IMC toolkit

Angle chasing — write every angle you can find on the diagram

HintFill in the ones you are not asked for as well; they are the stepping stones.

The whyMark each angle as you deduce it — angles on a straight line, in a triangle, on parallel lines — and the target usually falls out after three or four steps. A chain of small known facts is what competition geometry is.

Subtract the circle's area from the square's

HintTwo shapes you can handle separately, then one take-away.

The whyArea by subtraction is the workhorse of competition geometry: find the big shape, find the hole, take one from the other. A circle of radius 3 inside a square of side 6 leaves 36 − 9π, and the answer choices are usually left in terms of π so you never have to compute it.

13 — the 5, 12, 13 Pythagorean triple

HintThere is a famous trio that begins with these two numbers.

The whyCompetition papers reuse a handful of whole-number right-angled triangles: 3-4-5, 5-12-13, 8-15-17 and their multiples such as 6-8-10. Spotting one saves the square root entirely and often unlocks a length the question never stated.

1 : 9

HintArea is two-dimensional, so the scale factor is used twice.

The whyLengths scale by k, areas by k squared, volumes by k cubed. An enlargement by 3 makes every area nine times bigger — a fact that appears in IMC questions about shadows, maps and nested shapes far more often than the word "similar" does.

24

HintAdd one to each power, then multiply the results together.

The whyCounting factors from a prime factorisation is a competition staple: each prime can appear in a factor anywhere from zero times up to its full power, so the choices multiply. It also tells you a number is a perfect square exactly when every power is even.

Pair them up: 100 × 101 ÷ 2 = 5050

HintMatch the first with the last, the second with the second-last, and notice what each couple adds to.

The whyFifty pairs each adding to 101 gives 5050 — the formula n(n + 1) ÷ 2 for the sum of 1 to n. These triangular numbers turn up disguised as handshakes, dots in a triangle and the number of matches in a league.

Work out the first few terms and spot the pattern

HintSmall versions of the problem reveal the rule; then jump to the big one.

The whyTry small cases is the most reusable competition strategy there is. Three or four terms are usually enough to see the rule — but check one more before you trust it, and watch the off-by-one trap between "term 0" and "term 1".

The laws of indices

HintKeep everything as powers of 2 and combine the exponents rather than multiplying out.

The whyWhen every option is a power of the same base, never evaluate: 4³ × 8² becomes 2⁶ × 2⁶ = 2¹², and the exponent alone gives the answer. Add to multiply, subtract to divide, multiply for a power of a power.

Test each option with an extreme or awkward case

HintTry to break every statement using a value at the edge — zero, one, a negative, a huge number.

The why"Must be true" questions are won by counterexamples: one edge value that breaks a statement rules it out. Zero and one behave unusually under multiplication, negatives flip inequalities, and very large numbers show which term dominates.

1

HintLook at how the last digits of 3, 9, 27, 81, 243 behave.

The whyPowers of 3 end 3, 9, 7, 1, 3, 9, 7, 1 … a cycle of four. 100 ÷ 4 leaves remainder 0, which is the fourth position in the cycle, so the units digit is 1 — the same as 3⁴ = 81. Remainder 0 means the end of the cycle, not the start: the classic off-by-one trap.

1

HintThe lighter of the two penalties.

The whyMarks: 5 each for 1–15, 6 each for 16–25, blanks 0, with 1 lost for a wrong 16–20 and 2 lost for a wrong 21–25. A blind guess on 21–25 loses marks on average — leave it blank unless you can rule out at least two of the five options (three left is already worth a guess). On 16–20 the penalty is small enough that even a blind guess comes out slightly ahead on average.

180(n − 2)

HintSplit the shape into triangles from one corner and count them.

The whyEvery polygon splits into n − 2 triangles from one vertex, each worth 180°, so a hexagon's angles total 720° and each angle of a regular hexagon is 120°. The exterior angles of any polygon always add to 360° — often the faster route.