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Spatial & Visual Reasoning — Fold it, turn it, see it from the other side

About this topic

Here is the strange thing about spatial problems: the kids who are ‘good at picturing things’ are not the ones who win them. The winners stop picturing the whole thing at once. They grab one piece — one face, one corner, one beam of light — and follow only that.

Try it right now. Hold a closed fist in front of you. That is a cube. Without moving your hand, which face is hidden against your palm side? You answered instantly — not by rotating the cube in your head, but by tracking one face. That is the entire skill, scaled up.

Two habits carry every chapter here. Redraw and label: copy the figure onto scratch paper and mark what you know, so the picture becomes your memory instead of your worry. Track one piece: never ask ‘what does the whole shape do?’ Ask ‘where does this one part go?’ The rest always follows.

The chapters climb a ladder: nets and cube counts first, then mirrors and dice, then tiling, grids, and the counting tricks that crack the hard problems. Start low. The optional For older kids blocks are there when you want them, not before.

CHAPTER 1

Cube nets — which flat shape folds up?

THEORY

Cut a cereal box along its edges and flatten it. What you get is a net — a flat shape that folds back into the solid. A cube has 6 square faces, so a cube net is always 6 squares joined edge to edge. But here is the catch that traps everyone: not every arrangement of 6 squares folds into a cube.

Fold one in your head and you feel why. As the squares swing up, each one has to land on its own face — floor, ceiling, or one of the four walls. If two squares both try to cover the same wall, the fold jams. Six squares is necessary. It is not enough.

The faster question: which faces end up opposite?

Most net problems do not ask ‘is this a cube?’ They ask ‘after folding, which face sits across from which?’ You could fold every time. There is a rule that saves you.

Look at any straight row of squares in the net. Walk along the row. Faces pair up by skipping one: the 1st with the 3rd, the 2nd with the 4th. Two squares that touch can never be opposite — on a real cube, neighbours share an edge, and opposite faces share nothing.

12341 & 3 opposite2 & 4 opposite

When four faces sit in a row, they wrap around the cube like a belt, and that belt hands you two opposite pairs at once (1–3 and 2–4). The two squares poking out of the belt are the last pair, top and bottom. All six faces sorted, no folding needed.

THE MOVE: opposite faces are the two with exactly one square between them in a straight line. Skip one to find the partner.

🎯 Try it
A cube net has a straight row of four squares labelled, in order, 2, 5, 1, 4 (two more squares are stuck on as top and bottom). Using the skip-one rule, which number ends up opposite the 5?
Walkthrough: walk the row and skip one. Position 1 (the 2) partners position 3 (the 1). Position 2 (the 5) partners position 4 (the 4). So the face opposite the 5 is 4. The two squares poking out as top and bottom are the cube’s last opposite pair.
THE TRICK

For a ‘does this fold into a cube?’ question, pick any square as the floor and fold the rest up one at a time, ticking off the 6 faces as you cover them: floor, ceiling, and four walls. The instant two squares fight for the same face, that shape is out. For a ‘which one does NOT fold?’ question you only need to catch the single clash.

WATCH OUT

Counting 6 squares and declaring ‘cube net, done’. A plus-sign of 5 squares with one stuck wrong, an L of 6 in the wrong bend — both have 6 squares and both fail. Always fold-check.

Bogus solution

In an L-shaped net, two faces sit at the far ends, pointing away from each other — one at the top of the column, one at the right of the row. They point opposite directions on the flat page, so they must be the opposite faces of the cube.

Why it breaks: ‘points opposite ways on the paper’ is not the same as ‘opposite on the cube’ — once you fold, those two end squares swing up and meet along an edge, so they are actually neighbours. Two faces that share an edge can never be opposite.

The fix: ignore how the flat squares point. Walk a straight run and skip one: that is the only way to read true opposites. If there is no straight run of three through a pair, fold-check by hand.

WORKED EXAMPLE
PROBLEM · 2015 #4

The diagram shows the net of a cube whose faces are numbered. Sascha adds the numbers that are on opposite faces of the cube. Which three results does he get?

Figure for Math Kangaroo 2015 Problem 4
A) 4, 6, 11 B) 4, 5, 12 C) 5, 6, 10 D) 5, 7, 9 E) 5, 8, 8

The net’s faces are numbered. Sascha folds it into a cube and adds each pair of opposite faces. Which three sums does he get? Do it without folding — use the skip-one rule.

Find the straight band first. Faces 1, 2, 3, 4 sit in a row, so they belt the cube: 1 is opposite 3, and 2 is opposite 4. That uses four faces. The leftover squares 5 and 6 have to be the top and bottom, so 5 is opposite 6. Now add the pairs: 1 + 3 = 4, 2 + 4 = 6, 5 + 6 = 11. The three sums are 4, 6, 11 — choice A. You never folded anything; you read the partners straight off the row.

Answer: A — 4, 6, 11
RULE OF THUMB

6 squares is necessary but not enough — always fold-check. There are exactly 11 different cube nets. To find opposite faces, skip one along a straight row.

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2015 · #7 The diagram shows the net of a three-sided prism. Which line of the diagram forms an edge of the prism together with line UV when the...

The diagram shows the net of a three-sided prism. Which line of the diagram forms an edge of the prism together with line UV when the net is folded up?

Figure for Math Kangaroo 2015 Problem 7
Show answer
Answer: CXY
Show hints
Hint 1 of 2
Fold the net into the prism and watch where the endpoints of UV land.
Still stuck? Show hint 2 →
Hint 2 of 2
When folded, the edge along UV meets another edge of the net; find which labelled segment touches it.
Show solution
Approach: fold the net and see which edge meets UV
  1. The net of the triangular prism wraps the rectangular faces around the two triangular ends.
  2. When it is folded up, segment UV is brought together with the segment that shares its endpoints after folding.
  3. Tracing the fold, that segment is XY.
  4. So the answer is XY (C).
2015 · #21 Nina wants to make a cube from the paper net. You can see there are 7 squares instead of 6. Which square(s) can she remove from the net,...

Nina wants to make a cube from the paper net. You can see there are 7 squares instead of 6. Which square(s) can she remove from the net, so that the other 6 squares remain connected and from the newly formed net a cube can be made?

Figure for Math Kangaroo 2015 Problem 21
Show answer
Answer: D — only 3 or 7
Show hints
Hint 1 of 2
A cube net needs the remaining 6 squares to stay connected AND to fold up without two squares landing on the same face.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each candidate removal: most leave a shape that overlaps when folded; only certain end squares work.
Show solution
Approach: test which removals leave a connected, foldable 6-square net
  1. Removing a square must keep the other six joined and able to fold into a cube with no doubled-up face.
  2. Taking out an interior square breaks the net or makes two squares fold onto the same face, so those fail.
  3. Removing square 3 works, and removing square 7 works, while no other single removal does — so the answer is only 3 or 7.
2017 · #6 (figure problem)
Figure for Math Kangaroo 2017 Problem 6
Show answer
Answer: D
Show hints
Hint 1 of 2
Each fold doubles the holes by reflecting them across the fold line.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the fold line that maps the two visible holes onto each other.
Show solution
Approach: reverse the fold by reflection
  1. Unfolding mirrors each punched hole across every fold line, so before unfolding the holes must be symmetric about the fold.
  2. Among the choices, only one dotted line is an axis of symmetry for the two holes shown.
  3. That fold line is choice D.
2011 · #20 The dark line halves the surface area of the die shown on the right. Which of the drawings A–E could represent the net of this die?

The dark line halves the surface area of the die shown on the right. Which of the drawings A–E could represent the net of this die?

Figure for Math Kangaroo 2011 Problem 20
Show answer
Answer: A
Show hints
Hint 1 of 2
Imagine folding each net back into the cube and follow where the drawn line goes.
Still stuck? Show hint 2 →
Hint 2 of 2
The correct net is the one whose line becomes the single curve that halves the cube surface.
Show solution
Approach: fold each net and check the dark line splits the cube's surface in half
  1. The dark line on the die separates its surface into two equal-area parts.
  2. When a net is folded into the cube, that same line must close up into one continuous halving curve.
  3. Only net A folds so the marked line cleanly divides the surface into two equal halves.
CHAPTER 2

Reading a 3D shape — count by layers

THEORY

A drawing of a 3D pile is a liar. It shows you the front and the top and hides everything behind and underneath. Try to count the whole pile at once and you will always miss the cubes in the back. So do not count the pile. Count slices.

