IMC 2015

360°

HintWalk once round the shape: by the time you face the way you started, how far have you turned?

The whyOne full turn, whatever the number of sides. A regular polygon with n sides therefore has an exterior angle of 360 ÷ n. The interior sum grows with n; the exterior sum never does.

108°

HintExterior angle first: divide a full turn by five, then take that off a straight line.

The whyA regular pentagon has exterior angles of 72° and interior angles of 108° — a pair worth knowing cold. Stand one on a straight line with a side on the line and the next side rises at 180 − 108 = 72°, which is where most "pentagon on a line" angle chases start. Two pentagons cannot meet at such a vertex, though: 108° + 108° = 216° is more than the 180° a straight line has to give. So a row of pentagons on a line is spaced out, and the gap between two of them closes at 180 − 72 − 72 = 36° — the same 36° that three pentagons round a point leave over (360 − 324), which is why regular pentagons will not tile.

6

HintWhich whole numbers, when squared, land between ten and ninety-nine?

The whyThey are 16, 25, 36, 49, 64 and 81 — the squares of 4 to 9. A question about two-digit squares is a question about six numbers, so just list them.

two

HintThink about what 1 is missing that 2 has.

The whyThe number 1 has only one factor, so by definition it is not prime — a convention chosen so that every number has a single prime factorisation. Competition questions lean on this: an expression that comes out as 1 does not count as prime, and a sequence that must begin with a prime cannot begin with 1.

four; twelve; five

HintThree solids, three different counts — picture each one in turn.

The whyFaces, edges and vertices are the three counts every solids question turns on: the tetrahedron has 4 faces, 4 vertices and 6 edges; the cube 6, 8 and 12; the square-based pyramid 5, 5 and 8. For every one of them, faces + vertices − edges = 2.

Their total: mean × count

HintUndo the averaging: instead of dividing, do the opposite.

The whyMean questions are almost always total questions in disguise. Once the total is fixed, "what is the largest one of them could be?" becomes "make the others as small as the rules allow" — and if they must be different positive whole numbers, the smallest others are 1, 2, 3 and so on.

Exactly half

HintDrop a vertical line from the top vertex and look at the two pieces separately.

The whyWherever the top vertex sits along the opposite side, the triangle is half the rectangle: each piece of the rectangle is cut in two by a diagonal. So a row of shaded triangles across a rectangle shades exactly half of it.

Multiply the choices together

HintIndependent decisions do not add up; think of a grid with one row per first choice.

The whyThe product rule: p options for the starter and q for the main give p × q meals. It extends to any number of positions — and a position locked in by earlier choices contributes a factor of 1, not another full set of options.

120

HintEvery one of them could come first — and that is only the start.

The why5 × 4 × 3 × 2 × 1 = 120, written 5!. n! grows fast: 6! = 720, 7! = 5040 — so a question about arranging a handful of things is usually about hundreds or thousands of cases, never a few.

Add or subtract the equations, scaling as needed

HintYou cannot pin down each unknown, but combining the two lines often produces exactly the mix wanted.

The whyWith fewer equations than unknowns you cannot solve for everything — and you are not asked to. Subtracting (or adding) the equations, then scaling, builds the combination the question wants. Check the coefficients you are aiming for and work towards them rather than towards single values.

half

HintAverage the two parallel sides first.

The whyArea = ½(a + b)h. An L-shape is a rectangle minus a rectangle; a trapezium is the average of its parallel sides times the distance between them. Shaded regions inside a square are usually quickest by subtraction: the whole square minus the pieces you can name.

Areas in the ratio of their bases

HintHalf base times height — one of those factors is shared.

The whyIf two triangles have the same height, dividing one base by the other divides one area by the other. A point one third of the way along a side therefore cuts off a triangle worth one third of the whole, without any lengths or heights being calculated.

The decimal points

HintPad the shorter figure with trailing zeros so every column matches.

The whyOnce the points line up, every column holds digits of the same place value, so you can add or subtract straight down exactly as with whole numbers. A quick chain of additions and subtractions across shrinking place values (tenths, then hundredths, then thousandths) can often be tracked digit by digit without writing out the full column sum.

Round both to nearby 'nice' numbers

HintSwap the messy figures for tidy multiples that divide cleanly into each other.

The whyRounding both the top and bottom of a division to numbers that share factors — for example treating 2977 ÷ 29 as roughly 3000 ÷ 30 — usually gives an answer close to the exact one, because the two roundings partly cancel. It turns an awkward division into one you can do in your head, which is exactly what an approximate 'how many per minute/per unit' question is really testing.

denominator

HintFind a shared multiple that every bottom number divides into exactly.

