. Every domino covers one dark and one light square, wherever it lies.
    4. Opposite corners have the same colour, so 30 squares of one colour remain and 32 of the other. 31 dominoes would need 31 of each. Impossible.
    Looking back. Gomory's theorem is the converse: remove any one dark and one light square and a tiling always exists. “Tile it” builds that tiling from a cycle through all 64 squares: removing two squares of opposite colours cuts the cycle into two paths of even length, each covered pair by pair. A colouring is an invariant designed to fit the moves.

    In physics: frustrated antiferromagnets

    Problem. In an antiferromagnet each bond wants its two spins to point opposite ways. On which lattices can every bond be satisfied?
    Click a spin to flip it.
    1. “Neighbours opposite on every bond” is exactly a 2-colouring of the lattice: up = dark, down = light.
    2. The square lattice is bipartite like a chessboard, so the Néel state satisfies every bond.
    3. The triangular lattice is full of triangles, odd cycles that cannot be 2-coloured: each triangle has at least one frustrated bond, whatever you do.
    4. This geometric frustration leaves a huge number of equally good ground states (Wannier, 1950), the same mechanism behind the residual entropy of water ice.

    Pigeonhole principle

    “If more than n objects go into n boxes, some box holds two. The art is choosing the boxes.”

    Problem. Five points are placed anywhere in a unit square. Show that two of them are at most √2/2 ≈ 0.707 apart.

    Drag the points; try to push them all apart.

    The heuristic at work

    1. Five points, and we want two that are close: treat the points as pigeons.
    2. Make four holes by cutting the square into four ½ × ½ squares.
    3. Five points in four boxes: some box holds two.
    4. Two points in a ½ × ½ square are at most its diagonal, √2/2, apart.
    Looking back. Cutting into k² boxes puts two of k² + 1 points within √2/k (move the slider). The principle itself is trivial; all the work is in choosing holes whose size matches the claim. A point on a boundary can be assigned to either box.

    In physics: Poincaré recurrence

    Problem. A deterministic, reversible system has finitely many states. Started from some state and left alone, must it eventually come back to exactly that state?
    1. Arnold's cat map scrambles an N × N picture by (x, y) → (2x + y, x + y) mod N; it looks like chaos within a few steps.
    2. There are only finitely many arrangements, so by the pigeonhole principle some arrangement must occur twice.
    3. The map is reversible (determinant 1), so the first repeat must be the starting arrangement itself: step back from any repeat and you get an earlier repeat.
    4. Poincaré's recurrence theorem for real mechanics replaces counting by phase-space volume, conserved by Liouville's theorem.

    Extremal principle

    “Look at the object that is largest, smallest or best by some measure. Its extremeness often forces the property you want.”

    Problem. n red and n blue points lie in the plane, no three on a line. Prove that you can pair each red point with a blue point so that none of the n segments cross.

    The heuristic at work

    1. There are finitely many red–blue matchings (n! of them), so one has the least total length. Look at that one.
    2. Suppose two of its segments, r₁b₁ and r₂b₂, crossed at a point X.
    3. Swap partners. By the triangle inequality, r₁b₂ + r₂b₁ < (r₁X + Xb₂) + (r₂X + Xb₁) = r₁b₁ + r₂b₂.
    4. The swap gives a shorter matching, contradicting minimality. So the shortest matching has no crossings.
    Looking back. The extremal object did all the work; no construction was needed. Uncrossing pairs one at a time (the buttons) also terminates, because total length strictly drops at each swap. It always reaches a crossing-free matching, though not always the shortest one.

    In physics: Fermat's principle of least time

    Problem. Light travels from A in air to B in glass. Which path does it take across the surface?
    Drag A, B and P.
    1. Look at the extremal path: of all paths A → P → B, the one of least time exists, since the time varies continuously with P.
    2. At that minimum, moving P slightly changes nothing to first order: dT/dx = 0.
    3. Differentiating T = AP/v₁ + PB/v₂ gives sin θ₁/v₁ = sin θ₂/v₂, which is Snell's law.
    4. The same extremal reasoning underlies the principle of least action, from which all of classical mechanics can be derived.

