Geometry & Measurement — Draw the picture. Then look for symmetry.

Picture two gardens. One is 20 feet by 45 feet. The other is 25 feet by 40 feet. Same shape family, different proportions. Which one needs more fence? Which one grows more vegetables? Those are two completely different questions — and mixing them up is the single most common slip in contest geometry.

Perimeter is the fence: walk all the way around the edge and add up the steps. Area is the dirt inside: how much surface you cover, counted in little squares. Fence is a length. Dirt is squares. Keep them in separate boxes in your head.

(That garden problem is real: 20×45 = 900 and 25×40 = 1000, so the second garden is bigger by 100 square feet — even though both use the same total fence. Shape matters.)

The one trap that catches everyone. The height in the triangle, parallelogram, and trapezoid formulas is the perpendicular distance — straight up from the base, like a plumb line. It is almost never the slanted side. Grabbing the slanted side instead is the most common geometry mistake there is.

Count along the grid — don’t reach for a formula

Some figures sit on a grid of unit squares. The lengths and areas are sitting right there in the squares. You don’t need a formula — you need to count carefully.

Say a thick zig-zag line traces the edges of grid squares, and each square has area 4. First turn area into length: a square of area 4 has side √4 = 2. Now count how many square-edges the line follows — say 9 of them. Length = 9 × 2 = 18. The whole problem is “find one edge, then count edges.”

For the perimeter of a shape glued from unit squares, there’s a faster count than walking the whole outline. Start from the squares apart: 6 squares have 6 × 4 = 24 unit edges. Every place two squares touch, they hide 2 unit edges (one from each). Count the touching pairs and subtract: 6 touches → perimeter = 24 − 2 × 6 = 12.

Substitution — use one length to pin down another

Some figures are built from several copies of the same rectangle, tucked together. You’re given the outside size and asked for one piece. There aren’t enough numbers to plug in directly. So you make some up — carefully.

Call one small rectangle w wide and h tall. Now read the picture: each outside length is built from a few w’s and h’s laid end to end. That gives you equations. Solve them, substitute back.

Here’s the cleanest version (it’s a real Kangaroo problem, kangaroo-2015-kadett-02). A big rectangle is built from 4 equal small rectangles. The short side of the big one is 10. Two small rectangles stand upright, spanning the full height — so a small rectangle’s long side is 10. The other two lie flat and stack to fill that same 10, so each small rectangle’s short side is 10 ÷ 2 = 5. The long side of the big rectangle = flat-length(10) + flat-length(10) = 20.

You never measured the long side directly. You named the small rectangle’s sides, found them from the one number you were given, then added them back up.

A close cousin is the sum-constraint: a total is split into named pieces. A board of length 6 has a 3-long middle; the two equal ends share what’s left, so each end = (6 − 3) ÷ 2 = 1.5. Total = sum of parts — write that equation and the unknown drops out.

Your instinct is to wait for more numbers. Resist it. Name the unknown piece, read each given length as a sum of named pieces, then substitute.

The midsegment of a trapezoid — just average the two bases

A trapezoid has a short top and a long bottom. Slice it horizontally exactly halfway up, joining the midpoints of the two slanted sides. That cut — the midsegment — has a length you can read off without measuring: it’s the plain average of the top and bottom.

Top 4, bottom 10 → midsegment = (4 + 10) ÷ 2 = 7. It always lands right in the middle of the two, which is where your eye expects it.

Framing inspired by Competition Math for Middle School (AoPS).

The quadrilateral family — who contains whom

Square, rectangle, rhombus, parallelogram, trapezoid — they sound like five separate shapes. They aren’t. They nest inside each other like measuring cups, and seeing the nesting tells you instantly which properties carry over.

