What to do when you have no idea how to start

There are two ways to be stuck, and they need opposite responses. One is knowing exactly what you want to do and being unable to do it. That is a technical problem, and it is a good place to be, because the difficulty has an address.

The other is the blank page. You have read the statement, you understand every word of it, and nothing happens. No move suggests itself. Ten minutes go by in which the only thing you produce is a growing certainty that you are not good enough at this.

That second kind of stuck is not a knowledge failure. It is a procedure failure, and it is fixable in an afternoon, because the fix is not learning more mathematics. It is having a fixed list of things to do to a problem that you run before you have any idea what the solution looks like.

Below is that list. It is ordered by cost, cheapest first, because the whole point is that you are allowed to start moving without a plan.

Move 1: shrink it until it is trivial

Replace every parameter in the problem with the smallest number that keeps the question interesting. Then solve that. Then go up by one.

This is the cheapest move in existence and it is skipped constantly, because solving $n = 2$ feels like avoiding the problem. It is not avoiding the problem. It is the only way to see the mechanism, and on a large fraction of olympiad problems the mechanism at $n = 2$ is the entire mechanism at $n = 1000$.

Take any sequence of $n^2+1$ distinct reals contains a monotone subsequence of length $n+1$. At $n = 2$ that reads: any 5 distinct numbers contain a monotone run of 3. That is small enough to check by hand, and small enough to check exhaustively if you are patient, and once you have watched it work on 5 numbers you know what the boxes have to be.

The shrunk case does something else too, which is worth more than the solution: it tells you whether the bound is tight. Try 4 numbers. The sequence $2, 1, 4, 3$ has no monotone run of 3 in either direction. So $n^2$ genuinely is not enough and $n^2+1$ is not a number someone chose for aesthetics. Now you know the proof must use every one of those $n^2+1$ terms, which is a real constraint on what the proof can be.

Move 2: count both sides of the sentence

If the problem says two of something must coincide, stop and count how many of that something there are, and how many distinct values they could possibly take. Just the two numbers. No cleverness.

From ten distinct two-digit numbers, two disjoint subsets with equal sums, IMO 1972 Problem 1, is the purest case of this working. Count the things: a ten-element set has $2^{10} - 1 = 1023$ non-empty subsets. Count the values: every subset sum is at least 10, and at most $90 + 91 + \cdots + 99 = 945$, so there are at most 936 possible sums.

And there it is. 1023 objects, 936 slots. The problem is finished in one line and you got there without a single idea, only arithmetic.

What makes this move so cheap is that both counts are mechanical. You never had to be creative; you had to be willing to write down two numbers and compare them. The reason it works so often is structural: the phrase "two of them must" is a claim about a collision, and a collision claim is nearly always settled by counting the objects against the slots.

Move 3: start from the answer and walk backwards

Ask what the last line of the solution looks like, then ask what would have to be true one line above it. Working backwards is a full technique in its own right, but as an opening move it is used more narrowly: you are not trying to construct the solution in reverse, you are trying to find out what shape the answer has.

Look at the nested radical equation from IMO 1959. You are asked for which real $x$ a sum of two square roots equals $A$, for three values of $A$. Before doing anything, ask what an answer to a question like that can look like. Two options: a list of points, or a set.

That question pays immediately. Squaring the equation collapses the cross term to $2|x - 1|$, so the whole left side is $2x + 2|x - 1|$, which for $x$ between $\tfrac12$ and 1 is identically 2. So the case $A = \sqrt2$ has an entire interval of solutions, not a point. The case $A = 2$ has the single solution $x = \tfrac32$. And $A = 1$ has none at all, because $2x + 2|x-1|$ never drops below 2 anywhere in the domain.

A student who assumed the answer would be a list of numbers will find the interval and think they made an error. Asking what shape the answer has, before starting, is what stops that.

Move 4: do something with the symmetry

If swapping two of the letters leaves the problem unchanged, that is not a decoration, it is a free hypothesis. Symmetry buys you an ordering at zero cost: you may assume $a \ge b \ge c$ and lose nothing.

The IMO 1964 triangle inequality is fully symmetric in its three sides. So order them, and immediately the triangle condition stops being three inequalities and becomes one, since $a \ge b \ge c$ makes $b + c > a$ the only one that can fail. That is a third of the hypotheses cleared away before any work.

The complementary version is worth as much: when a problem is symmetric, guess that the extreme case is the symmetric one, and check it. Here the two sides are equal exactly when $a = b = c$, and nowhere else. Knowing where equality lives tells you how tight the argument has to be, and any step you write that is loose at $a = b = c$ is a step that cannot appear in a correct proof.

Move 5: solve a problem you are allowed to fail

Deliberately weaken the statement until you can prove something, and prove that instead. Drop a hypothesis. Prove a weaker bound. Prove it for integers rather than reals. Prove one direction of the equivalence.

This feels like cheating and it is the single most productive habit on this list, because the weakened problem usually uses the same mechanism as the real one and shows it to you without the difficulty. If you can prove the bound with a 4 where the problem wants a 3, you now know which step is lossy, and the actual work is upgrading exactly one line rather than inventing a proof from nothing.

It also converts a blank forty minutes into a partial answer. On a graded olympiad paper a proved special case is worth points. A blank page is worth none of them.

The rule for when to switch

The list only works if you actually cycle through it, and the failure is always the same: you commit to move 1, it does not immediately pay, and you spend thirty minutes making it pay because switching feels like admitting defeat.

Set a budget before you start. Give each move ten minutes and a fixed patch of paper. When the ten minutes are up and there is nothing on the paper, write one line at the bottom saying what you tried and why it stalled, and go to the next move. That one line is the important part. It is what stops you from silently running move 1 again in twenty minutes' time under a different name, and it is what makes the attempt useful later when you read the solution and want to know where your reasoning left the road.

Five moves at ten minutes each is fifty minutes, which is longer than you would ever have on a single AMC question and shorter than a single olympiad problem. Neither number is arbitrary: an exam that gives you three hours for three problems is telling you it expects several dead ends per problem.

Getting reps at the opening

The moves are cheap to learn and expensive to make automatic, and reading about them does approximately nothing for the second part. What works is separating the opening from the solving: take problems you have no intention of finishing, and for each one spend five minutes producing only the first move and one line saying why.

Lemma is built around that split. Working backwards and symmetry are taught as named, reusable moves with the recognition cues attached rather than as tricks you happen to meet twice, and the full technique index is readable without an account. The problem archive carries the statements above with difficulty and expected time on each, so you can pick something you are meant to struggle with rather than something you are meant to finish. And the daily problem posts one a day in three tiers, which is about the right dose for practising openings and nothing else.