    Monovariants

    “Find a quantity that every move strictly decreases. If it can't decrease forever, the process must stop.”

    Problem. In a parliament, each member has at most three enemies, and enmity is mutual. Show that the members can be split into two houses so that each has at most one enemy in their own house.

    Red lines are enemy pairs inside the same house. Ringed members are unhappy.

    The heuristic at work

    1. Start with any split. Call a member unhappy if they have two or more enemies in their own house.
    2. Move an unhappy member to the other house. They had at least 2 enemies at home, so at most 1 in the other house.
    3. Let E be the number of enemy pairs sitting in the same house. The move removes at least 2 such pairs and creates at most 1, so E drops by at least 1.
    4. E is a whole number and never negative, so it cannot drop forever. When the moves stop, nobody is unhappy.
    Looking back. E is a monovariant: unlike an invariant it changes, but only in one direction. Every proof that a loop terminates uses one (a loop variant), and so does the uncrossing on the previous page.

    In physics: energy as a Lyapunov function

    Problem. A pendulum swings with friction. Prove that, from any start, it settles at rest.
    Click in the phase plane to start from there.
    1. Track the energy E = ½ω² − cos θ (contours of constant E are drawn).
    2. Along the motion dE/dt = −γω² ≤ 0: E never increases.
    3. E is bounded below by −1, so it must level off, and it can only stop falling where the pendulum is permanently at rest: an equilibrium.
    4. This is Lyapunov's method: one monovariant proves stability without solving the equations.

    Symmetry and reflection

    “Use a symmetry of the situation: reflect, rotate or swap, and see what the transformation preserves.”

    Problem. A and B are on the same side of a straight river. You walk from A to the river for water, then on to B. Where should you touch the river to make the walk as short as possible? (Heron of Alexandria.)
    Drag A, B and P. The curve in the water is the walk's length as P moves.

    The heuristic at work

    1. The path bends at P, which makes lengths hard to compare.
    2. Reflect B in the river to get B′. For every P on the bank, PB = PB′, so AP + PB = AP + PB′.
    3. AP + PB′ is a path from A to B′, shortest when it is straight.
    4. So the best point P* is where the segment AB′ meets the river. There the two angles with the bank are equal: the law of reflection.
    Looking back. Fermat later derived Snell's law of refraction from the same least-time idea. Reflection tricks also solve billiard paths and the shortest path on the surface of a box (unfold it).

    In physics: the infinite resistor grid

    Problem. In an infinite square grid of identical resistors R, what is the resistance between two adjacent nodes?
    1. Feed a current I in at node A and let it flow out at infinity. By the four-fold symmetry of A, exactly I/4 flows through each of its four resistors, including the one to B.
    2. Separately, let I flow in from infinity and out at B. By symmetry again, I/4 flows from A to B.
    3. Superpose the two: I in at A and out at B, with I/2 through the AB resistor. So V_AB = R · I/2 and the resistance is R/2.
    4. The plot solves finite grids numerically; they converge to ½. A direct infinite calculation needs Fourier integrals.

    Change of representation

    “Translate the problem into a different language (a graph, an equation, a picture) where the structure is easier to see.”

    Problem. Königsberg had seven bridges joining two islands and the two river banks. Can you walk through the city crossing every bridge exactly once? (Euler, 1736.)
    Click a district to start, then click bridges to cross them.

    The heuristic at work

    1. Try the walk on the map. You always get stuck.
    2. Euler's change of representation: shapes and distances are irrelevant. Shrink each land mass to a point and each bridge to a line, giving a graph. Switch on the graph view.
    3. Each time a walk passes through a land mass it uses two bridges, one in and one out. So every land mass except the start and the end needs an even number of bridges.
    4. In Königsberg all four have an odd number (5, 3, 3, 3). At most two may be odd, so no such walk exists.
    Looking back. Euler's representation founded graph theory. Add an eighth bridge and only two odd land masses remain; then the walk exists, starting at one of them and ending at the other.