Cut the shape into flat layers — floor layer, next layer up, next — and count one easy 2D slice at a time. Write each layer’s number down. A solid n×n×n cube is n identical layers of n×n cubes, which is why it holds n³ small cubes: a 3×3×3 is three layers of 9, so 27.

top floor: 9middle: 9floor: 99 + 9 + 9 = 27

See why the back cubes hide? On the page you only meet the front edge of each floor — the other eight cubes of every floor sit behind it. Counting floor-by-floor catches all nine each time; eyeballing the drawing never does.

Front, top & side views

Kangaroo loves showing a stack and asking what it looks like from the front, the top, or the side — or running it backwards: here are the three flat views, how many cubes are there?

One idea unlocks all of them. A view records the tallest stack along each line you look down. From the front you see the highest cube in each column. From the top you see the footprint — which floor squares are covered at all. From the side, the highest cube in each row.

the stack (front)top view (footprint)

To rebuild a cube count from views, lay the footprint out and raise each column to the tallest height the front and side views allow. If the problem asks for the maximum, push every column as high as it is permitted; for the minimum, raise only as much as the views force.

THE MOVE: slice into layers and count flat. Each view is the tallest cube along your line of sight.

🎯 Try it
You are building a solid 3×3×3 cube. So far you have stacked the bottom two full layers, plus 1 lonely cube on the top layer. How many more small cubes do you still need to finish it?
Walkthrough: a full 3×3×3 needs 3 × 3 × 3 = 27 cubes. You have two full layers of 9 (that is 18), plus 1 more, so 19 are in place. Full minus placed: 27 − 19 = 8. (Same as: the top layer needs 9 and you have 1, so 8 are still missing.)
THE TRICK

Two clean tactics. (1) Subtract from full: for ‘how many more to finish the big cube?’, take the finished cube’s total minus the cubes already placed. (2) Count by layers: tally each slice and add. Both beat eyeballing every time.

WATCH OUT

Counting only the cubes you can see and forgetting the hidden ones at the back and bottom. Before you write a number, ask: what is sitting behind this front face that I cannot see?

Bogus solution

A drawing shows a stack of small cubes. I count carefully around the picture — every cube whose face I can see — and reach 14, so there are 14 cubes.

Why it breaks: the drawing only ever shows the front, top, and one side; a whole column of cubes can hide directly behind a front cube and never show a single face. Counting visible faces always lands too low.

The fix: count by layers, floor by floor, including the cubes you cannot see, and write each layer’s total down. Or, when the shape is heading toward a full block, take the full total and subtract the gap.

WORKED EXAMPLE
PROBLEM · 2013 #2

Nathalie wants to build a large cube out of small cubes (the complete cube is shown on the left). How many small cubes are missing from the shape on the right so that it would form the large cube?

Figure for Math Kangaroo 2013 Problem 2
A) 5 B) 6 C) 7 D) 8 E) 9

Nathalie wants to build a big cube. The picture shows how far she has gotten. How many small cubes are still missing? Use the subtract-from-full move.

The finished cube is 3 wide, 3 tall, 3 deep, so it needs 3 × 3 × 3 = 27 small cubes. Now count what is already there — and do it by layers, not by eye, so the hidden back cubes are not lost. Layer by layer the built-up shape holds 20 cubes. Missing = 27 − 20 = 7, choice C. Notice you never had to picture the gap as a 3D blob; you counted a full cube and subtracted a layered count.

Check it a second way. Instead of subtracting, count the gap straight on: go layer by layer and tally only the empty spots in each, and they add to 7 as well. Two different roads to the same number is the surest sign you have it right — a spatial answer you reach two ways is one you can trust. Make that your habit on every cube-count.

Framing inspired by AoPS Prealgebra.

Answer: C — 7
RULE OF THUMB

Layers turn a scary 3D count into a few easy 2D counts. Write each layer down so you cannot lose one. For ‘how many more’: full total − placed.

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2009 · #6 How many faces does the object shown have? (It is a prism with a hole through it.)

How many faces does the object shown have? (It is a prism with a hole through it.)

Figure for Math Kangaroo 2009 Problem 6
Show answer
Answer: D — 8
Show hints
Hint 1 of 2
Count the outside surfaces first, then remember the hole adds new inside surfaces too.
Still stuck? Show hint 2 →
Hint 2 of 2
A triangular tube has 3 inner walls plus 3 outer walls, and the two ends are still faces.
Show solution
Approach: count outer and inner faces
  1. The two triangular ends are now frames, but each is still one face: 2 faces.
  2. The three outer rectangular sides: 3 faces.
  3. Drilling the triangular hole creates three inner rectangular walls: 3 more faces.
  4. Total = 2 + 3 + 3 = 8 faces — answer D.
2023 · #4 Nine steps of a staircase winding around a cylinder can be seen, starting at the bottom and leading all the way to the top. All the...

Nine steps of a staircase winding around a cylinder can be seen, starting at the bottom and leading all the way to the top. All the steps are equally high. How many steps cannot be seen?

Figure for Math Kangaroo 2023 Problem 4
Show answer
Answer: D — 12
Show hints
Hint 1 of 3
From the front you only see the steps facing you; the rest of the staircase keeps winding around the hidden back of the cylinder.
Still stuck? Show hint 2 →
Hint 2 of 3
The picture shows the front-facing steps spiralling up; figure out how high the whole tower climbs, then take away the nine you can already see.
Still stuck? Show hint 3 →
Hint 3 of 3
Each level of the spiral has steps on both the front and the back, so the hidden back steps roughly mirror the visible front ones, plus a few more for the extra height.
Show solution
Approach: see that the spiral has front and back steps at each level, then count the hidden ones
  1. The nine steps you can see are the ones facing you as the staircase spirals up the front of the cylinder.
  2. As the spiral turns, the same number of steps run around the hidden back at each level, and the tower keeps climbing past where the front steps stop.
  3. Counting the back steps level by level all the way to the top gives twelve steps that are turned away and cannot be seen.
  4. So the number that cannot be seen is D, 12.
2022 · #21 A building is made of cubes of the same size. The three pictures show it from above (von oben), from the front (von vorne) and from the...

A building is made of cubes of the same size. The three pictures show it from above (von oben), from the front (von vorne) and from the right (von rechts). What is the maximum number of cubes that could be used to make this building?

Figure for Math Kangaroo 2022 Problem 21
Show answer
Answer: B — 19
Show hints
Hint 1 of 2
The top view fixes which columns can hold cubes; the front and side views cap each column's height.
Still stuck? Show hint 2 →
Hint 2 of 2
For the maximum, make every column as tall as its views allow.
Show solution
Approach: raise each column to the height its views permit
  1. The top view shows which floor positions are occupied.
  2. The front and right views give the largest height allowed for each row and column.
  3. Stacking each column to its maximum allowed height totals 19 cubes.
  4. So the answer is B.
2020 · #22 John built a structure of equal-sized wooden cubes whose front, right-side and top views are shown, using as many cubes as possible. His...

John built a structure of equal-sized wooden cubes whose front, right-side and top views are shown, using as many cubes as possible. His sister Ana wants to remove as many cubes as she can without changing any of these three views. At most, how many cubes can she remove?

Figure for Math Kangaroo 2020 Problem 22
Show answer
Answer: B — 12
Show hints
Hint 1 of 2
The three views (front, side, above) must all stay the same after removing cubes.
Still stuck? Show hint 2 →
Hint 2 of 2
Keep only the cubes forced by all three silhouettes; count how many of the fullest build can be taken away.
Show solution
Approach: compare the fullest build with the minimum that keeps all views
  1. Build the most cubes giving those three views, then strip out any cube not needed by all three silhouettes.
  2. Each removed cube must leave the front, side and top outlines unchanged.
  3. The largest number she can remove while preserving every view is 12.
★ MINI-QUIZ

Fold & count check

One net, one cube count. Fold in your head with the skip-one rule, then count by layers.

2015 · #4 The diagram shows the net of a cube whose faces are numbered. Sascha adds the numbers that are on opposite faces of the cube. Which...

The diagram shows the net of a cube whose faces are numbered. Sascha adds the numbers that are on opposite faces of the cube. Which three results does he get?

Figure for Math Kangaroo 2015 Problem 4
Show answer
Answer: A — 4, 6, 11
Show hints
Hint 1 of 2
Fold the net into a cube in your head and see which faces end up opposite each other.
Still stuck? Show hint 2 →
Hint 2 of 2
On a band of four faces in a row, opposite faces skip one; then pair the two faces sticking out.
Show solution
Approach: identify the three pairs of opposite faces
  1. Faces 1, 2, 3, 4 form a band around the cube, so 1 is opposite 3 and 2 is opposite 4.
  2. The remaining faces 5 and 6 are top and bottom, so 5 is opposite 6.
  3. The three opposite-face sums are 1+3 = 4, 2+4 = 6, and 5+6 = 11.
  4. So Sascha gets 4, 6, 11 (A).
2009 · #6 How many faces does the object shown have? (It is a prism with a hole through it.)

