STEP 1 OF 22 · Lesson Opening

Today: 3D Coordinates & 3D Pythagoras

From the flat page into space. One coordinate axis added, one extra Pythagoras applied — and a third of every AIMO 3D solid problem dissolves into algebra.

📌 What you will learn today

Topic: Place a point as $(x, y, z)$; compute distance $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$; cube space diagonal $= s\sqrt{3}$; cuboid space diagonal $= \sqrt{a^2 + b^2 + c^2}$; unfold a surface path with a 2D Pythagoras.
Category: Geometry · 3D (G3D) — sub-topic Solids · Pass 1 · 3D Coordinates + 3D Pythagoras (Universal Tool).
Solves these AIMO problems: 2018 Q5
Plus dozens of cube / cuboid / tetrahedron warm-ups in the past 20 years — every time you see "find the longest segment inside this box", today's tools close the question in two lines.
AoPS Reference: Introduction to Geometry by Richard Rusczyk, Chapter 18 (3D Geometry).
Why this matters: 3D Pythagoras is just 2D Pythagoras applied twice. Once you see that, every "space diagonal", "longest insect path on a cube", or "vertex-to-vertex distance in a cuboid" question collapses to one calculation. This is the single biggest leverage move in the 3D unit — Pass 1 universal tool.
Time required: About 70–90 minutes for the full lesson, plus practice problems.

How this lesson is structured

Phase 0 (Steps 2–3): Prerequisites — 2D Pythagoras review and "what is a coordinate" in 3D.
Phase 1 (Steps 4–6): Visual intuition — axes, cube body diagonal, two points in space.
Phase 1.5 (Step 7): Formula handbook (3D distance, cube diagonal, cuboid diagonal).
Phase 2 (Steps 8–10): Three guided derivations (distance formula, cube diagonal, cuboid diagonal).
Phase 3 (Steps 11–15): Five worked examples ⭐⭐ → ⭐⭐⭐⭐⭐.
Phase 4 (Step 16): Five practice problems (P1–P5) with Hint / Answer / Solution.
Phase 5 (Step 17): One real AIMO past paper — AIMO 2018 Q5 (full Observe / Strategy).
Phase 5.5 (Step 18): Synthesis problem — insect on a cube (unfold + 2D Pythagoras).
Step 19: Atomic-skill matrix + 8 micro-validations.
Steps 20–21: Cheat sheet, ⭐ self-assessment.
Step 22: Error book + bridge to Part 3 (Solids — Volume & Surface Area).

Pedagogy: Each worked example and AIMO problem is wrapped in the same 5-step Observe template — keywords / known / unknown / intermediate / hidden constraint — so the next unseen 3D problem in the contest hall hands you a checklist instead of a guess.

STEP 2 OF 22 · Phase 0 · Prerequisite

Prerequisite ① — 2D Pythagoras must be a reflex

Every 3D distance formula on this page is just 2D Pythagoras applied twice. If 2D isn't automatic, the rest won't lock in.

The one rule

$a^2 + b^2 = c^2$

Pythagoras in 2D

$c$ is always the hypotenuse — the side opposite the right angle.

Three classic right triangles you should know cold:

$3$–$4$–$5$ ($9 + 16 = 25$)
$5$–$12$–$13$ ($25 + 144 = 169$)
$8$–$15$–$17$ ($64 + 225 = 289$)

Mini-problem (verify you remember)

Right triangle, legs $5$ and $12$. Hypotenuse?

Right triangle, hypotenuse $25$, one leg $7$. The other leg?

🔑 Speed trick. In an AIMO 3D problem, the moment you see $9$ and $16$ floating in your arithmetic, expect a $3$–$4$–$5$ to appear. The moment you see $25$ and $144$, expect $5$–$12$–$13$ — and that is exactly the cuboid in WE 3.

STEP 3 OF 22 · Phase 0 · Prerequisite

Prerequisite ② — A 3D point has three numbers $(x, y, z)$

Two numbers locate a point on a page. Three numbers locate a point in a room.