Read it from the inside out: a square is the overlap — it’s a rectangle AND a rhombus at once. A rectangle (four right angles) and a rhombus (four equal sides) are both parallelograms (two pairs of parallel sides). And every parallelogram is a quadrilateral. In symbols:

square ⊂ rectangle & rhombus ⊂ parallelogram ⊂ quadrilateral

Why this matters on a contest: a property true of every parallelogram is automatically true of rectangles, rhombuses, and squares (they live inside). But a special rhombus fact — like “the diagonals cross at right angles” — need not hold for a general parallelogram. The nesting tells you exactly how far a property travels.

Framing inspired by AoPS Prealgebra.

Here’s a shape with a chunk bitten out of one corner — an L. There’s no “area of an L” formula. So what do you do? You cut. And the satisfying part: there is almost always more than one good place to cut, and every cut gives the same answer. Watch two different kids attack the same L and land on 36.

So the skill isn’t a formula — it’s deciding where to cut. Here’s how to choose fast.

Where the formulas come from — one cut turns it into a rectangle

The area formulas for a parallelogram and a triangle aren’t magic. Each one is a rectangle in disguise. Cut once, slide, and look.

Parallelogram → rectangle. A parallelogram looks like a rectangle that got shoved sideways. Slice the slanted triangle off one end and slide it to the other end — it fits perfectly, and you’re holding a plain rectangle. Same base, same height, same area. So A = base × height — and the height is the straight-up distance, never the slanted side.

Triangle = half a rectangle. Take any triangle. Make a second identical copy, flip it upside down, and snug it against the first. The two together form a parallelogram (a rectangle, if the triangle is right) on the same base and height. So one triangle is exactly half of it: A = ½ × base × height. The “½” isn’t a rule to memorize — it’s the second copy you didn’t draw.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

A triangle inside a rectangle — cut differently

A pentagon with vertices (0,0), (4,0), (5,3), (2,5), (0,3). Cut into a rectangle and two triangles:

• Bottom rectangle: width 4, height 3 → area 12.
• Bottom-right triangle: base 1, height 3 → ½ · 1 · 3 = 1.5.
• Top triangle: base 5, height 2 → ½ · 5 · 2 = 5.

Total: 12 + 1.5 + 5 = 18.5 square units.

Rearrangement — slide a piece, don’t measure it

Sometimes a shape has slanted or bumpy edges that look painful to measure. Before you reach for the Pythagorean theorem, ask: did the slants just move area around?

Look at a shape where the bottom edge slopes down on the left and the top edge slopes up by the exact same amount. The slant on top took a triangle of area off one corner; the slant on the bottom handed back an identical triangle somewhere else. Net change: zero. Slide the pieces back and you’re left with a plain rectangle. If that rectangle is 2 by 3, the area is 2 × 3 = 6 — no slants to measure at all.

This is the same idea behind the L-shape perimeter trick in chapter 1 — pieces that cancel. Here it’s area, not perimeter.

The strong version shows up with curves too. If a star is built by bending the four arcs of a circle inward, no area was created or destroyed — the star and the leftover “bites” are the same arcs as the circle. Find a square the pieces tile exactly, and the area comes from subtraction, not from arc formulas.

Your instinct is to measure every slanted edge. Resist it. Look for matching pieces that cancel, slide them into a plain shape, and read the area off that.

You want the area of a picture mat — the cardboard border around a photo. You could measure all four strips and add them. Or you could do this: outer rectangle minus the photo. One subtraction. Done.

That is the frame trick, and it is the most common single move in contest geometry. Whenever the region you want is defined by what got removed — a hole, a cut, a leftover — don’t trace its messy boundary. Measure the big thing and the missing thing, and subtract.

Three flavors:

• A square with a circle removed: area = square − circle.
• A photo inside a frame: frame area = big rectangle − inner photo.
• The leftover after cutting: original area − cut piece.

You never measure the shaded boundary directly. You measure the big rectangle and the hole, then subtract.