How many faces does the object shown have? (It is a prism with a hole through it.)

Figure for Math Kangaroo 2009 Problem 6
Show answer
Answer: D — 8
Show hints
Hint 1 of 2
Count the outside surfaces first, then remember the hole adds new inside surfaces too.
Still stuck? Show hint 2 →
Hint 2 of 2
A triangular tube has 3 inner walls plus 3 outer walls, and the two ends are still faces.
Show solution
Approach: count outer and inner faces
  1. The two triangular ends are now frames, but each is still one face: 2 faces.
  2. The three outer rectangular sides: 3 faces.
  3. Drilling the triangular hole creates three inner rectangular walls: 3 more faces.
  4. Total = 2 + 3 + 3 = 8 faces — answer D.
CHAPTER 3

Tiling & tessellation — cover with no gaps

THEORY

Look at a brick wall, a bathroom floor, a honeycomb. Copies of one shape, fitted edge to edge, with no gaps and no overlaps. That is a tiling, and Kangaroo turns it into two questions: can this shape cover the region, and what is the fewest (or most) pieces that do it.

Reach for area first — it is free. If the region has area A and each tile covers area t, then you need at least A ÷ t tiles, and never more than that with a single tile size. That one division kills half the answer choices before you place a single piece.

area 24, each tile area 4 → need 24 ÷ 4 = 6 tiles

Area gives the ceiling. Then you must check the shape actually reaches it, because thin arms, odd corners, or a parity clash can stop you one short. A quick checkerboard colouring is the classic tie-breaker: if every tile must cover an equal number of black and white squares but the region has unequal black and white, a perfect cover is impossible.

THE MOVE: divide areas for the count, then confirm a real arrangement (or a colouring argument) reaches it.

🎯 Try it
You want to cover a 4×6 rectangle exactly with L-shaped pieces that each cover 4 squares. What is the most pieces that could possibly fit (the area ceiling)?
Walkthrough: the rectangle holds 4 × 6 = 24 unit squares. Each L covers 4 squares, so at most 24 ÷ 4 = 6 pieces. (That is only the ceiling — you would still sketch one real layout to be sure 6 truly fit, and L-tetrominoes do tile a 4×6.)
THE TRICK

To use the fewest tiles, cover the big open stretches with your biggest tiles first, then mop up the awkward tips and corners with the small ones — those forced small pieces set the real limit. To use the most, do the reverse.

WATCH OUT

Trusting the area ceiling and stopping. Area says ‘at most this many’; it never promises the pieces fit. A shape can fail the cover even when the areas divide perfectly. Always sketch one real placement.

WORKED EXAMPLE
PROBLEM · 2025 #22

Julio wants to make the shape shown in the top picture on the right. He has several of each of the five tiles shown in the bottom picture on the right. The tiles must be placed next to each other without overlapping. What is the smallest number of tiles he must use?

Figure for Math Kangaroo 2025 Problem 22
A) 11 B) 12 C) 13 D) 15 E) 17

Julio builds a plus-shaped cross out of five tile types and wants the fewest tiles. What is the smallest number he can use? Big tiles for the body, small ones forced at the tips.

Fewest tiles means each tile should swallow as much area as it can, so fill the fat straight body of the cross with the biggest tiles — the long rectangle and the large triangle. Now the four slanted arm-tips are too thin and pointy for any big tile, so each tip has to be finished with the small triangles. Those small tips are unavoidable, and they are what stop the count from dropping further. Pack the body big and the tips small, no overlaps, and the whole cross is covered in 13 tiles — choice C. No arrangement does it in fewer, because the tips force those last small pieces.

Answer: C — 13
RULE OF THUMB

Area gives the ceiling; a drawing or a checkerboard colouring confirms whether you reach it. Big tiles on open space, small tiles forced at the tips.

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2017 · #3 Anna has four identical building blocks that each look like the one shown (a straight strip of three squares). Which of the shapes in...

Anna has four identical building blocks that each look like the one shown (a straight strip of three squares). Which of the shapes in the options can she not form with them?

Figure for Math Kangaroo 2017 Problem 3
Show answer
Answer: E
Show hints
Hint 1 of 2
Each block covers three squares in a straight line; four of them cover twelve squares.
Still stuck? Show hint 2 →
Hint 2 of 2
A shape can be built only if it can be cut into straight 1×3 pieces — try tiling each one.
Show solution
Approach: tile each shape with straight triominoes
  1. The block is a straight strip of three squares, so four blocks cover 12 squares total.
  2. Each pictured shape has 12 squares, so the test is whether it splits into four straight 1×3 strips.
  3. Four of the shapes can be cut into such strips; the remaining one cannot be tiled this way.
  4. That shape is (E).
2013 · #4 Melanie has a square piece of paper with a 4×4 grid drawn on it. She cuts along the gridlines, cutting out several shapes that each look...

Melanie has a square piece of paper with a 4×4 grid drawn on it. She cuts along the gridlines, cutting out several shapes that each look like the one pictured or its mirror image. How many squares are left over if she cuts out as many shapes as possible?

Figure for Math Kangaroo 2013 Problem 4
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
The shape is a 4-square piece; the grid holds 16 squares in total.
Still stuck? Show hint 2 →
Hint 2 of 2
Try to fit as many copies (or mirror images) as you can without overlap, then count the leftovers.
Show solution
Approach: tile and count leftovers
  1. The 4×4 grid has 16 unit squares; each cut-out piece uses 4 of them.
  2. These S/Z-shaped pieces cannot fill the 4×4 square completely.
  3. The best packing fits 3 pieces (12 squares), leaving 4 squares uncovered.
  4. So 4 squares are left over.
2024 · #24 Tiler Teri wants to cover a square floor with a regular pattern (see diagram) using six-sided and three-sided tiles. She estimates that...

Tiler Teri wants to cover a square floor with a regular pattern (see diagram) using six-sided and three-sided tiles. She estimates that she will need about 3000 six-sided tiles for the whole floor. About how many three-sided tiles will she need?

Figure for Math Kangaroo 2024 Problem 24
Show answer
Answer: D — 6000
Show hints
Hint 1 of 2
Focus on one hexagon and count how many triangles touch it, then notice each triangle is shared between hexagons.
Still stuck? Show hint 2 →
Hint 2 of 2
Pick out the small repeating block of the pattern and count hexagons versus triangles inside it to get the fixed ratio.
Show solution
Approach: use the fixed triangle-to-hexagon ratio of the repeating pattern
  1. In this regular pattern, six triangles ring each hexagon, but every triangle is shared by three hexagons, so each hexagon effectively owns 6 ÷ 3 = 2 triangles.
  2. That means the tiling always has 2 triangles for every hexagon.
  3. With about 3000 hexagons, the number of triangles is about 2 × 3000 = 6000.
CHAPTER 4

Mirrors, reflections & rotations

THEORY

Hold a word up to a bathroom mirror. AMBULANCE on the front of a van comes out backwards — which is why real ambulances print it reversed, so it reads right in your rear mirror. That is a reflection: left and right swap across a line. A rotation is different — the whole figure turns around a point, like a clock hand.

Both go wrong the same way: you try to flip or spin the entire shape in your head and lose your place. Do not. Track one feature. Pick a single corner, one arrow, one beam, and ask only where it lands. Across a mirror, a point hops to the other side, the same distance from the line. The rest of the figure obeys.

RRmirror line: each point hops across, same distance

Two facts handle most of these. The back view of something is its front view flipped left-to-right — so reading a wall from behind reverses it. And a beam that hits a mirror bounces off at an equal angle: trace it one straight segment at a time.

Reflect to straighten a bent path

Here is reflection used as a tool, not a question. A frog at A must hop down to touch a straight bank, then up to B. What route is shortest? The bent two-leg path is hard to measure — so get rid of the bend. Mirror B across the bank to a copy B′ on the far side. Every down-then-up route to B is exactly as long as the matching straight-across route to B′ (the second leg is just flipped over the line). The shortest of those is the dead-straight line A→B′, and where it crosses the bank is the touch point.

ABB′touch here

The same straighten-the-bend idea reads a bouncing beam or a fold: replace the reflected leg with its mirror image and the whole path snaps into one straight line you can measure.