The 3D coordinate system uses three mutually perpendicular axes:

$x$ — how far right from the origin
$y$ — how far back from the origin
$z$ — how far up from the origin

A point is written $(x, y, z)$. Origin is $(0, 0, 0)$. The point $(2, 5, 3)$ sits $2$ right, $5$ back, $3$ up.

Convention. In every AIMO 3D problem on this page we use the right-handed system — the same orientation a maths textbook uses. The answer to a distance problem does not depend on which orientation you choose, so don't waste time worrying about it.

Mini-problem

A cube has one vertex at the origin and edges of length $4$ along each axis. What are the coordinates of the vertex diagonally opposite the origin?

A point sits on the floor (the $xy$-plane). What is its $z$-coordinate?

STEP 4 OF 22 · Phase 1 · Visual

Visual ① — The 3D coordinate axes

Three perpendicular axes. A single point pinned in space.

Walk the dashed path: 3 along $x$, then 4 along $y$, then 5 up along $z$ to reach $P$. Three numbers, one point.

🔑 Mental routine. Whenever a problem hands you a vertex of a 3D shape, ask: "what are its $(x, y, z)$?" If you cannot answer that quickly, put the origin at a clever vertex of the solid and read the rest off the edge lengths. Coordinates are chosen, not given.

STEP 5 OF 22 · Phase 1 · Visual

Visual ② — A cube and its space diagonal

The single most-tested 3D segment in AIMO.

Cube with edge $s$. The teal dashed segment $AC$ is a face diagonal (length $s\sqrt{2}$). The red segment $AG$ is the space diagonal (length $s\sqrt{3}$). The triangle $ACG$ is right-angled at $C$ — and that single right triangle is the whole proof.

STEP 6 OF 22 · Phase 1 · Visual

Visual ③ — Two points $P, Q$ and the segment $PQ$ in 3D

Imagine a box around the segment with edges parallel to the axes. The segment is the box's space diagonal.

The red segment $PQ$ is the space diagonal of the axis-aligned box with edge lengths $|\Delta x| = 4$, $|\Delta y| = 3$, $|\Delta z| = 3$. Length $PQ = \sqrt{16 + 9 + 9} = \sqrt{34}$.

STEP 7 OF 22 · Phase 1.5 · Handbook

Formula handbook — what to memorise this week

All four formulas you will use today, in one place. Memorise the boxed surd values too.

① 3D Distance Formula

$d \;=\; \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

From $P(x_1, y_1, z_1)$ to $Q(x_2, y_2, z_2)$

Squared form: $d^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$. AIMO usually asks for $d^2$ — leave the square root for the very last step.

② Cube space diagonal

$\text{diag}_{\text{cube}}(s) \;=\; s\sqrt{3}$

Special case of ① with $\Delta x = \Delta y = \Delta z = s$

Memorise the squared form: $\text{diag}^2 = 3s^2$.

③ Cuboid space diagonal

$\text{diag}_{\text{cuboid}}(a, b, c) \;=\; \sqrt{a^2 + b^2 + c^2}$

Watch for "Pythagorean quadruples": $1,2,2 \to 3$; $\;2,3,6 \to 7$; $\;3,4,12 \to 13$; $\;4,4,7 \to 9$.

④ 2D projections onto coordinate planes

$P(x,y,z) \;\;\to\;\; (x,y,0) \;\text{ on the } xy\text{-plane}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\to\;\; (x,0,z) \;\text{ on } xz; \;\;\;\to\;\; (0,y,z) \;\text{ on } yz$

Setting one coordinate to zero "drops" $P$ onto the plane.