When pieces overlap, subtract the double-count

The frame trick has a twin for overlapping shapes. If you slap two regions together and they overlap, just adding their areas counts the overlap twice. Fix it by subtracting the overlap once:

union = A + B − overlap

Grown-ups call this inclusion–exclusion. You call it “pay back the part you counted twice.” It works for any number of pieces: add everything, then subtract each overlap once.

Before we tackle right triangles next chapter, we need one tool from arithmetic — because you meet the square root sign the moment you touch Pythagoras: legs 2 and 3 give a hypotenuse of \(\sqrt{13}\). But what is that? Most kids treat \(\sqrt{\ }\) as a scary button on a calculator. It isn’t. It’s just the question “what number, times itself, gives this?” run backwards.

Squaring takes 5 to 25. The square root walks back: 25 to 5. That’s the whole idea — it undoes squaring.

So \(\sqrt{25}=5\), \(\sqrt{49}=7\), \(\sqrt{144}=12\). A number like 25 or 144 that comes out whole is a perfect square. The little hook symbol \(\sqrt{\ }\) has a name — the radical.

Simplifying a root — pull out the perfect-square factor

What about \(\sqrt{50}\)? It’s not whole — 50 isn’t a perfect square. But hiding inside 50 is a perfect square: \(50 = 25 \times 2\). And a root splits across a product:

\(\sqrt{a \times b} = \sqrt{a}\,\times\,\sqrt{b}\)

So \(\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\times\sqrt{2} = 5\sqrt{2}\). The 25 walks out as a 5; the 2 has no perfect-square factor left, so it stays under the radical. That’s simplified form: a whole number times a root with nothing square left inside.

Hunt for the biggest perfect-square factor and you finish in one step. Miss it and you still get there in two: \(\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18} = 2\sqrt{9 \times 2} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}\). Same answer, just more steps.

Framing inspired by AoPS Prealgebra.

Multiplying and adding roots — two different rules

Multiplying is friendly. Roots multiply straight across: \(\sqrt{2}\times\sqrt{8} = \sqrt{16} = 4\). Whatever’s inside combines under one radical.

Adding is fussy. You can only add roots that are already the same root — like terms. \(2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}\) (count the \(\sqrt{3}\)s), exactly like \(2\) apples \(+\) \(5\) apples. But \(\sqrt{2} + \sqrt{3}\) just stays \(\sqrt{2} + \sqrt{3}\) — different roots, can’t combine.

The payoff: simplify first, and unlike-looking roots often turn out alike. \(\sqrt{50} - \sqrt{18} = 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2}\). They were both \(\sqrt{2}\) in disguise.

Estimating a root — trap it between two integers

A hypotenuse comes out \(\sqrt{40}\) and the answer choices are plain numbers. About how big is it? You don’t need a calculator — you need the two perfect squares it sits between.

The rule that makes this work: bigger inside means bigger root. Since \(36 < 40 < 49\), taking roots keeps the order: \(\sqrt{36} < \sqrt{40} < \sqrt{49}\), so \(6 < \sqrt{40} < 7\). And 40 is much closer to 36 than to 49, so \(\sqrt{40}\) is a touch above 6 — about 6.3.

Three numbers worth more than any formula in this lesson: 3, 4, 5. Build a triangle with sides 3, 4, 5 and the corner between the 3 and the 4 comes out exactly square. Carpenters have used this for thousands of years to make right angles with nothing but a knotted rope. On a contest, the moment you spot a 3, a 4, and a 5 (or a 5 and a 12, or an 8 and a 15), you can write the missing side with zero arithmetic.

These magic trios are Pythagorean triples: right triangles whose three sides are all whole numbers. They obey the most famous equation in geometry.

See WHY it’s true — the four-triangle puzzle

Most kids are told a² + b² = c² and asked to believe it. You don’t have to. Here is a picture that proves it, with no algebra — only cutting and counting.

Take four copies of the same right triangle (legs a and b, hypotenuse c). Pack them, pinwheel-style, into the corners of a big square whose side is a + b. The four slanted hypotenuses fence off a tilted square in the middle, and that square’s side is exactly c — so its area is c².