    In physics: phasors for AC circuits

    Problem. A series R, L, C circuit is driven by V₀ cos ωt. What current flows, and with what phase?
    1. In the time domain this is a second-order differential equation.
    2. Change representation: let each sinusoid be a rotating arrow in the complex plane, a phasor. Differentiation becomes multiplication by iω.
    3. Kirchhoff's law becomes vector addition (left): V = I(R + iωL + 1/iωC) = IZ.
    4. At ω = 1/√(LC) the inductor and capacitor arrows cancel and Z = R: resonance, read straight off the picture.

    Counting two ways

    “Count the same set in two different ways, and set the answers equal.”

    Problem. Find 1 + 2 + 3 + … + n.

    The heuristic at work

    1. Draw the sum as a staircase of dots: row k has k dots.
    2. Take a second copy, turn it half a turn, and fit it onto the first.
    3. Together they make an n × (n + 1) rectangle. Counted by rows and columns that is n(n + 1) dots; counted as two staircases it is 2S.
    4. So S = n(n + 1)/2.

    The same move, again

    Problem. At a party, everyone counts how many hands they shook. Why is the total of those counts always even, and why is the number of people with an odd count always even?

    Count handshakes two ways. Each handshake adds 1 to two people's counts, so the sum of all counts is 2 × (handshakes), which is even. The even counts add up to something even, so the odd counts must come in pairs.

    Looking back. Double counting proves a great many identities, such as Σ C(n,k) = 2ⁿ (count subsets by size, or by in/out choices). The handshake lemma is also why odd-degree vertices, like Königsberg's, always come in pairs.

    In physics: pressure from kinetic theory

    Problem. A gas is a swarm of molecules bouncing off the walls. What pressure does it exert?
    1. Count the impulse delivered to one wall in time t in two ways.
    2. As a force: impulse = F t.
    3. Molecule by molecule: each molecule returns every 2L_x/|v_x| and delivers 2m|v_x| each time, so in time t it delivers m v_x² t / L_x.
    4. Equate: F = Σ m v_x² / L_x, so PV = N m⟨v_x²⟩. With the identification m⟨v_x²⟩ = kT, this is the ideal gas law.

    Complementary counting

    “If the event is complicated, compute what it is not.”

    Problem. How many people must be in a room before the chance that two share a birthday exceeds ½? (Ignore 29 February.)

    One room's birthdays. Red days are shared.

    The heuristic at work

    1. “At least two share” splits into a mess of cases: one pair, two pairs, a triple, and so on.
    2. The complement, “all birthdays differ”, is one clean case.
    3. Place people one at a time: P(all differ) = (365/365)(364/365)⋯((365 − n + 1)/365).
    4. P(a shared birthday) = 1 − P(all differ), which passes ½ at n = 23.
    Looking back. Why so few people? 23 people form C(23, 2) = 253 pairs, each with a 1/365 chance. Any “at least one” question is worth trying through “none”.

    In physics: Babinet's principle

    Problem. What diffraction pattern does a thin wire, or a hair, of width a cast in a laser beam?
    1. The wire is the complement of a slit of the same width: wire + slit = an unobstructed beam.
    2. Diffraction amplitudes add, so U_wire = U_beam − U_slit.
    3. A smooth laser beam's own far field is a narrow spike around θ = 0; everywhere else U_beam ≈ 0, so I_wire = I_slit. (A hard-edged beam would add its own fringes, which is why lasers make the trick work.)
    4. So a hair gives the familiar single-slit pattern, with minima at sin θ = mλ/a: an easy way to measure it with a laser pointer.

    Change the point of view

    “Look at the problem from another vantage point: a different unknown, a different frame, a different time.”

    Problem. Two trains start 100 miles apart on the same track and approach each other at 50 mph each. A fly starts at the front of one train, flies at 75 mph to the other, turns back, and keeps zigzagging until the trains meet. How far does the fly travel?