Surd table (memorise — used every problem)

$\sqrt{2} \approx 1.414$ — face diagonal of unit square
$\sqrt{3} \approx 1.732$ — space diagonal of unit cube
$\sqrt{5} \approx 2.236$ — appears in the ant-on-cube synthesis
$\sqrt{6} \approx 2.449$ — appears in regular tetrahedron heights

STEP 8 OF 22 · Phase 2 · Derivation

Derivation ① — 3D distance from two applications of 2D Pythagoras

Read this once carefully. Every other derivation today is a special case of this one.

Setup. Two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. Drop the foot $F$ of $Q$ onto the horizontal plane through $P$ — so $F = (x_2, y_2, z_1)$.

Step A (in the horizontal plane). The points $P$ and $F$ have the same $z$, so they lie on the same horizontal plane. Their horizontal distance is the 2D Pythagoras of $\Delta x$ and $\Delta y$:

$PF^2 \;=\; (\Delta x)^2 + (\Delta y)^2$

Step B (vertical right triangle). $F$ sits directly below $Q$, so $FQ$ is vertical with length $|\Delta z|$. The triangle $PFQ$ has a right angle at $F$ (horizontal meets vertical). One more 2D Pythagoras:

$PQ^2 \;=\; PF^2 + FQ^2 \;=\; (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$

Take the square root and you have the formula. The whole thing is just 2D Pythagoras applied twice — once in the plane, once vertically.

🔑 Why "general". $P$ and $Q$ are arbitrary. The construction never used numbers — only that the foot $F$ is directly below $Q$. So the formula works for any two points in space. Every cube, cuboid and tetrahedron problem in the rest of this lesson is a special case.

STEP 9 OF 22 · Phase 2 · Derivation

Derivation ② — Cube space diagonal $= s\sqrt{3}$

A concrete special case of the distance formula.

Place a cube of edge $s$ at the origin. Opposite vertices are $A = (0,0,0)$ and $G = (s,s,s)$.

Apply the 3D distance formula:

$AG \;=\; \sqrt{(s-0)^2 + (s-0)^2 + (s-0)^2} \;=\; \sqrt{3s^2} \;=\; s\sqrt{3}.$

Or, by the two-step picture: $AC$ is the bottom-face diagonal, length $s\sqrt{2}$ (2D Pythagoras on a square). Then $\triangle ACG$ has a right angle at $C$, with $CG = s$ vertical, so

$AG^2 \;=\; AC^2 + CG^2 \;=\; 2s^2 + s^2 \;=\; 3s^2.$

🔑 Memorise: cube edge $s$ → face diagonal $s\sqrt{2}$ → space diagonal $s\sqrt{3}$. The square roots of $1, 2, 3$ for edge, face, space.

STEP 10 OF 22 · Phase 2 · Derivation

Derivation ③ — Cuboid space diagonal $= \sqrt{a^2 + b^2 + c^2}$

Same proof, three different edge lengths.

A cuboid with edges $a, b, c$ has opposite vertices $(0,0,0)$ and $(a, b, c)$. Apply the 3D distance formula directly:

$d^2 \;=\; a^2 + b^2 + c^2.$

Famous Pythagorean quadruples (memorise — they appear constantly):

$1^2 + 2^2 + 2^2 = 9 \Rightarrow d = 3$
$2^2 + 3^2 + 6^2 = 49 \Rightarrow d = 7$
$1^2 + 4^2 + 8^2 = 81 \Rightarrow d = 9$
$3^2 + 4^2 + 12^2 = 169 \Rightarrow d = 13$ (this is WE 3)
$4^2 + 13^2 + 16^2 = 441 \Rightarrow d = 21$

Common slip. In an AIMO question, edges may be given as face diagonals rather than edges. Read the wording carefully. If a face diagonal $f$ is given for a square face of side $s$, then $f = s\sqrt{2}$ and $s = f/\sqrt{2}$ — substitute back before using formula ③.

STEP 11 OF 22 · Phase 3 · Worked Example 1

WE 1 — Plug-in to the 3D distance formula (⭐⭐)

Find the distance between $A(1, 2, 3)$ and $B(4, 6, 3)$.

Solution

Compute the three differences first.