Here is the trick. The big square has a fixed area. The four triangles never change. So whatever is left over after you place the triangles must always be the same amount of leftover — no matter how you arrange them.

Arrange them the pinwheel way (left): leftover = the tilted square = c². Now slide the same four triangles to hug two corners instead (right): the leftover splits into two upright squares — one of side a (area a²) and one of side b (area b²). Same big square, same four triangles, so the leftovers match:

c² = a² + b².

No formula handed down — you watched the same blank space get re-counted two ways. That is a proof.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

You can always compute the third side with this. But the triples let you skip the computing. Memorize these three:

The three most common Pythagorean triples: 3-4-5, 5-12-13, 8-15-17.

Other triples worth knowing:

• 7-24-25, 9-40-41, and (rarely) 20-21-29.
• Any multiple of a triple is also a triple: 6-8-10, 9-12-15, 30-40-50, 10-24-26, …

The two special right triangles — where the √2 and √3 come from

Triples give whole-number triangles. But two right triangles show up constantly with non-whole sides — and you don’t memorize them, you build them from shapes you already know.

The 45-45-90 is half a square. Slice any square along its diagonal. You get two identical right triangles with equal legs and two 45° angles. If each leg is 1, Pythagoras gives the diagonal: 1² + 1² = 2, so the hypotenuse is √2. That’s the whole rule — hypotenuse = leg · √2, ratio 1 : 1 : √2. (This is exactly why a square of side s has diagonal s√2.)

The 30-60-90 is half an equilateral triangle. Take an equilateral triangle with side 2 and drop a straight-down line from the top. It splits into two matching right triangles. The base was 2, now halved to 1 (the short leg). The hypotenuse is still a full side, 2. Pythagoras fills in the tall leg: it’s √(2² − 1²) = √3. So ratio 1 : √3 : 2 — short leg, long leg, hypotenuse.

Framing inspired by AoPS Prealgebra.

The 13-14-15 triangle — the one non-right triangle worth memorizing

It’s not a right triangle, but its sides are all whole numbers and so is its area:

Drop the altitude to the side of length 14. It splits into a 5-12-13 right triangle and a 9-12-15 right triangle (a 3-4-5 tripled). Both share the altitude 12, and the bases add to 5 + 9 = 14 ✓.

Fold a square in half. The two halves land exactly on top of each other. That fold line is a line of symmetry, and it’s a cheat code: anything you learn about one half, you get the other half for free. Contest figures are loaded with symmetry, and the kids who hunt for it first do a quarter of the work.

Here’s a fast pattern to lock in. Count the fold lines of these shapes — the number always matches the number of equal sides:

• Equilateral triangle → 3 axes.  Square → 4 axes (two through midpoints, two diagonals).
• Rectangle → 2 axes (through midpoints — not the diagonals).  Regular hexagon → 6 axes.
• Circle → infinitely many — any line through the center.

There’s a second kind: rotational symmetry, where a figure looks the same after you spin it. A square looks identical after a quarter turn; a hexagon after a sixth of a turn.

The spinning-wedge trick

Here’s a figure contests love. Take a square. Pin the corner of a right angle (two perpendicular rays) at the center of the square, and spin that right angle to any angle. What fraction of the square does the wedge cover?

It looks like it should depend on the angle. It doesn’t. The wedge always covers exactly one quarter.

Why: spin the right angle until its two rays point straight along the diagonals (or straight to two opposite sides) — now the covered region is plainly one of the four matching corners, a quarter. Rotate away from that and whatever sliver you lose on one side, symmetry hands back on the other. The four quarter-turns of the right angle about the center are identical, so each covers the same area, and together they fill the whole square. One quarter each.

When a right angle turns about the center of a square, the slice it cuts stays constant — rotate it to the easy position and read off the fraction.