Strategy framing inspired by Posamentier, The Art of Problem Solving (teacher resource).

🎯 Try it
Point A sits 2 units above a straight line; point B sits 4 units above it, 8 units across from A. Touch the line once and go from A to B. Reflect B to B′ (4 units below the line) and measure the straight line A–B′: what is the shortest total distance?
Walkthrough: after reflecting B down to B′, the straight line A–B′ is the hypotenuse of a right triangle. Its horizontal leg is the 8 across; its vertical leg is how far A is above B′, which is 2 + 4 = 6. So the distance is \(\sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10\). The bent path is straightened into a clean 6–8–10 right triangle.

Why a turn is not a flip

This is the mix-up that costs the most points. A rotation keeps the shape itself unchanged — same handedness, only pointed a new way, the way your right hand is still a right hand after you wave it around. A reflection flips handedness — your right hand’s mirror image is a left hand, and no amount of turning will ever make them match. Watch one letter prove it:

bstartturn(rotate 180°)qflip(mirror)d

Same letter, two fates: turn the b and you get q; mirror it and you get d. They land in different places because a turn and a flip are genuinely different moves. When a problem says turn / rotate, spin it; when it says flip / mirror / back view, reverse it.

For symmetry questions (does this picture have a mirror line?), test the pieces, not the whole. A stacked word is vertically symmetric only if every single letter is. One bad letter kills it.

THE MOVE: follow one point, arrow, or beam through the flip or turn — never the whole figure.

🎯 Try it
In the word BRAINY, how many of the six letters look the same in a vertical mirror (a mirror line running straight up and down)?
Walkthrough: check each letter against a vertical mirror, one at a time. B — no. R — no. A — yes. I — yes. N — no (it mirrors to a backwards N). Y — yes. The ones that pass are A, I, Y, so the count is 3.
THE TRICK

For a reflection, send your tracked point straight across the mirror line to the same distance on the far side. For a rotation, fix the centre and turn that one feature alone by the given angle. For a bouncing beam, redraw each reflected segment at an equal angle and walk it to the edge.

WATCH OUT

Confusing a reflection with a rotation. A flipped b becomes a d (mirror); a turned b becomes a q (rotation). They are not the same picture. Read the problem: ‘flip / mirror / back view’ means reflect; ‘turn / rotate’ means rotate.

WORKED EXAMPLE
PROBLEM · 2018 #2

If the letters of the word MAMA are written one underneath another, the word has a vertical axis of symmetry. For which of these words is that also true?

A) ADAM B) BAUM C) BOOT D) LOGO E) TOTO

Write a word with its letters stacked in a column, like MAMA. It has a vertical mirror line. For which of ADAM, BAUM, BOOT, LOGO, TOTO is that also true? Test one letter at a time.

A stacked word has a vertical mirror line only if every letter is symmetric left-to-right on its own. So scan each word for a single bad letter and you can throw it out instantly. ADAM has D — D is not vertically symmetric, out. BAUM and BOOT both start with B — B fails the vertical mirror, out. LOGO has L and G, both lopsided, out. That leaves TOTO: T is symmetric across a vertical line, O is symmetric, T again, O again — every letter passes. So the answer is TOTO, choice E. You never pictured the whole word in a mirror; you checked letters one by one and the first failure ended each candidate.

Answer: E — TOTO
RULE OF THUMB

Reflection = left/right flip across a line; rotation = turn about a point. Back view = front flipped. Track one point or one beam, not the whole figure.

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2016 · #1 Which of the following road signs has the most axes of symmetry?

Which of the following road signs has the most axes of symmetry?

Figure for Math Kangaroo 2016 Problem 1
Show answer
Answer: C — The no-entry sign.
Show hints
Hint 1 of 3
Imagine folding each sign along a straight line so the two halves land exactly on top of each other.
Still stuck? Show hint 2 →
Hint 2 of 3
Try both a left-right fold and a top-bottom fold on every sign, then count how many folds work.
Still stuck? Show hint 3 →
Hint 3 of 3
A plain horizontal bar inside a circle matches itself for both folds.
Show solution
Approach: fold each sign and count the lines that match
  1. An axis of symmetry is a fold line where one half lands perfectly on the other half.
  2. The arrow signs match only one fold (or none, once an arrowhead points a direction), and the car shape matches just its up-down fold.
  3. The no-entry sign (a horizontal bar in a circle) matches a left-right fold AND a top-bottom fold, so it has 2 folds.
  4. Two is the most of any sign, so the answer is the no-entry sign, choice (C).
2020 · #5 Flipping a card over its top edge, we see the photo of the kangaroo shown. If instead we flip the card over its right edge, what will appear?

Flipping a card over its top edge, we see the photo of the kangaroo shown. If instead we flip the card over its right edge, what will appear?

Figure for Math Kangaroo 2020 Problem 5
Show answer
Answer: D
Show hints
Hint 1 of 2
Flipping over the right edge is a mirror across a vertical line, not the same as flipping over the top.
Still stuck? Show hint 2 →
Hint 2 of 2
A right-edge flip mirrors the picture left-to-right; work out that orientation.
Show solution
Approach: apply a flip about the right edge (a horizontal mirror)
  1. Flipping the card over its top edge gives the shown kangaroo, a flip about a horizontal line.
  2. Flipping instead over the right edge is a flip about a vertical line, the left-to-right mirror of the original.
  3. Carrying out that vertical flip on the starting picture gives the kangaroo in option D.
2022 · #3 The two-sided mirrors reflect the laser beam as shown in the small picture on the left. At which letter does the laser beam leave the...

The two-sided mirrors reflect the laser beam as shown in the small picture on the left. At which letter does the laser beam leave the picture on the right?

Figure for Math Kangaroo 2022 Problem 3
Show answer
Answer: B — B
Show hints
Hint 1 of 2
Each diagonal mirror turns the beam by a right angle; the small example shows which way.
Still stuck? Show hint 2 →
Hint 2 of 2
Step the beam square by square, bouncing 90 degrees at every mirror, until it reaches an edge.
Show solution
Approach: trace the beam, reflecting 90 degrees at each mirror
  1. Use the small picture to learn how each slanted mirror deflects the beam.
  2. Starting from the entry arrow on the big grid, advance the beam and turn it a quarter-turn at every mirror it hits.
  3. Following the bounces, the beam leaves the grid at letter B.
CHAPTER 5

Dice & labelled cubes — use the hidden pairing

THEORY

Pick up a real die and add the dots on any two opposite faces. You get 7 every time: 1 across from 6, 2 across from 5, 3 across from 4. Three pairs, all summing to 7. Memorise them and a whole family of problems cracks open.

1+6=72+5=73+4=7

Two facts ride along with it. Faces that share an edge are neighbours, never opposite — so to find a face’s partner, list everything that touches it and the one number left over is the opposite. And all six faces add to 1+2+3+4+5+6 = 21, which is your budget for any ‘total of the visible faces’ problem: start from 21 per die and subtract whatever is hidden or glued away.

Beyond dice: numbered cubes

Some problems put your own numbers on a cube and ask about corner sums (add the three faces meeting at a corner). The quiet gift here: two corners at opposite ends of a space diagonal together touch all six faces once, so their corner sums always add to the same total. That lets you pair a known corner with an unknown one and solve in one line.

THE MOVE: opposite faces sum to 7, all six sum to 21, neighbours are never opposite. For corner sums, pair opposite corners.

🎯 Try it
A normal die sits on a table showing 2 on top. Add up the four side faces you can walk around (not the top, not the bottom). What is their total?
Walkthrough: all six faces total 21. The top is 2, so the bottom is its partner, 7 − 2 = 5. Top and bottom together take away 2 + 5 = 7. The four sides are what is left: 21 − 7 = 14. You never needed to know which side faces show which numbers.
THE TRICK

For glued-dice or stacked-cube totals, work from 21 per cube and subtract the faces pressed together or hidden. To make the visible total small, hide the big numbers; to make it large, hide the small ones. For a die rolling across a grid, forget the whole cube — track only the bottom face as it tips one square at a time.

WATCH OUT

Assuming any cube with dots is a real die. A problem may print an impossible die to bait you. Check it: opposite faces must total 7, and the three faces around a corner must match a genuine die’s arrangement. If not, reject that picture.

WORKED EXAMPLE
PROBLEM · 2015 #24

Maria writes a number on each face of the cube. Then, for each corner point of the cube, she adds the numbers on the faces which meet at that corner. (For corner B she adds the numbers on faces BCDA, BAEF and BFGC.) In this way she gets a total of 14 for corner C, 16 for corner D, and 24 for corner E. Which total does she get for corner F?