$\Delta x = 4 - 1 = 3$ $\Delta y = 6 - 2 = 4$ $\Delta z = 3 - 3 = 0$ — the two points share a $z$-coordinate $AB^2 = 3^2 + 4^2 + 0^2 = 9 + 16 = 25$ $AB = \sqrt{25} = 5$

Comment. Because $\Delta z = 0$, the two points lie in the same horizontal plane and the 3D formula collapses to 2D Pythagoras on the $3$–$4$–$5$ triangle. This is a common AIMO subcase — when one coordinate matches, drop a dimension.

Your answer — distance $AB$

Answer: $AB = 5$.

STEP 12 OF 22 · Phase 3 · Worked Example 2

WE 2 — Cube space diagonal (⭐⭐⭐)

A cube has edge length $6$. Find the square of the length of its space diagonal.

Solution

Place the cube with one vertex at $(0,0,0)$ and opposite vertex at $(6,6,6)$.

$d^2 = 6^2 + 6^2 + 6^2 = 3 \times 36 = 108$ $d = 6\sqrt{3}$

Or, use the formula directly: $d = s\sqrt{3} = 6\sqrt{3}$, so $d^2 = 108$.

AIMO note. AIMO answers must be integers $0$–$999$, so when an answer is a surd, the question always asks for $d^2$, $d^2 + k$, or some related integer. Stay on guard for this rewrite.

Your answer — $d^2$

Answer: $d^2 = 108$ (so $d = 6\sqrt{3}$).

STEP 13 OF 22 · Phase 3 · Worked Example 3

WE 3 — Cuboid $3 \times 4 \times 12$ (⭐⭐⭐⭐)

A box has dimensions $3 \times 4 \times 12$. Find the length of its space diagonal.

Solution

Use formula ③:

$d^2 = 3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169$ $d = \sqrt{169} = 13$

The hidden structure. Note $3^2 + 4^2 = 25$, so the floor diagonal is $5$. Then $5^2 + 12^2 = 169 = 13^2$ — the classic $5$–$12$–$13$ triangle reappears as the right triangle "floor-diagonal + vertical edge + space-diagonal". This is the canonical AIMO-friendly cuboid; expect to see it again.

Your answer — diagonal length

Answer: $d = 13$.

STEP 14 OF 22 · Phase 3 · Worked Example 4

WE 4 — Regular tetrahedron from a cube (⭐⭐⭐⭐)

A cube has edge length $4$. Inside it, pick four vertices that form a regular tetrahedron: $A(0,0,0)$, $B(4,4,0)$, $C(4,0,4)$, $D(0,4,4)$. Find $AB^2$.

Solution

Apply the squared 3D distance formula to $A$ and $B$:

$AB^2 = (4-0)^2 + (4-0)^2 + (0-0)^2$ $\;\;\;\;\;\; = 16 + 16 + 0 = 32$

Why this works. $A$ and $B$ are opposite corners of the bottom face — they are joined by a face diagonal of the cube. Face diagonal of a cube of edge $s$ has squared length $2s^2 = 2 \cdot 16 = 32$. Check: $\sqrt{32} = 4\sqrt{2}$.

Tetrahedron fact (bonus). Every pair from $\{A, B, C, D\}$ also gives squared distance $32$, so the tetrahedron has all edges $4\sqrt{2}$ — it is regular. This is the cleanest way AIMO embeds a regular tetrahedron in a problem.

Your answer — $AB^2$

Answer: $AB^2 = 32$ (so $AB = 4\sqrt{2}$).

STEP 15 OF 22 · Phase 3 · Worked Example 5

WE 5 — Insect on a cube (unfold) (⭐⭐⭐⭐⭐)

A cube has edge length $6$. An ant starts at one vertex and crawls along the surface of the cube to the diagonally opposite vertex. Find the square of the shortest surface distance.

Strategy — unfold

The ant cannot fly through the cube; the straight space-diagonal $6\sqrt{3}$ is unreachable. But any path along the surface that crosses exactly two faces can be straightened by unfolding those two faces into one flat rectangle.