(Rotating-square overlap adapted from Problem Solving via the AMC, Australian Maths Trust.)

Reflect a bent path straight — then it’s just one hypotenuse

Symmetry isn’t only for shaded fractions. It also straightens crooked paths. Whenever a route bends at a wall, a river, or a mirror — touch the line, then carry on — you can fold the second leg across the line. The bend opens flat into a single straight segment, and a straight segment is just the hypotenuse of a right triangle you can measure with Pythagoras.

Here’s the picture. A walker starts at A, must touch a straight river, then reach B on the same side. Reflect B across the river to its mirror image B′. Any path A → (river) → B has the exact same length as the matching path A → (river) → B′, because the second leg and its reflection are mirror copies — same length. So the bent trip equals a trip from A to B′, and the shortest one is the dead-straight line A–B′.

Now read off the right triangle for the straight line A–B′. A sits 3 above the river; B′ sits 5 below it, so the vertical leg is 3 + 5 = 8. A and B are 6 apart along the river, so the horizontal leg is 6. The straight distance is √(6² + 8²) = √100 = 10 — a 6-8-10 triangle, no slanted measuring at all.

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

Hold a die in your hand. It has 6 faces, 12 edges, and 8 corners — and that’s true for every box, no matter the size. Lock those three numbers in; contests ask for them constantly (a real AMC problem just adds them up: 6 + 12 + 8 = 26).

3D problems split into two questions. How much skin? is surface area — add up the faces. How much stuffing? is volume — how much fills the inside.

A net is a 3D shape unfolded flat — the pattern of faces you’d cut from paper and fold up. A cube has 11 different nets; the most common is a plus-sign cross of 6 squares. The key skill: given a net, tell which faces end up opposite each other after folding.

Fold F up as the front. T folds to the top, B to the bottom (T ↔ B). L folds left, R folds right (L ↔ R). K wraps around to the back (F ↔ K). On any cube net, opposite faces are the ones never sharing an edge — and a face one step away on a straight strip ends up opposite the face two steps along.

Surface area = unfold the box and add the flat pieces

Surface area sounds three-dimensional and scary. It isn’t. Unfold the box flat into its net and the whole thing becomes a flat-area problem you already know. Add up the rectangles. Done.

Take a box that is 4 long, 3 wide, 2 tall. Cut along the edges and lay it open. You get six rectangles in three matching pairs:

Add them: two tops/bottoms (12 + 12), two fronts/backs (8 + 8), two sides (6 + 6) = 52. That is exactly what the formula 2(ab + bc + ca) = 2(12 + 6 + 8) = 52 packs into one line — but the net shows you why: three pairs of matching rectangles, each pair counted twice because every face has a twin on the far side.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

Faces, vertices, edges — recover the formula from a cube

Every solid with flat faces obeys one tidy rule connecting its faces (F), corners or vertices (V), and edges (E): F + V − E = 2. It’s called Euler’s formula, and it holds for a pyramid, a prism, a soccer ball — anything.

Here’s the part worth stealing from the pros: don’t memorize which letters go where. Remember a formula exists, then rebuild it from the cube you already know cold — 6 faces, 8 vertices, 12 edges. Test: 6 + 8 − 12 = 2. ✓ Now you have it, and you can’t mix up the signs, because the cube checks itself.

This recovery habit beats raw memory all over 3-D geometry. Forget the cylinder’s surface area? Unroll it: two circles (2πr²) plus a rectangle as tall as the cylinder and as wide as the circle rolls (2πr · h). The picture rebuilds the formula every time.

Framing inspired by Competition Math for Middle School (AoPS).

Angle chasing is a game, and once you see it that way it gets fun. You start with one or two angles the problem hands you. You apply a rule — angles on a line, angles in a triangle — to find a new angle. You write it on the figure. Then that new angle unlocks the next one. Most angle problems fall in three or four moves, like dominoes.

You can’t play without the rulebook, so here it is. Know these cold.