Figure for Math Kangaroo 2015 Problem 24
A) 15 B) 19 C) 22 D) 24 E) 26

Maria writes a number on each face of a cube, then for each corner adds the three faces meeting there. She gets 14 at corner C, 16 at D, 24 at E. What does corner F get? Pair opposite corners.

Do not chase the six face values — you do not need them. A corner’s number is the sum of its three faces. Two corners at opposite ends of a space diagonal together cover all six faces exactly once, so any such pair adds to the same total S (the sum of all six faces). On this cube, C and E are an opposite pair, and D and F are the other, so C + E = D + F. Plug in: 14 + 24 = 16 + F, so 38 = 16 + F, giving F = 22, choice C. One pairing, one equation, done.

Answer: C — 22
RULE OF THUMB

Opposite faces sum to 7; all six sum to 21; touching faces are never opposite. To roll a die, follow only the bottom face. For corner sums, opposite corners share the same total.

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2019 · #5 (figure problem)
Figure for Math Kangaroo 2019 Problem 5
Show answer
Answer: E
Show hints
Hint 1 of 3
On a real die, opposite faces always add to 7 (1-6, 2-5, 3-4).
Still stuck? Show hint 2 →
Hint 2 of 3
Also, the faces 1, 2 and 3 meet at one corner in a fixed turning order on every standard die.
Still stuck? Show hint 3 →
Hint 3 of 3
Check each picture: its three visible faces must be able to sit around one corner of a standard die.
Show solution
Approach: check the visible faces against a standard die
  1. On an ordinary die opposite faces sum to 7 (1-6, 2-5, 3-4), and the three faces around one corner follow a fixed orientation.
  2. Reject any picture whose three visible faces could not all sit around one corner of a standard die.
  3. Only picture E shows three faces consistent with a genuine ordinary die.
2011 · #20 The dark line halves the surface area of the die shown on the right. Which of the drawings A–E could represent the net of this die?

The dark line halves the surface area of the die shown on the right. Which of the drawings A–E could represent the net of this die?

Figure for Math Kangaroo 2011 Problem 20
Show answer
Answer: A
Show hints
Hint 1 of 2
Imagine folding each net back into the cube and follow where the drawn line goes.
Still stuck? Show hint 2 →
Hint 2 of 2
The correct net is the one whose line becomes the single curve that halves the cube surface.
Show solution
Approach: fold each net and check the dark line splits the cube's surface in half
  1. The dark line on the die separates its surface into two equal-area parts.
  2. When a net is folded into the cube, that same line must close up into one continuous halving curve.
  3. Only net A folds so the marked line cleanly divides the surface into two equal halves.
★ MINI-QUIZ

Mirror & dice check

A reflection and a die. Track one feature; use opposite-faces-sum-to-7.

2020 · #5 Flipping a card over its top edge, we see the photo of the kangaroo shown. If instead we flip the card over its right edge, what will appear?

Flipping a card over its top edge, we see the photo of the kangaroo shown. If instead we flip the card over its right edge, what will appear?

Figure for Math Kangaroo 2020 Problem 5
Show answer
Answer: D
Show hints
Hint 1 of 2
Flipping over the right edge is a mirror across a vertical line, not the same as flipping over the top.
Still stuck? Show hint 2 →
Hint 2 of 2
A right-edge flip mirrors the picture left-to-right; work out that orientation.
Show solution
Approach: apply a flip about the right edge (a horizontal mirror)
  1. Flipping the card over its top edge gives the shown kangaroo, a flip about a horizontal line.
  2. Flipping instead over the right edge is a flip about a vertical line, the left-to-right mirror of the original.
  3. Carrying out that vertical flip on the starting picture gives the kangaroo in option D.
2019 · #5 (figure problem)
Figure for Math Kangaroo 2019 Problem 5
Show answer
Answer: E
Show hints
Hint 1 of 3
On a real die, opposite faces always add to 7 (1-6, 2-5, 3-4).
Still stuck? Show hint 2 →
Hint 2 of 3
Also, the faces 1, 2 and 3 meet at one corner in a fixed turning order on every standard die.
Still stuck? Show hint 3 →
Hint 3 of 3
Check each picture: its three visible faces must be able to sit around one corner of a standard die.
Show solution
Approach: check the visible faces against a standard die
  1. On an ordinary die opposite faces sum to 7 (1-6, 2-5, 3-4), and the three faces around one corner follow a fixed orientation.
  2. Reject any picture whose three visible faces could not all sit around one corner of a standard die.
  3. Only picture E shows three faces consistent with a genuine ordinary die.
CHAPTER 6

Building & cutting solids — count what changes

THEORY

Hand a kid a few odd pieces and ask which ones snap together into a clean shape, and the instinct is to shove pieces around at random until something looks right. Resist it. There is a calmer way in, and it starts with counting before fitting.

Whether you are joining pieces or slicing a solid, ask first: what number must stay fixed, and what changes? If three blocks combine into a cube building, the building must have exactly as many little cubes as the three blocks together — count cubes and cross out every choice with the wrong total before you fit anything. If you cut or drill a solid, count how each cut adds or removes faces, edges, and corners.

1 corner……cut → 2 new corners here

When you do fit pieces, match a bump to a notch: a piece’s stick-out has to drop into another piece’s matching dent, no gap and no overlap. Find the bump and the dent it fills, slot them, and trace the outside edge — if it comes out clean, that is your pair.

For face counts, do not forget the surfaces a hole or tunnel creates inside. A block with a tube drilled through it gains the inner walls of that tube as brand-new faces.

THE MOVE: count the fixed quantity first (cubes, area, corners), then fit by matching bump to notch.

🎯 Try it
You slice off all 8 corners of a cube. Each cut opens a small triangular face where a corner used to be. How many corners does the new solid have?
Walkthrough: a cube has 8 corners. Each cut deletes one old corner but creates a triangle, and a triangle has 3 corners. So every cut turns 1 corner into 3 new ones. With 8 corners cut: 8 × 3 = 24 corners. (Same move as the tetrahedron worked example, now starting from 8 corners instead of 4.)

One more ‘count what changes’ trap, away from solids: a straight log gets cuts, and each cut adds one more piece — so pieces are always one ahead of cuts. Sketch it and you see why the count is off by one.

🎯 Try it
It takes a saw 4 minutes to cut a log into 3 pieces (each cut takes the same time). At that rate, how many minutes to cut a log of the same length into 4 pieces?
Walkthrough: count cuts, not pieces. Three pieces take only 2 cuts (sketch it: two slices make three chunks). So 4 minutes ÷ 2 cuts = 2 minutes per cut. Four pieces need 3 cuts: 3 × 2 = 6 minutes. The trap is reading ‘3 pieces’ as 3 cuts — pieces are always one more than cuts.

Framing inspired by AoPS Prealgebra.

THE TRICK

Cube count is your fast filter: total the little cubes in the given pieces and reject any target with a different count. For truncations and drillings, track corners and faces one cut at a time: slicing off a corner removes that single vertex but adds the small new face’s corners in its place.

WATCH OUT

Forgetting the faces or corners a cut creates. Slicing a corner does not only delete a point — it opens a new little face with new corners of its own. Counting only what you removed gives the wrong total.

Bogus solution

A tetrahedron has 4 corners. We slice off all 4 corners. Each cut removes one corner, so 4 corners are gone and the new solid has 4 − 4 = 0 corners.

Why it breaks: it only counts what each cut takes away and never what the cut opens up. Every slice replaces the old point with a small triangular face — and that triangle brings 3 brand-new corners.

The fix: sketch one cut and watch the new triangle appear — three faces met at that vertex, so the flat cut is a triangle with 3 corners. Each of the 4 cuts trades 1 corner for 3, so 4 × 3 = 12 corners. Count the corners and faces a cut creates, not only the ones it deletes.

WORKED EXAMPLE
PROBLEM · 2024 #4

Julio cuts off all four corners of a regular tetrahedron (see picture). How many corners does the object have now?

Figure for Math Kangaroo 2024 Problem 4
A) 8 B) 9 C) 11 D) 12 E) 15

Julio cuts off all four corners of a regular tetrahedron. How many corners does the new solid have? Track what each cut creates.