Unfold the front face $6 \times 6$ and the top face $6 \times 6$ side-by-side. They form a rectangle $6 \times 12$. The start vertex sits at one corner; the end vertex sits at the opposite corner of this rectangle.

Two adjacent faces unfolded into a $6 \times 12$ rectangle. The shortest surface path is the straight red diagonal.

Solve the flat rectangle by 2D Pythagoras

$d^2 = 6^2 + 12^2 = 36 + 144 = 180$ $d = \sqrt{180} = 6\sqrt{5}$

Why this is shortest. Any two-face path can be unfolded — straight lines beat any bent path in the plane. Three-face paths are also possible, but they unfold to a longer rectangle (e.g. $6 \times 18$ gives $d = \sqrt{360}$, worse than $\sqrt{180}$).

Your answer — $d^2$

Answer: $d^2 = 180$ (so $d = 6\sqrt{5}$).

STEP 16 OF 22 · Phase 4 · Practice

Five practice problems — P1 to P5

Mixed difficulty. Try each cold before clicking Hint. Then check the Answer, and only open the Solution if you got it wrong or need to compare technique.

P1 · ⭐⭐

Find the distance between $A(0,0,0)$ and $B(1,2,2)$.

Plug into the squared 3D distance formula. Note $1^2 + 2^2 + 2^2$ is a famous Pythagorean quadruple.

$d = 3$

$d^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 \Rightarrow d = 3$. This is the $(1,2,2)\to 3$ quadruple.

P2 · ⭐⭐

A cube has edge $10$. Find the square of its space diagonal.

Cube formula: $d^2 = 3s^2$.

$d^2 = 300$

$d^2 = 3 \times 10^2 = 300$, so $d = 10\sqrt{3}$.

P3 · ⭐⭐⭐

Cuboid $2 \times 3 \times 6$. Find its space diagonal.

$d^2 = a^2 + b^2 + c^2$. The arithmetic is a famous quadruple.

$d = 7$

$d^2 = 4 + 9 + 36 = 49 \Rightarrow d = 7$. This is the $(2,3,6) \to 7$ quadruple.

P4 · ⭐⭐⭐

Points $P(2, -1, 5)$ and $Q(6, 2, -7)$. Find $PQ$.

Careful with negative coordinates — square the difference, not the values. $\Delta z = -7 - 5 = -12$, and $(-12)^2 = 144$.

$PQ = 13$

$\Delta x = 4, \Delta y = 3, \Delta z = -12$. $PQ^2 = 16 + 9 + 144 = 169 \Rightarrow PQ = 13$. (The $3$-$4$-$12$-$13$ cuboid in disguise.)

P5 · ⭐⭐⭐⭐

A rectangular room measures $5 \times 6 \times \text{height}$. Its space diagonal is $\sqrt{86}$. Find the height.

Set up $5^2 + 6^2 + h^2 = 86$ and solve for $h$.

$h = 5$

$25 + 36 + h^2 = 86 \Rightarrow h^2 = 25 \Rightarrow h = 5$. (Negative height is rejected.)

STEP 17 OF 22 · Phase 5 · AIMO Past Paper

AIMO 2018 · Q5 — Diagonals inside a cube-derived polyhedron

Real AIMO problem · 3 marks · officially classified G3D. We use 3D coordinates to model and count.

AIMO 2018 · Q5 · 3 marks

Statement

Each edge of a cube is marked with its trisection points. Each vertex $v$ of the cube is cut off by a plane that passes through the three trisection points closest to $v$. The resulting polyhedron has $24$ vertices. How many diagonals joining pairs of these vertices lie entirely inside the polyhedron?

Observe — 5-step breakdown

Observe template (do this every AIMO problem)

① Keywords

"cube", "trisection points", "cut off", "polyhedron", "24 vertices", "diagonals", "entirely inside". The phrase entirely inside is the load-bearing word — diagonals on a face don't count.