The exterior-angle rule earns its keep more than any other. Extend one side of a triangle past a vertex; the angle that sticks out equals the sum of the two far interior angles (the ones it doesn’t touch). It lets you skip a whole step.

Why a polygon’s angles add to (n−2)·180° — just fan it into triangles

You know a triangle’s angles sum to 180°. Every other polygon’s angle sum is built straight from that one fact — no new rule to trust.

Stand at one corner of the polygon and draw a line to every other corner you can reach. The polygon fans out into triangles. Count them: a pentagon (5 sides) splits into 3 triangles; a hexagon (6 sides) into 4. Every triangle’s angles pile up at the polygon’s corners and nowhere else — so the polygon’s total angle is just (number of triangles) × 180°.

The pattern: an n-sided polygon always fans into n − 2 triangles (you skip the two corners next to you). So the angle sum is (n − 2) · 180°. For a regular polygon, share that total equally among the n corners.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

The hidden gem: a polygon’s OUTSIDE angles always add to 360°

Here is the fact almost no one remembers, and it’s the easiest one of all. Walk all the way around the edge of any convex polygon, like a tiny car driving the outline. At each corner you turn by the exterior angle — the bit of turn the corner forces. By the time you’re back where you started, facing the way you began, you have made exactly one full turn. One full turn is 360°. So no matter how many sides — triangle, hexagon, 100-gon — the exterior angles always sum to 360°.

That single fact makes regular polygons trivial. A regular polygon splits its 360° of turning evenly among n corners, so each exterior angle = 360° ÷ n — and the interior angle is 180° − that. A regular hexagon: each exterior = 360 ÷ 6 = 60°, so each interior = 180 − 60 = 120°. No (n−2)·180 needed.

Framing inspired by AoPS Prealgebra.

Parallel lines + a transversal — the Z, F, and C angle pairs

Draw two parallel lines, then a third line slicing across both (a transversal). You get eight angles — but only two different sizes, and they alternate. Slide one crossing up the transversal until it lands on the other: every angle drops right onto a matching one. That’s why the pattern repeats.

Three quick shapes catch all the pairs you need. Trace the letter and the two angles it hugs are related:

So mark one angle and every other angle in the figure is either that same number or its supplement (180° minus it). The deeper power: this runs both ways — if a pair of corresponding angles turns out equal, the two lines must be parallel. That’s how a problem proves lines are parallel without a ruler.

Framing inspired by AoPS Prealgebra.

Casework — when the picture isn’t pinned down

Some angle problems don’t hand you a fixed figure. “An isosceles triangle has an angle of 70°” — but which angle? The 70° could be the odd one out, or one of the matching pair. Each guess is a different triangle.

Don’t pick one and hope. List every case, solve each, then collect the answers.

Two angles of an isosceles triangle are 70° and x°. The equal pair could be any of three things:

• Case 1: x matches the 70° → x = 70.
• Case 2: the two 70° angles are the pair → x = 180 − 70 − 70 = 40.
• Case 3: two x° angles are the pair → 2x + 70 = 180 → x = 55.

All three are real triangles, so all three count. Sum: 70 + 40 + 55 = 165.

The trap is stopping at the first case you draw. The phrase “sum of all possible values” is a flashing sign that more than one case survives. When the figure isn’t forced, split into cases, check each is valid, then combine.

The triangle inequality — a side can’t be too long

Take two sticks, 3 and 4 long, hinged at one end. Swing them open. The gap between the far ends can be almost anything — but only up to a point. Stretch the hinge flat and the gap maxes out at 3 + 4 = 7. Snap it shut and the gap shrinks to 4 − 3 = 1. So the third side of any triangle is always squeezed between those two numbers: bigger than the difference, smaller than the sum.

This one looks too obvious to bother with — and that is exactly why it catches good students. A problem hands you “a triangle with sides 2, 3, and 8” and you start computing, never noticing 2 + 3 = 5 can’t reach 8. There is no such triangle.