Start from what is fixed: a regular tetrahedron has 4 corners. Now follow one cut. Slicing off a single corner deletes that one vertex, but the flat cut opens a small triangular face right where the point was — and a triangle has 3 corners. So one corner gone is replaced by 3 new ones. The old vertices all vanish; only the new triangle-corners survive. Four corners cut, each giving 3 new corners: 4 × 3 = 12 corners, choice D. The whole problem is ‘what does each cut create?’ not ‘what does it remove?’.

Answer: D — 12
RULE OF THUMB

Count the fixed quantity first (cubes / area / corners). When you cut a solid, count the faces and corners the cut creates, not only what it takes away.

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2025 · #23 Tina wants to combine the three building blocks shown in the picture to form a cube building. Which one of the following cube buildings...

Tina wants to combine the three building blocks shown in the picture to form a cube building. Which one of the following cube buildings could she make? (The three blocks and the five choices A–E are pictured with the question.)

Figure for Math Kangaroo 2025 Problem 23
Show answer
Answer: D
Show hints
Hint 1 of 2
First just count: how many little cubes are in the three blocks all together? The answer building must use exactly that many cubes.
Still stuck? Show hint 2 →
Hint 2 of 2
Throw out any choice with the wrong cube count, then check the survivors by mentally snapping the three blocks together.
Show solution
Approach: count cubes first, then fit the blocks
  1. Each of the three building blocks is made of small cubes; counting them gives a fixed total number of cubes that the finished building must contain.
  2. Count the cubes in each answer building and cross out the ones with the wrong total — the right building must have exactly as many cubes as the three blocks combined.
  3. Among the buildings with the correct cube count, only (D) can actually be assembled from those three particular blocks fitting together with no gaps, so the answer is (D).
2011 · #5 (figure problem)
Figure for Math Kangaroo 2011 Problem 5
Show answer
Answer: E — Piece E.
Show hints
Hint 1 of 2
Picture the empty gap in the cuboid: count how the missing cubes are arranged.
Still stuck? Show hint 2 →
Hint 2 of 2
Match that exact 3-D arrangement of cubes to one of the five pieces.
Show solution
Approach: match the missing block shape to a choice
  1. The cuboid is missing a chunk of small cubes in a particular 3-D arrangement.
  2. Compare the shape of that gap with each offered piece.
  3. Only piece E has cubes arranged so it fits the gap exactly and completes the cuboid.
2009 · #6 How many faces does the object shown have? (It is a prism with a hole through it.)

How many faces does the object shown have? (It is a prism with a hole through it.)

Figure for Math Kangaroo 2009 Problem 6
Show answer
Answer: D — 8
Show hints
Hint 1 of 2
Count the outside surfaces first, then remember the hole adds new inside surfaces too.
Still stuck? Show hint 2 →
Hint 2 of 2
A triangular tube has 3 inner walls plus 3 outer walls, and the two ends are still faces.
Show solution
Approach: count outer and inner faces
  1. The two triangular ends are now frames, but each is still one face: 2 faces.
  2. The three outer rectangular sides: 3 faces.
  3. Drilling the triangular hole creates three inner rectangular walls: 3 more faces.
  4. Total = 2 + 3 + 3 = 8 faces — answer D.
2025 · #4 Grey squares of equal size are glued onto a cube (see picture). All surfaces of the cube then look the same. How many grey squares were...

Grey squares of equal size are glued onto a cube (see picture). All surfaces of the cube then look the same. How many grey squares were used in total?

Figure for Math Kangaroo 2025 Problem 4
Show answer
Answer: D — 18
Show hints
Hint 1 of 2
A cube has 6 faces, and the puzzle says every face ends up looking exactly the same.
Still stuck? Show hint 2 →
Hint 2 of 2
So count the grey squares on just one face, then multiply by 6.
Show solution
Approach: count one face, then multiply by six
  1. Because all six faces look identical, you only need to count the grey squares on a single face and then multiply.
  2. Each face carries 3 grey squares in its diamond design.
  3. With 6 matching faces that makes 3 × 6 = 18 grey squares in total, so the answer is (D) 18.
CHAPTER 7

Grids — locate by row and column, not by eye

THEORY

Grid problems look like easy counting, and that is the trap. Every single time, the danger is the same: you miss a cell or count one twice because you trusted your eye. The cure is to count in an order you cannot lose track of.

If a grid is filled with numbers in a steady march — 1 to 8 across the first row, 9 to 16 across the next, and so on — you never have to read every cell. Pin down one cell you are sure of, then step to the others by the grid’s rhythm: the cell to the right is ‘plus 1’, and the cell directly below is ‘plus one whole row’.

1220+8 (one row down)

With eight per row, the cell under any number is exactly 8 more. So once you place a single anchor, you know precisely which numbers a puzzle piece would sit on — and you match the piece to those positions, not to a fuzzy picture.

For ‘how many little squares / rectangles’ problems, count by size class: all the 1×1s, then all the 2×2s, and so on, so nothing slips past. For path problems, count by which cell you step through.

THE MOVE: anchor one cell you are sure of, then reach every other by ‘plus a row, plus a column’.

🎯 Try it
A grid is filled with 1, 2, 3, … going 10 numbers per row (1–10, then 11–20, and so on). Which number sits directly below 17?
Walkthrough: moving straight down one cell is moving one whole row forward, which is +10 when there are 10 per row. So the cell under 17 is 17 + 10 = 27. No need to redraw the grid — add one row’s worth.
THE TRICK

Find a landmark — a corner, a known number, a marked star — and compute every other cell’s position by counting rows and columns from it. Never decide a shape’s position by squinting at the drawing.

WATCH OUT

Matching a puzzle piece to the picture by its outline alone. Two pieces can look almost identical; only the actual numbers (or coordinates) the cells carry tell them apart. Compute the cells, then match.

WORKED EXAMPLE
PROBLEM · 2023 #1

Holger writes the numbers up to 40 into the table in the same way as shown. Which of the pieces A to E can he then cut out from the table?

Figure for Math Kangaroo 2023 Problem 1

Holger fills a table with 1, 2, 3, …, eight numbers per row. Which puzzle piece can be cut from a real patch of the table? Anchor a number, then step by the row size.

Eight per row is the rhythm. Take the number 12: the first row is 1–8, so the second row is 9, 10, 11, 12 — that puts 12 in row 2, column 4. The cell directly under 12 is one full row down, so add 8: that is 20. Next to it, 21. Go one more row under the 21, add 8 again: 29. So a real patch reads 12 on top, 20 and 21 in the middle, and 29 hanging beneath the 21. Only one piece’s outline holds exactly those four cells in those spots — that is piece C. You computed which numbers sit where; you never trusted the picture.

Answer: C
RULE OF THUMB

Count grids in a fixed order; locate cells by ‘plus a row, plus a column’ from one anchor, never by eye. Count squares by size class.

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2023 · #1 The diagram shows a grid made of vertical and horizontal lines. Which part was cut from the grid?

The diagram shows a grid made of vertical and horizontal lines. Which part was cut from the grid?

Figure for Math Kangaroo 2023 Problem 1
Show answer
Answer: E
Show hints
Hint 1 of 2
Look at the hole in the grid and count the rows and columns of small cells it spans.
Still stuck? Show hint 2 →
Hint 2 of 2
Match the exact pattern of lines inside the hole — not just its outline — to one of the five pieces.
Show solution
Approach: match the cut-out's internal line pattern to a choice
  1. The missing region has a fixed size and a specific arrangement of internal horizontal and vertical lines.
  2. Check each option for the same number of crossing lines and the same dimensions as the hole.
  3. Only piece E reproduces that exact line pattern.
  4. So the part cut from the grid is E.
2023 · #3 Which of the shapes cannot be split into two triangles using a single straight line?

Which of the shapes cannot be split into two triangles using a single straight line?

Figure for Math Kangaroo 2023 Problem 3
Show answer
Answer: A
Show hints
Hint 1 of 2
A single straight cut makes two pieces; to get two triangles each piece must end up with exactly three sides.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the sides: a four-sided shape can be cut corner-to-corner into two triangles, but a six-sided one cannot.
Show solution
Approach: see which shapes a single cut can split into two triangles
  1. The rectangle, trapezoid and square each have four sides, so a diagonal cut turns them into two triangles.
  2. The triangle can be split into two triangles by a line from a vertex to the opposite side.
  3. The hexagon has six sides; one straight cut cannot reduce it to two three-sided pieces.
  4. So the shape that cannot be split is the hexagon, A.
2025 · #20 Joanna divides the figure into five equal-sized, same-shaped parts, each of which consists of three squares. Which of the letters is in...

Joanna divides the figure into five equal-sized, same-shaped parts, each of which consists of three squares. Which of the letters is in the part with the star?