② Known

A cube has 8 vertices. After cutting each corner with a plane through the 3 nearest trisection points, each vertex becomes a small triangular face. Each of the original 6 cube-faces becomes a regular octagon (its 4 corners are cut off — leaving 8 sides). So the polyhedron has 8 triangular + 6 octagonal faces = 14 faces, and $8 \times 3 = 24$ vertices.

③ Unknown

The number of internal diagonals — pairs of vertices whose connecting segment lies strictly in the interior, never on a face.

④ Intermediate quantities

Total pairs $= \binom{24}{2} = 276$. Pairs that share a face (either a triangular face or an octagonal face) — those don't count as interior diagonals. Total interior = $\binom{24}{2}$ − (edges) − (face diagonals on each face).

⑤ Hidden constraint

Every vertex sits on exactly $3$ faces — one triangle and two octagons. The two octagons at a vertex share one edge (a triangular face). So the number of "face-neighbours" of a vertex is $2 + 7 + 6 = $ … careful here: the 2 in the triangle, then $7$ more on one octagon, then $6$ new on the other octagon (the two octagons share 2 vertices already counted in the triangle).

Strategy

Count from a single vertex, multiply by $24$, divide by $2$.

From a vertex $v$, there are $23$ other vertices.
$7$ lie on the same octagonal face as $v$ (octagon has 8 vertices total).
$6$ more lie on the other octagonal face at $v$ (the two octagons share 2 vertices — the triangle-edge).
$2$ lie on the triangle at $v$ — but those are already counted on the two octagons.
So $v$ shares a face with $7 + 6 = 13$ other vertices.
Remaining $23 - 13 = 10$ vertices are reached by an interior diagonal from $v$.
Total interior diagonals $= \dfrac{24 \times 10}{2} = 120$ (divide by 2 because each diagonal joins two vertices).

Why this method is general

The "count from one vertex × $n$ / $2$" technique works for any polyhedron's diagonal count: classify the other vertices into (face-sharing) and (interior), use symmetry to do it from one vertex only, then halve. No coordinates were strictly needed — but if you weren't sure about the face-sharing counts, you could place the cube at the origin with edge $3$ (so trisection points are integers) and read off coordinates directly. That's the 3D coordinate fallback.

Your answer (integer 0–999)

Skill link. Atomic skills D2-A2 (3D distance / coordinate placement) and D2-A3 (3D Pythagoras for face vs space diagonal classification). Used as a fallback to verify which vertex-pairs lie on a common face by checking whether their coordinates satisfy a face equation.

💡 Hints — click to open

For each vertex, classify the other 23 vertices into (shares a face with $v$) vs (interior diagonal from $v$). Sum over all $v$, divide by 2.

From $v$, count vertices on its triangle face (2) and on its two octagonal faces (7 + 6, with overlaps already counted). Net face-sharing = 13.

All 24 vertices are equivalent under the symmetry of the polyhedron — same count from every vertex. So total interior diagonals = $24 \times 10 / 2 = 120$.

Last resort:

Full solution

The polyhedron has $14$ faces ($8$ triangles + $6$ octagons) and $24$ vertices. Each vertex is on exactly one triangle and two octagons, the two octagons sharing the triangle-edge.

From a fixed vertex $v$:

total other vertices $= 23$ vertices on the same octagon (#1) as $v$ $= 7$ vertices on the other octagon (#2) as $v$ that aren't already counted $= 6$ vertices on the triangle at $v$ already lie on octagons #1 or #2 — counted face-sharing neighbours $= 7 + 6 = 13$ interior-diagonal partners $= 23 - 13 = 10$

By symmetry, every vertex donates $10$ interior diagonals; each diagonal joins $2$ vertices, so total $= 24 \cdot 10 / 2 = 120$.

Answer: $\boxed{120}$.

