STEP 1 OF 22 · Lesson Opening
Today: 3D Foundations — Cube, Cuboid, Tetrahedron, Sphere
Five atomic 3D skills — the workhorse toolkit that unlocks AIMO solids questions. Volume, surface area, nets and symmetry, all in one lesson.
What you will learn today
How this lesson is structured
- Phase 0 (Step 2): Prerequisites — cube/cuboid parts, tetrahedron parts, Pythagoras, sphere formulas.
- Phase 1 (Step 3): Visual intuition — three SVG diagrams: unfolded cube net, tetrahedron projection, sphere with great circle.
- Phase 1.5 (Step 4): Formula handbook — five formulas as one cheat sheet.
- Phase 2 (Steps 5–7): Three guided derivations — cuboid SA from net, tetrahedron edge → volume, sphere formulas.
- Phase 3 (Steps 8–12): Five Worked Examples (⭐ → ⭐⭐⭐⭐⭐), each fully written.
- Phase 4 (Step 13): Five practice problems with hints + solutions.
- Phase 5 (Steps 14–15): Two real AIMO past papers in exam format with the 5-step Observe template.
- Phase 5.5 (Step 16): One synthesis problem (cube + inscribed sphere).
- Steps 17–21: Atomic skill matrix + cheat sheet bridge.
- Step 22: Summary, ⭐ self-rating, error-book preview.
Pedagogical note: The hardest part of 3D is matching the formula to the figure. Read the cue words: edge of cube → \(s^3\); three different side lengths → \(abc\); regular tetrahedron → \(s^3/(6\sqrt 2)\); radius → \(\tfrac{4}{3}\pi r^3\); unfolded → net.
STEP 2 OF 22 · Phase 0 · Prerequisites
Phase 0 — Prerequisites you must own
Five micro-skills. If any of these feel slow, pause and revise before pushing on.
P0.1 — Cube parts: 12 edges, 6 faces, 8 vertices
A cube of edge \(s\) has \(12\) edges (all length \(s\)), \(6\) congruent square faces (each area \(s^2\)), and \(8\) vertices. Three edges meet at every vertex, mutually perpendicular.
Check
How many edges does a cube have?
P0.2 — Cuboid (rectangular box) parts
A cuboid with edges \(a, b, c\) has three pairs of opposite faces with areas \(ab\), \(bc\), \(ca\). The space (interior) diagonal has length \(\sqrt{a^2+b^2+c^2}\). The three face diagonals have squared lengths \(a^2+b^2\), \(b^2+c^2\), \(a^2+c^2\).
Check
For a cuboid \(2 \times 3 \times 6\), what is the squared space diagonal \(a^2+b^2+c^2\)?
P0.3 — Regular tetrahedron parts
A regular tetrahedron has \(4\) congruent equilateral-triangle faces, \(6\) equal edges, and \(4\) vertices. All edges have the same length \(s\); each face has area \(\tfrac{\sqrt 3}{4}s^2\).
Check
How many faces does a regular tetrahedron have?
P0.4 — 2D Pythagoras still rules
Even in 3D, lengths are found by repeated 2D Pythagoras. The space diagonal of a \(a \times b \times c\) cuboid is just two Pythagoras steps: first the face diagonal \(\sqrt{a^2+b^2}\), then combine with \(c\).
Check
Cube of edge \(6\): what is the space diagonal? (Give a decimal.)
P0.5 — Sphere of radius \(r\)
Volume \(V = \tfrac{4}{3}\pi r^3\); surface area \(S = 4\pi r^2\). A great circle (slicing the sphere through its centre) has radius \(r\) and area \(\pi r^2\).
Check
Sphere of radius \(3\): what is the surface area \(S\), divided by \(\pi\)? (Enter just the number multiplying \(\pi\).)
P0.6 — Equilateral triangle area
The equilateral triangle of side \(s\) has area \(\tfrac{\sqrt 3}{4}s^2\). You will use this for every tetrahedron face. Memorise: \(\tfrac{\sqrt 3}{4}\) is the multiplier.