Figure for Math Kangaroo 2025 Problem 20
Show answer
Answer: E — E
Show hints
Hint 1 of 2
All five pieces are the same shape made of three squares, so figure out that shape first from a corner that can only be filled one way.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you know the piece shape, build outward and watch which three squares end up grouped with the star.
Show solution
Approach: find the repeating 3-square piece, then read off the star’s group
  1. Since every piece is the same three-square shape, start at a corner of the figure where only one shape can fit; that fixes what the repeating piece looks like.
  2. Lay that same piece again and again to tile the whole figure with no gaps or overlaps — there is only one way it all fits together.
  3. The piece that ends up covering the starred square also covers the square labelled E, so the answer is (E).
CHAPTER 8

Overcount, then subtract — the counting fix for hard spatial problems

THEORY

The hardest spatial problems are not about seeing — they are about counting without missing or doubling. And the single most powerful counting move sounds backwards: count too much on purpose, then take back the extra.

Picture marking every little cube that touches a drawn line, going face by face. A cube tucked on an edge or corner of the big cube belongs to two faces, so you count it twice. That is fine — as long as you then subtract those doubles back out. Counting loose and correcting beats trying to count each thing exactly once while juggling overlaps in your head.

count Acount BA + B counts the middle twice → subtract it once

This is the same idea as the overlapping circles above: add the two counts, then remove the shared middle once so it is not counted twice. On cubes it is the cubes on shared edges; on tunnels it is the cubes where two tunnels cross; on grids it is a cell two shapes both claim.

The other half of clean counting is casework: split the situation into cases that do not overlap and do not leave a gap, and count each. Keep a written tally so a case never silently disappears.

Counting figures hidden inside a figure: sort by size, then by direction

‘How many triangles in this picture?’ is the classic miss-or-double trap. Pointing at random never works. The fix is a two-key sort: group every triangle first by its size, then within each size by which way it points (up or down). A cut-out cardboard triangle you slide around the figure makes sure none escapes.

Take a triangle split into a 3-row grid of nine small triangles. Sort and tally:

updown
sizepointing uppointing down
small (1 cell)63
medium (4 cells)30
large (whole)10

Add the column entries: 6 + 3 + 3 + 1 = 13 triangles. The down-pointing ones are exactly the ones beginners forget — sorting by direction forces you to look for them. When a figure is symmetric (a six-point star, say), up-count equals down-count, so you tally one and double, halving the work.

Strategy framing inspired by Posamentier, The Art of Problem Solving (teacher resource).

🎯 Try it
In that 3-row triangle above, count the triangles yourself by the size-and-direction sort: 6 small-up, 3 small-down, 3 medium-up, 1 large. What is the total?
Walkthrough: add the sorted tallies: small 6 up + 3 down = 9, medium 3 up + 0 down = 3, large 1. Total 9 + 3 + 1 = 13. The win is the 3 small down-pointing ones — the sort makes you hunt for them instead of missing them.

THE MOVE: count everything loosely, then subtract exactly the things you counted more than once.

🎯 Try it
On a 7×7 grid you draw both diagonals (corner to corner, each way). How many little squares get a line through them? (Watch the middle.)
Walkthrough: overcount first. One diagonal crosses 7 squares; the other crosses 7 too, so loosely that is 7 + 7 = 14. But on an odd grid both diagonals pass through the single centre square, so it got counted twice. Subtract that one overlap: 14 − 1 = 13. Count loose, then take back the double-count.
THE TRICK

Do the easy per-piece count first — per face, per line, per tunnel — as if nothing overlapped. Then hunt for what got counted twice (shared edges, crossings, corners) and subtract precisely those. Always write the tally down.

WATCH OUT

Either forgetting to subtract the overlaps (too big) or subtracting them more than once (too small). Count the shared items, then remove each one exactly once.

Bogus solution

Nine straight tunnels are drilled through a 5×5×5 cube. Each tunnel takes out a row of 5 little cubes, so the number removed is 9 × 5 = 45.

Why it breaks: deep inside the block the tunnels cross, and a cube sitting at a crossing was counted once for every tunnel running through it. The loose total of 45 counts those shared cubes more than once.

The fix: keep the overcount, then subtract each crossing cube back out so it is counted exactly once. After removing the double-counts the true total drops to 39 — the loose count was only the starting point, never the answer.

WORKED EXAMPLE
PROBLEM · 2017 #27

Mike has 125 small, equally big cubes. He glues some of them together in such a way that one big cube with exactly nine tunnels is created (see diagram). The tunnels go all the way straight through the cube. How many of the 125 cubes is he not using?

Figure for Math Kangaroo 2017 Problem 27
A) 52 B) 45 C) 42 D) 39 E) 36

A big cube of 125 little cubes (5×5×5) has nine straight tunnels drilled through it. How many little cubes are removed? Overcount the tunnels, then subtract the crossings.

Each tunnel is a straight line punched all the way through, so it removes a row of 5 little cubes. Nine tunnels, counted loosely, would remove 9 × 5 = 45 cubes. But that is too many: deep inside the cube the tunnels cross each other, and a cube sitting at a crossing got counted once for each tunnel through it — so it was double-counted. Subtract those shared crossing cubes back out, and the true number removed comes down to 39, choice D. That is the whole chapter in one problem: count loose (45), then take back the overlaps to land on 39.

Answer: D — 39
RULE OF THUMB

Overcount on purpose, then subtract the double-counts. Or split into non-overlapping cases. Keep a written tally so nothing is missed.

MORE LIKE THIS
2019 · #3 A 3 × 3 × 3 cube is made up of small 1 × 1 × 1 cubes. Then the middle cubes from front to back, from top to bottom and from right to...

A 3 × 3 × 3 cube is made up of small 1 × 1 × 1 cubes. Then the middle cubes from front to back, from top to bottom and from right to left are removed (see diagram). How many 1 × 1 × 1 cubes remain?

Figure for Math Kangaroo 2019 Problem 3
Show answer
Answer: C — 20
Show hints
Hint 1 of 2
Start from all 27 small cubes and figure out exactly which ones get drilled away.
Still stuck? Show hint 2 →
Hint 2 of 2
The three tunnels all pass through the very middle, so the centre cube is removed only once.
Show solution
Approach: count removed cubes, watch the shared centre
  1. A 3×3×3 block has 27 unit cubes.
  2. Each of the three tunnels (front-back, top-bottom, left-right) removes the 3 cubes down its middle line.
  3. All three lines share the single centre cube, so together they remove the centre plus 6 face-centre cubes = 7 cubes.
  4. 27 − 7 = 20 cubes remain.
2024 · #22 Tarek wants to colour two more cells of the 4×4-square black so that the pattern of white and black cells has exactly one axis of...

Tarek wants to colour two more cells of the 4×4-square black so that the pattern of white and black cells has exactly one axis of symmetry. In how many ways can he do that?

Figure for Math Kangaroo 2024 Problem 22
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Answer: E — 6
Show hints
Hint 1 of 3
A 4×4 grid can be symmetric about four possible lines: vertical, horizontal, and the two diagonals.
Still stuck? Show hint 2 →
Hint 2 of 3
For each candidate axis, reflect the two black cells already there and see what the two new cells would have to be.
Still stuck? Show hint 3 →
Hint 3 of 3
Be careful to keep exactly one axis — a placement that accidentally creates a second axis does not count.
Show solution
Approach: make the picture symmetric about each axis in turn and count the valid two-cell completions
  1. Reflecting the two given black cells across the vertical axis forces one extra colouring that is symmetric only about that vertical line.
  2. The horizontal axis similarly forces one valid colouring.
  3. The two diagonal axes are more flexible and give the remaining colourings, for a total of four diagonal-symmetric ways.
  4. Adding them up, 1 vertical + 1 horizontal + 4 diagonal = 6 ways.
2025 · #20 Joanna divides the figure into five equal-sized, same-shaped parts, each of which consists of three squares. Which of the letters is in...

Joanna divides the figure into five equal-sized, same-shaped parts, each of which consists of three squares. Which of the letters is in the part with the star?

Figure for Math Kangaroo 2025 Problem 20
Show answer
Answer: E — E
Show hints
Hint 1 of 2
All five pieces are the same shape made of three squares, so figure out that shape first from a corner that can only be filled one way.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you know the piece shape, build outward and watch which three squares end up grouped with the star.
Show solution
Approach: find the repeating 3-square piece, then read off the star’s group
  1. Since every piece is the same three-square shape, start at a corner of the figure where only one shape can fit; that fixes what the repeating piece looks like.
  2. Lay that same piece again and again to tile the whole figure with no gaps or overlaps — there is only one way it all fits together.
  3. The piece that ends up covering the starred square also covers the square labelled E, so the answer is (E).
⬢ FINAL TEST

Spatial stretch test

Five harder spatial problems — folding, cube-views, tiling, counting tricks.