Where it earns its keep: counting the possible whole-number lengths of a missing side. Say two sides are 5 and 9. The third side x must satisfy 9 − 5 < x < 9 + 5, so 4 < x < 14. The whole numbers from 5 to 13 work — that’s 9 of them.

Framing inspired by Competition Math for Middle School (AoPS).

Every circle hides the same secret number. Roll any circle along the floor for one full turn, then measure how far it traveled: that distance is always almost exactly 3.14 times the width across. Big wheel, tiny coin — same ratio, every time. That number is π, and it’s the only thing standing between you and circle problems.

Where π comes from — and why area = πr²

First, what is π? Wrap a string once around any circle, then lay it flat against the width across. The string is always about 3.14 widths long. That number — the roll-distance divided by the diameter — is π, and it’s the same for a coin and a stadium. So C = π d isn’t a rule to memorize; it’s the definition of π rearranged.

Now the area. A = π r² looks like it fell from the sky. It didn’t. Slice the disk into thin pizza wedges and re-lace them.

The wedges, laid point-up then point-down in a row, settle into a near-rectangle. Its height is the radius r (the length of each wedge). Its base is half the rim — the top arcs and bottom arcs split the full roll 2πr in two, so the base is πr. Thinner wedges make a cleaner rectangle, and the area is just base × height:

A = (πr)(r) = π r².

That’s the formula — not memorized, but rebuilt from a sliced-up pizza.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

On a contest you can usually leave π in your answer — the choices will too. The arc and sector formulas aren’t new facts; they’re just “take this fraction of the whole circle.” A quarter circle is ¼ of everything; a 90° pie slice is ¼ of the area.

Circles packed into squares show up constantly — know these two by heart:

• Circle inscribed in a square of side s: the diameter equals s, so radius = s/2.
• Square inscribed in a circle of radius r: the diagonal equals 2r, so side = r√2.

Two circle facts that unlock a LOT of problems

Give a triangle three corner coordinates and there’s a way to get its area without drawing anything, without finding a base or a height — just plug the numbers into one formula and turn a crank. It’s called Shoelace, because of the criss-cross pattern you multiply in, like lacing a shoe.

Try it on a triangle with corners A=(0,0), B=(4,0), C=(0,3):

1. List the coordinates in order, repeating the first at the bottom.
2. Multiply down-right (x₁·y₂, x₂·y₃, x₃·y₁) and add: 0·0 + 4·3 + 0·0 = 12.
3. Multiply down-left (y₁·x₂, y₂·x₃, y₃·x₁) and add: 0·4 + 0·0 + 3·0 = 0.
4. Subtract, take the absolute value, halve: |12 − 0| / 2 = 6.

(Matches ½ · base · height = ½ · 4 · 3 = 6 ✓.) Shoelace works for ANY polygon, even concave ones.

One shortcut before you even Shoelace: if the triangle has a horizontal or vertical side, use that side as the base. Its length and the height are both just coordinate differences — you may not need the formula at all.

Pick’s Theorem — the magical dot-counting formula

Here’s a totally different way to find a polygon’s area when its corners sit on grid dots (lattice points). No coordinates, no multiplying. You just count two kinds of dots.

Count the dots below, then plug in.

● boundary dots (B = 8)  ·  ● interior dots (I = 5)

Plug in: Area = 5 + 8/2 − 1 = 5 + 4 − 1 = 8. No coordinates, no Shoelace — just dot-counting.

Which to use? Easy coordinates (especially off-grid) → Shoelace. A clean grid where you can SEE the dots → Pick’s. Both always agree.

Take a square sheet of paper. Fold it in half, then in half again so it’s a quarter the size. Now snip the folded corner with scissors and open it up. One little snip became four matching holes — arranged with perfect symmetry. That surprise is the whole topic: a fold is a mirror, and whatever you do on one side happens to every layer it stacks.