Check
Equilateral triangle side \(4\): area? (Decimal, two places.)
P0.7 — Volume of a pyramid \(= \tfrac{1}{3} B h\)
For any pyramid with base area \(B\) and perpendicular height \(h\), the volume is \(\tfrac{1}{3} B h\). The regular tetrahedron is a special case where the base is equilateral and the apex sits above the centroid.
Check
A square pyramid with base side \(6\) and height \(5\) has volume \(?\)
Why these prerequisites matter. Each row of the cheat sheet you will see at Step 18 is built on top of one or two of these prerequisites. Pythagoras, equilateral-triangle area, and pyramid volume are the three calculations that recur in every derivation. If they feel slow, no advanced trick will save time on the exam.
STEP 3 OF 22 · Phase 1 · Visual Intuition
Phase 1 — Three pictures that unlock 3D
If you can sketch these three diagrams from memory, half the AIMO 3D problems become routine.
Picture 1 — Cube unfolded into its net (cross of six squares)
A cube's surface is six squares; unfolded, they form the cross above. Surface area of the cube of edge \(s\) is therefore \(6s^2\).
Picture 2 — Regular tetrahedron (3D projection)
All four faces are equilateral triangles; all six edges have length \(s\); four vertices, three meeting at every corner. Volume \(V = s^3 / (6\sqrt 2)\).
Picture 3 — Sphere with a great circle
Sphere: \(V = \tfrac{4}{3}\pi r^3\), \(S = 4\pi r^2\). The great circle has area \(\pi r^2\) — useful for cross-section problems.
STEP 4 OF 22 · Phase 1.5 · Formula Handbook
Phase 1.5 — Five atomic formulas, one cheat sheet
Read each formula and immediately think "what triggers this one?".
D-A1 · Cube and cuboid
Cube: \(V = s^3\), \(\;\; \text{SA} = 6s^2\)
Cuboid \(a \times b \times c\)
\(V = abc, \;\; \text{SA} = 2(ab + bc + ca)\)
Decode the variables
s— edge length of the cube (all 12 edges are equal).a, b, c— the three distinct edge lengths of the cuboid.SA= sum of the six face areas (or use the net).
D-A2 · Regular tetrahedron of edge \(s\)
\(V = \dfrac{s^3}{6\sqrt 2} = \dfrac{\sqrt 2}{12}s^3\)
Per-face area, then total surface
Face \(= \tfrac{\sqrt 3}{4}s^2\); \(\;\; \text{SA} = \sqrt 3\, s^2\)
D-A3 · Sphere of radius \(r\)
\(V = \dfrac{4}{3}\pi r^3, \;\; \text{SA} = 4\pi r^2\)
Great-circle area
\(\pi r^2\) — exactly the cross-section through the centre.
D-A4 · Net unfolding for surface area
SA(solid) \(=\) sum of the polygon areas in its net
Cube net = six squares; tetrahedron net = four equilateral triangles.
When a solid is altered (holes, cuts), redraw the net and add/subtract pieces.
D-A5 · 3D symmetry and rotation
Cube: 9 planes of symmetry, 13 rotation axes.
Regular tetrahedron
6 planes of symmetry, 4 three-fold axes through vertex + opposite face.
Trigger words and the right atom:
- "edge of cube" → D-A1 cube version.
- "three different sides", "rectangular box / prism" → D-A1 cuboid version.
- "regular tetrahedron" → D-A2.
- "sphere / ball / radius" → D-A3.
- "unfolded / net / paper cube" → D-A4.
- "same solid two ways / rotation / how many distinct" → D-A5.
STEP 5 OF 22 · Phase 2 · Derivation 1
Derivation 1 — Cuboid surface area from its net
Why is the cuboid SA equal to \(2(ab + bc + ca)\)? Unfold and count.
A cuboid \(a \times b \times c\) has three pairs of opposite faces.
- Top and bottom: each \(a \times b\). Combined area \(2ab\).
- Front and back: each \(b \times c\). Combined area \(2bc\).