The whyA common denominator (often the lowest common multiple of the individual denominators) lets every fraction be rewritten with matching bottoms, so the numerators can simply be added. Skipping this step and combining fractions some other way — say, adding numerators and denominators separately — gives a wrong, and usually much smaller, result.

a to the power mn

HintDon't add the two indices — multiply them instead.

The whyThis is why a number can be a hidden square even when it is not written as a familiar one: any number of the form (a to the power m) to the power 2, for any whole number m, is automatically a square, because doubling the outer index just multiplies the whole exponent by 2. Spotting a number written this way is often faster than trying to recognise it from its ordinary value.

Equilateral triangles (same side length as the hexagon)

HintJoin the centre to every vertex and look at the angle each wedge makes.

The whyThe angle at the centre of each wedge is a full turn shared six ways, 360° ÷ 6 = 60°. The two sides running from the centre to the hexagon's corners are equal (a hexagon happens to have that radius-equals-side property), so the third angle must also be 60° — giving an equilateral triangle. That is the quickest route into comparing a regular hexagon's area with any other shape built from equilateral triangles of the same side length.

4

HintWrite the odd number as 2k + 1, multiply through by 4, then split off a full multiple of 8.

The why4 × (2k + 1) = 8k + 4, which is a multiple of 8 plus exactly 4 more — so the remainder is always 4, whatever odd number is chosen. This is the standard move for remainder questions built from a product: pull out every factor of 2 you can find, then look at what is left over.

No — check each shared edge individually

HintThink about where the board has a jagged or non-rectangular outline.

The whyOn a full rectangular grid, colour strictly alternates with every step, so after an even number of moves you are guaranteed back on the starting colour. Cut the grid into an irregular shape — an L, a cross, or anything with a 'dent' in its outline — and two squares of the same colour can end up sharing an edge at the join, breaking that guarantee. A problem that looks like a simple parity puzzle can hide exactly this kind of irregular boundary, so sketch the actual shape rather than trusting the general rule.

They are all equal, to the mean of the first two terms

HintAdding a new value that already matches the running average never changes that average.

The whyCall the first two terms a and b. The third term works out to be (a + b) ÷ 2, and once that value is in the list, adding another term equal to the current mean can never shift the mean — so the fourth, fifth and every later term is stuck at that same (a + b) ÷ 2, no matter how far down the sequence you look. So being told one distant term of such a sequence actually tells you the mean of the very first two.

Creates a smaller triangle similar to the original one

HintThe angles stay identical; only the overall scale changes.

The whyBecause the two triangles are similar, every length inside the smaller one scales by the same factor as its base compares with the original base — so an unknown height, base or side in the smaller triangle can be found from one ratio, without measuring the whole figure directly. This is the standard route into "strip" problems, where a shape is sliced into equal-width bands and each slice needs its own dimensions worked out.

Twice (once from each adjoining face)

HintPicture two neighbouring panels meeting along a seam — that seam belongs to both of them.

The whySo to count the actual distinct edges (or joins, seams, boundaries — anything shared between two neighbours), add up each piece’s own edge-count and divide by 2. It is the same idea behind counting handshakes at a party: if everyone reports how many hands they shook, the total is double the real number of handshakes, because each handshake got reported by both people involved in it.

overlap

HintAdding them separately counts the shared part twice — take one copy of it back off.

The whyThis is the addition rule for overlapping regions, the same idea behind a two-circle Venn diagram: total covered = first area + second area − overlap. Miss the subtraction and the shared part gets double-counted, which is exactly the trap in any "total shaded area" question built from two overlapping shapes.

√2 times that step

HintThink of it as the diagonal cutting across a square, then relate the diagonal to one side using Pythagoras.

The whyThis follows straight from Pythagoras: if both the horizontal and vertical legs equal d, the hypotenuse is the square root of (d² + d²), which simplifies to d√2. It is the fastest way to find the length of a diagonal segment built from equal horizontal and vertical steps, including diagonals threaded across a grid of identical small squares, without measuring anything directly.

Check which value fits every constraint from the diagram

HintOne of the two answers could make a stated measurement negative or impossibly small — rule it out.

The whyQuadratics often produce two mathematically valid roots, but geometry adds real-world constraints: a length can't be negative, a point has to lie inside the shape, or one value must exceed some other given measurement. Always test both roots against the diagram's conditions before choosing an answer, rather than assuming the 'nicer' number is correct.