Reflections on a grid

A reflection is the same mirror idea on coordinates: one number flips, the other holds still. Reflect across a vertical mirror (like the y-axis) and the x flips while y stays put. Reflect across a horizontal mirror (the x-axis) and y flips while x stays.

Watch one point. Reflect (3, 2) across the y-axis: the 3 walks to the other side and becomes −3, the 2 doesn’t move — you land on (−3, 2). Reflect the same point across the x-axis instead and it’s the 2 that flips: (3, −2). Do every corner of a shape this way and the whole figure comes out mirror-reversed — left and right swap, the way letters read backwards in a mirror.

Rolling problems — two turns at once

When a shape rolls around the outside of another, its total spin comes from two sources, and forgetting one is the classic mistake:

• Rolling along edges = (distance traveled) ÷ (rolling shape’s perimeter) × 360°.
• Pivoting at corners = at each corner it tips over the exterior angle. Around a hexagon (exterior 60°), six corners add 6 · 60° = 360°.

Add both parts for the total rotation.

Blow up a photo to twice the width and you’d guess it uses twice the ink. It uses four times. Double every length and area doesn’t double — it quadruples. This single surprise is behind a whole family of contest problems, and most kids get it wrong the first time.

Two shapes are similar if they’re the same shape at different sizes — one is an enlarge-or-shrink copy of the other. Every matching angle is equal, every matching pair of sides shares the same ratio. Call that ratio the scale factor k.

Feel it with two squares. Side 3 and side 6, so k = 2:

• Perimeters: 12 and 24 — ratio 2 (that’s k).
• Areas: 9 and 36 — ratio 4 (that’s k²).

Sides double → area quadruples. Always 2² = 4, never 2.

See k, k², k³ in the tiles

Why does area scale by k² and volume by k³? Don’t take it on faith — count the little boxes. Take a 1×1 square. Double every length (k = 2) and the new square isn’t one big square — it’s a 2-by-2 grid of 4 copies of the original. Triple it (k = 3) and you get a 3-by-3 grid of 9. The grid is k tiles wide and k tiles tall, so it holds k × k = k² tiles. That’s the “four times the ink” surprise, made of countable squares.

Now go 3-D. Double a cube’s edge and it fills with a 2×2×2 stack of 8 little cubes — k × k × k = k³. Same counting, one dimension up. That is the whole story:

• Lengths × k — one row of tiles.
• Area × k² — a square grid of tiles.
• Volume × k³ — a cube stack of boxes.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

When are two triangles similar?

You don’t check all six things (3 angles + 3 ratios). Any ONE of these is enough:

AA is the workhorse. Two triangles sharing an angle, or formed by parallel lines cut by a transversal, are similar — set up the side-ratio equation and solve. And the classic: a line parallel to one side of a triangle cuts off a smaller, similar triangle.

Same height → area ratio = base ratio (no height needed)

This one isn’t about similar shapes — it’s about triangles that share a height but sit on different bases. It saves you from ever finding the height.

Take a triangle. Put a point on the bottom that splits it into a piece of length 2 and a piece of length 5. Draw a line from that point up to the top corner. Now you have two skinny triangles side by side.

The slow way. Pick a height — say 6. Left triangle: ½ · 2 · 6 = 6. Right: ½ · 5 · 6 = 15. Ratio = 6 : 15 = 2 : 5. Notice the 6 showed up in both and cancelled.

The fast way. Both triangles reach the same top corner, so they share a height. Same height means area is just ½ · base · (same height) — so the areas land in the exact ratio of the bases. 2 : 5. Done.

One famous case: a median (corner to the midpoint of the far side) splits the base into two equal halves, so it splits the triangle into two equal-area pieces every time.

Two triangles with the same height have areas in the ratio of their bases — skip the height entirely.

(Same-altitude area ratio adapted from Problem Solving via the AMC, Australian Maths Trust.)