STEP 18 OF 22 · Phase 5.5 · Synthesis

Synthesis — Ant on a cube edge $12$

Combine: unfolding + 2D Pythagoras + the squared-form-for-AIMO trick.

Synthesis

A cube has edge $12$. An ant starts at one vertex and walks along the surface of the cube to the diagonally opposite vertex. Find the square of the shortest surface distance.

Strategy

The straight-line space diagonal $12\sqrt{3}$ is interior — unreachable by the ant.
Any surface path crossing exactly two faces unfolds into a flat rectangle.
Unfold one face plus an adjacent face: rectangle $12 \times 24$.
Straight line corner-to-corner is the shortest path (any bent path in a plane is longer).

Compute

unfolded rectangle: $12 \times 24$ $d^2 = 12^2 + 24^2 = 144 + 576 = 720$ $d = \sqrt{720} = 12\sqrt{5}$

Verify shortness

Three-face unfolding gives rectangle $12 \times 36$, $d^2 = 144 + 1296 = 1440$ — worse. Any unfolding crossing the diagonal edge gives the same $12 \times 24$ rectangle by symmetry. So $720$ is the unique minimum.

Your answer — $d^2$

Answer: $d^2 = 720$ (so $d = 12\sqrt{5}$).

STEP 19 OF 22 · Atomic Skills + Micro-validation

Atomic-skill matrix + 8 micro-validations

The four skills you must own before moving to Part 3. Two quick validations each.

Code	Skill	What it sounds like
`D2-A1`	3D coordinate placement	Place a cube/cuboid vertex at the origin and write down the coordinates of every other vertex.
`D2-A2`	3D distance formula	$d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$ — compute $d$ or $d^2$ from two points.
`D2-A3`	3D Pythagoras / space diagonal	Cube $\to s\sqrt{3}$; cuboid $\to \sqrt{a^2+b^2+c^2}$. Recognise face diagonal vs space diagonal in a triangle.
`D2-A4`	Unfold a surface path	Two adjacent faces of a cube/cuboid unfold to a rectangle; shortest surface path is the straight diagonal in that rectangle.

Micro-validations

① D2-A1: A cube of edge $5$ has one vertex at the origin. What is the coordinate of the diagonally opposite vertex? (type as 5,5,5)

② D2-A2: Distance from $(0,0,0)$ to $(2,3,6)$?

③ D2-A2: Squared distance from $(1,1,1)$ to $(4,5,1)$?

④ D2-A3: Cube edge $5$. Squared space diagonal?

⑤ D2-A3: Cuboid $1 \times 4 \times 8$. Space diagonal length?

⑥ D2-A3: Cuboid $4 \times 13 \times 16$. Space diagonal length?

⑦ D2-A4: Cube edge $3$. Ant walks on surface from one vertex to the opposite vertex. Square of the shortest path?

⑧ D2-A4: Cuboid $1 \times 1 \times 4$. Ant walks from $(0,0,0)$ on the surface to $(1,1,4)$. Unfold the two adjacent $1 \times 4$ faces — square of the shortest path?

STEP 20 OF 22 · Cheat Sheet

One-page cheat sheet

Take a screenshot. Re-skim it on Sunday morning before the mock exam.

① 3D Distance

$d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$

Three differences, three squares, one square root. AIMO usually asks for $d^2$ — leave the root for last. Watch the sign of each difference but remember squaring removes it: $(-12)^2 = 144$, same as $12^2$.

Vertex-coordinate table for a cube of edge $s$

Place the cube with one vertex at the origin and edges along the axes — then the 8 vertices are:

$A_1 = (0,0,0)$, $A_2 = (s,0,0)$, $A_3 = (s,s,0)$, $A_4 = (0,s,0)$
$A_5 = (0,0,s)$, $A_6 = (s,0,s)$, $A_7 = (s,s,s)$, $A_8 = (0,s,s)$

Edges: pairs that differ in exactly one coordinate. Face diagonals: differ in exactly two. Space diagonals: differ in all three.