    The heuristic at work

    1. Following the fly gives an infinite series of shrinking legs (listed above). Each leg is a fixed fraction of the one before.
    2. Change the point of view: watch the trains instead. The gap closes at 100 mph, so they meet after exactly 1 hour.
    3. The fly is flying for that whole hour at 75 mph.
    4. Distance: 75 miles. No series needed.
    Looking back. The story goes that John von Neumann answered instantly, and when told that most people try to sum the series, replied that he had summed it. The space-time diagram shows both views at once: the zigzag is the series, the closing triangle is the clock.

    In physics: collisions in the centre-of-mass frame

    Problem. Two balls of masses m₁ and m₂ collide head-on. What are their velocities afterwards?
    1. In the lab frame you must solve momentum and energy conservation together, a quadratic.
    2. Change the point of view: ride along with the centre of mass. There the total momentum is zero.
    3. An elastic collision can then only reverse both velocities (an inelastic one also scales them by e).
    4. Transform back: v = V − e(u − V) for each ball, where V is the CM velocity. No quadratic required.

    Extreme cases

    “Push a parameter to its limit. If the answer can't depend on it, the extreme case gives the answer for free.”

    Problem. A cylindrical hole 6 inches long is drilled straight through the centre of a sphere. What volume remains? (The napkin ring problem.)

    The heuristic at work

    1. The sphere's radius isn't given. If the problem has an answer, the answer cannot depend on it.
    2. So take the extreme case: a sphere of diameter exactly 6, where the hole has zero width.
    3. Nothing is removed. The volume is the whole sphere: (4/3)π · 3³ = 36π ≈ 113.1 cubic inches.
    4. Check it. Slice at height y: the ring's cross-section is an annulus of area π[(R² − y²) − (R² − 9)] = π(9 − y²), the same as the slice of the 6-inch sphere. By Cavalieri's principle the volumes agree for every R.
    Looking back. The extreme case found the answer; the slice argument justified the assumption that there was one. A larger sphere leaves a ring that is thinner but wider, with exactly the same volume.

    In physics: testing candidate laws of velocity addition

    Problem. In relativity, how do two velocities u and v combine? Several plausible formulas compete.
    1. You need no derivation to eliminate candidates; push the parameters to their limits.
    2. At everyday speeds the answer must be Galileo's u + v.
    3. Light combined with anything must still travel at c, since c is the same in every frame.
    4. Two speeds below c must combine to a speed below c. Only (u + v)/(1 + uv/c²) survives every test in the table.

    Dimensional analysis

    “Check what units the answer must carry. Often only one combination of the data has the right dimensions.”

    Problem. How does the period of a pendulum depend on its length L, the bob's mass m, gravity g and the release angle θ₀?

    The heuristic at work

    1. List the dimensions: T in s, L in m, m in kg, g in m/s², θ₀ dimensionless.
    2. Mass is the only quantity containing kg, and nothing could cancel it, so it cannot appear.
    3. The only combination of L and g measured in seconds is √(L/g).
    4. So T = √(L/g) · f(θ₀). Only f needs real physics, and f → 2π for small swings. Change m and nothing happens; change L or g and T scales exactly as predicted.
    Looking back. Dimensional analysis gives the form of the law without solving the equation of motion: Galileo's √L and the independence from mass fall out for free. The exact f(θ₀) = 2π / AGM(1, cos(θ₀/2)) needs the full solution; the right-hand plot shows it.

    In physics: G. I. Taylor and the first atomic bomb

    Problem. In 1950 Taylor estimated the energy of the 1945 Trinity explosion from a declassified sequence of photographs of the fireball, without access to the classified yield. How?
    1. In the early blast the radius R can depend only on the energy E, the air density ρ and the time t.
    2. The only length that can be built from them is (Et²/ρ)^(1/5), so R = S (E/ρ)^(1/5) t^(2/5) with S a pure number (about 1.03 for air).
    3. The photographs (red points) do follow slope 2/5 on a log-log plot, confirming the form.
    4. Fitting gives E ≈ 7 × 10¹³ J, about 17 kilotons of TNT; the official yield, kept secret at the time, was about 21 kilotons.