2015 · #21 Nina wants to make a cube from the paper net. You can see there are 7 squares instead of 6. Which square(s) can she remove from the net,...

Nina wants to make a cube from the paper net. You can see there are 7 squares instead of 6. Which square(s) can she remove from the net, so that the other 6 squares remain connected and from the newly formed net a cube can be made?

Figure for Math Kangaroo 2015 Problem 21
Show answer
Answer: D — only 3 or 7
Show hints
Hint 1 of 2
A cube net needs the remaining 6 squares to stay connected AND to fold up without two squares landing on the same face.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each candidate removal: most leave a shape that overlaps when folded; only certain end squares work.
Show solution
Approach: test which removals leave a connected, foldable 6-square net
  1. Removing a square must keep the other six joined and able to fold into a cube with no doubled-up face.
  2. Taking out an interior square breaks the net or makes two squares fold onto the same face, so those fail.
  3. Removing square 3 works, and removing square 7 works, while no other single removal does — so the answer is only 3 or 7.
2022 · #21 A building is made of cubes of the same size. The three pictures show it from above (von oben), from the front (von vorne) and from the...

A building is made of cubes of the same size. The three pictures show it from above (von oben), from the front (von vorne) and from the right (von rechts). What is the maximum number of cubes that could be used to make this building?

Figure for Math Kangaroo 2022 Problem 21
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Answer: B — 19
Show hints
Hint 1 of 2
The top view fixes which columns can hold cubes; the front and side views cap each column's height.
Still stuck? Show hint 2 →
Hint 2 of 2
For the maximum, make every column as tall as its views allow.
Show solution
Approach: raise each column to the height its views permit
  1. The top view shows which floor positions are occupied.
  2. The front and right views give the largest height allowed for each row and column.
  3. Stacking each column to its maximum allowed height totals 19 cubes.
  4. So the answer is B.
2025 · #22 Julio wants to make the shape shown in the top picture on the right. He has several of each of the five tiles shown in the bottom...

Julio wants to make the shape shown in the top picture on the right. He has several of each of the five tiles shown in the bottom picture on the right. The tiles must be placed next to each other without overlapping. What is the smallest number of tiles he must use?

Figure for Math Kangaroo 2025 Problem 22
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
To use as few tiles as possible, you want each tile to cover as much of the cross as it can, so reach for the biggest tiles first.
Still stuck? Show hint 2 →
Hint 2 of 2
The straight parts of the cross are easy to cover with the large rectangle and big triangle; the pointy arm-tips are what force you to use the small triangles.
Show solution
Approach: cover the big areas with big tiles, the tips with small ones
  1. Fewer tiles means each tile should cover as much as possible, so fill the wide straight parts of the cross with the largest tiles (the long rectangle and the big triangle).
  2. The four slanted arm-tips are too thin for the big tiles, so each tip has to be finished with the small triangle pieces — these are unavoidable and set the limit on how low the count can go.
  3. Packing the big tiles in the body and the small triangles at the tips, with no overlaps, covers the whole cross in 13 tiles, and no arrangement does it in fewer, so the answer is (C) 13.
2015 · #24 Maria writes a number on each face of the cube. Then, for each corner point of the cube, she adds the numbers on the faces which meet at...

Maria writes a number on each face of the cube. Then, for each corner point of the cube, she adds the numbers on the faces which meet at that corner. (For corner B she adds the numbers on faces BCDA, BAEF and BFGC.) In this way she gets a total of 14 for corner C, 16 for corner D, and 24 for corner E. Which total does she get for corner F?

Figure for Math Kangaroo 2015 Problem 24
Show answer
Answer: C — 22
Show hints
Hint 1 of 2
Two corners at opposite ends of a space diagonal use all six faces between them, so their corner-sums add up to the same total every time.
Still stuck? Show hint 2 →
Hint 2 of 2
Pair the given corner with the unknown one along a space diagonal, and pair the other two the same way.
Show solution
Approach: use that opposite corners of the cube share the same total of all six faces
  1. A corner's number is the sum of its three meeting faces. Two corners on opposite ends of a space diagonal together touch all six faces exactly once, so each such pair has the same sum S (the total of all six faces).
  2. Corners C and E are opposite, and corners D and F are opposite, so C + E = D + F.
  3. Thus 14 + 24 = 16 + F, giving F = 38 − 16 = 22.
2017 · #27 Mike has 125 small, equally big cubes. He glues some of them together in such a way that one big cube with exactly nine tunnels is...

Mike has 125 small, equally big cubes. He glues some of them together in such a way that one big cube with exactly nine tunnels is created (see diagram). The tunnels go all the way straight through the cube. How many of the 125 cubes is he not using?

Figure for Math Kangaroo 2017 Problem 27
Show answer
Answer: D — 39
Show hints
Hint 1 of 2
Count how many unit cubes are removed to make the nine straight tunnels.
Still stuck? Show hint 2 →
Hint 2 of 2
Tunnels share cubes where they cross inside the big cube — don't double-count.
Show solution
Approach: count removed cubes via inclusion-exclusion
  1. Each tunnel removes a straight line of 5 cubes; nine tunnels would remove 45, but the tunnels intersect inside the cube.
  2. Subtracting the cubes shared at the crossings leaves 39 cubes actually removed.
  3. So Mike does not use 39 cubes.
APPENDIX

Quick reference

Memorize these

SPATIAL FACTS

  • A cube has 6 faces, 12 edges, 8 vertices; opposite faces never touch.
  • A cube net is 6 squares — exactly 11 different ones exist. Six squares alone do not guarantee a net; fold-check. Opposite faces skip one along a straight row (1–3, 2–4); two squares that touch are neighbours, never opposite.
  • Count 3D by layers: an n×n×n cube is n³ small cubes; remember the hidden ones. Each flat view is the tallest cube along your line of sight.
  • Die: opposite faces sum to 7, all six sum to 21, neighbours are never opposite. Opposite corners of a cube share the same corner-sum.
  • Tiles needed ≥ region area ÷ tile area — then check it is achievable (checkerboard colouring).
  • Back view = front view flipped left-right. Reflection flips across a line; rotation turns about a point. To shorten a bent touch-the-line path, reflect the far point across the line and draw it straight.
  • Cuts on a line: pieces = cuts + 1 (so 3 pieces need 2 cuts). Slicing a corner creates a new face with its own corners — count what a cut adds, not only what it removes.
  • Count figures inside a figure by size, then by orientation (up vs down); the down-pointing ones are the ones beginners miss.
  • Overcount, then subtract the double-counts (shared edges, tunnel crossings). When a count comes out, redo it a second way — two roads to the same number means you can trust it.
Common traps

COMMON TRAPS

  • Counting only visible cubes and forgetting the hidden ones.
  • Assuming any 6-square shape folds into a cube.
  • On a net, reading two squares that point opposite ways on the paper as opposite faces — once folded they often share an edge, so they are neighbours.
  • Confusing a reflection (mirror flip) with a rotation (turn).
  • Trusting the area ceiling without checking the tiles actually fit.
  • Forgetting the faces or corners a cut creates, or trusting an impossible die.
  • Reading ‘3 pieces’ as 3 cuts — pieces are always one more than cuts.
  • Forgetting to subtract overlaps when counting — or subtracting them twice; missing the down-pointing figures in a count-the-triangles problem.

REACHING FOR 20+

  • View-reconstruction. Given front, top, and side views, rebuild the cube count by laying the footprint down and raising each column. For the maximum, push every column as high as the views allow; for the minimum, raise only what the views force.
  • Corner-sum pairing. On a numbered cube, two corners at opposite ends of a space diagonal touch all six faces once, so opposite-corner sums are equal — pair a known corner with the unknown and solve in one equation.
  • Inclusion–exclusion on solids. For tunnels or lines through a cube, overcount per tunnel, then subtract the cubes at crossings exactly once (a triple crossing is shared by three, so handle it carefully).
  • Reflect-to-straighten. Turn a bent touch-the-line (or bouncing-beam, or fold) path into one straight segment by reflecting the endpoint; then a right triangle gives the length.
  • Symmetry-halving. When a figure is symmetric, up-count equals down-count — tally one and double.