② Cube space diagonal

$d = s\sqrt{3}, \;\;\; d^2 = 3s^2$

Cube edge $s$. Memorise edge $\to$ face diag $s\sqrt{2}$ $\to$ space diag $s\sqrt{3}$.

③ Cuboid space diagonal

$d = \sqrt{a^2 + b^2 + c^2}$

Famous quadruples: $(1,2,2)\to 3$, $(2,3,6)\to 7$, $(1,4,8)\to 9$, $(3,4,12)\to 13$, $(4,13,16)\to 21$.

④ Surface path: UNFOLD

unfold $\to$ rectangle $\to$ 2D Pythagoras

Cube edge $s$, opposite-vertex ant path: unfold two adjacent faces $\to$ $s \times 2s$ rectangle $\to$ $d = s\sqrt{5}$.

Caution: the path must stay on the surface. Don't use $s\sqrt{3}$ (that's interior).

Surd quick-reference

$\sqrt{2} \approx 1.41$; $\sqrt{3} \approx 1.73$; $\sqrt{5} \approx 2.24$; $\sqrt{6} \approx 2.45$.

Coordinate-placement reflex

For any cube/cuboid problem: put one vertex at $(0,0,0)$, edges along the axes. Then every vertex has $0$s and edge-lengths as its coordinates. Solving becomes substitution.

Edge / Face-Diagonal / Space-Diagonal classifier table

For any pair of vertices $(P, Q)$ on a cube or cuboid placed at the origin, count how many of the three coordinates differ:

Coordinates differing	Segment type	Length on a cube of edge $s$
1 of 3	Edge	$s$
2 of 3	Face diagonal	$s\sqrt{2}$
3 of 3	Space diagonal	$s\sqrt{3}$

This single table classifies the $\binom{8}{2} = 28$ vertex-pairs of a cube into $12$ edges, $12$ face diagonals (2 per face × 6 faces) and $4$ space diagonals — a useful sanity check in any counting problem.

Common AIMO twists to expect

Twist 1. Coordinates given negative — the differences are unchanged after squaring. Don't waste time choosing "the right" orientation; squaring handles it.
Twist 2. Edge given as a face diagonal — divide by $\sqrt{2}$ to recover the edge before using formula ③.
Twist 3. Asked for $d^2$ instead of $d$ — AIMO's typical way to force an integer answer when the geometry produces a surd.
Twist 4. Surface path on a cube/cuboid — the answer is never $s\sqrt{3}$. Always unfold.
Twist 5. "Find a missing coordinate" — set up the distance squared as an equation and solve a quadratic in the unknown coordinate.

STEP 21 OF 22 · Self-assessment

⭐ Self-assessment — rate each atomic skill 1 to 5

Be honest. Anything below 3 stars needs a re-attempt today.

D2-A1 · Place vertices of a cube/cuboid in coordinates.

D2-A2 · 3D distance formula — plug-and-compute $d$ or $d^2$.

D2-A3 · 3D Pythagoras for cube and cuboid space diagonals.

D2-A4 · Unfold surface path → 2D Pythagoras.

STEP 22 OF 22 · Error Book + Wrap-up

📒 Error book + bridge to Part 3

Every question you missed is recorded here. Re-attempt before Sunday.

Your errors this session

No errors yet — submit some practice problems to populate this list.

Bridge → Part 3 (Solids: Volume & Surface Area). Today we measured straight lines inside 3D shapes. Next, we measure the space they enclose and the skin that wraps them — volume and surface area. The 3D Pythagoras you just learned is the foundation for slant heights of pyramids and cones, which is how almost every "find the volume" AIMO problem begins.

Before Sunday Mock:

Re-attempt every item in your error book until you get it right under 2 minutes cold.
Re-skim the Cheat Sheet (Step 20). Memorise the cube/cuboid surd table.
Make sure all four atomic skills are 3-star on your self-assessment.

▶ Next: Part 3 — Solids: Volume & Surface Area