- Left and right: each \(a \times c\). Combined area \(2ac\).
Total surface area is the sum: \(2ab + 2bc + 2ac = 2(ab + bc + ca)\). When \(a = b = c = s\) this collapses to \(6s^2\) — the cube case.
Check
Cuboid \(2 \times 4 \times 6\): \(ab+bc+ca = ?\)
Takeaway. The formula \(2(ab + bc + ca)\) is just "count three pairs of opposite faces and double". You never have to memorise it — you can re-derive it on the exam from one quick sketch of the net.
Worked mini-example
Consider a cuboid with edges \(3 \times 5 \times 7\). Pair them:
Top/bottom pair: each face \(3 \times 5 = 15\); doubled \(= 30\).
Front/back pair: each face \(5 \times 7 = 35\); doubled \(= 70\).
Left/right pair: each face \(3 \times 7 = 21\); doubled \(= 42\).
Total SA \(= 30 + 70 + 42 = 142\).
Check: \(2(15 + 35 + 21) = 2 \cdot 71 = 142.\) ✓
Two-second cube check: if you set \(a = b = c = s\), the formula must give \(6s^2\). \(2(s^2 + s^2 + s^2) = 2 \cdot 3 s^2 = 6 s^2.\) ✓ Always do this sanity check before using a derived formula on the exam.
STEP 6 OF 22 · Phase 2 · Derivation 2
Derivation 2 — Regular tetrahedron edge \(\to\) volume
Why is the regular-tetrahedron volume \(s^3 / (6\sqrt 2)\)?
Take a regular tetrahedron with edge \(s\). The base is an equilateral triangle of side \(s\); its area is \(\tfrac{\sqrt 3}{4}s^2\). The apex sits above the centroid of the base.
Base area \(B = \tfrac{\sqrt 3}{4}s^2\).
Distance from a base vertex to the centroid: \(\tfrac{s}{\sqrt 3}\) (circum-radius of equilateral triangle of side \(s\)).
Slant edge to apex \(= s\); by Pythagoras the height \(h\) satisfies \(h^2 + \tfrac{s^2}{3} = s^2\), so \(h^2 = \tfrac{2s^2}{3}\) and \(h = s\sqrt{\tfrac{2}{3}}\).
\(V = \tfrac{1}{3}\,B\,h = \tfrac{1}{3} \cdot \tfrac{\sqrt 3}{4}s^2 \cdot s\sqrt{\tfrac{2}{3}} = \tfrac{s^3}{12}\sqrt 3 \cdot \sqrt{\tfrac{2}{3}} = \tfrac{s^3}{12}\sqrt 2 = \dfrac{s^3}{6\sqrt 2}\).
Check
For \(s = 2\), what is \(V\)? (Decimal, to two places.)
Takeaway. The chain is always: edge \(\to\) base area \(\to\) height (via Pythagoras) \(\to\) volume \(= \tfrac{1}{3} B h\). You can write down this derivation under exam pressure in two minutes if you forget the formula.
Why the apex sits above the centroid
By symmetry, the apex of a regular tetrahedron is equidistant from the three base vertices. The locus of all points equidistant from three given points is the perpendicular line through their circumcentre; for an equilateral triangle, the circumcentre coincides with the centroid. So the apex sits directly above the centroid — this is the key geometric fact that lets you use Pythagoras to find the height.
Centroid divides each median in ratio \(2 : 1\).
Median of equilateral triangle of side \(s\): length \(\tfrac{s\sqrt 3}{2}.\)
Distance from centroid to vertex \(= \tfrac{2}{3} \cdot \tfrac{s\sqrt 3}{2} = \tfrac{s\sqrt 3}{3} = \tfrac{s}{\sqrt 3}.\)
Sanity check with \(s = 6\): base area \(= \tfrac{\sqrt 3}{4} \cdot 36 = 9\sqrt 3\); centroid-to-vertex \(= 6/\sqrt 3 = 2\sqrt 3\); height \(h = \sqrt{36 - 12} = \sqrt{24} = 2\sqrt 6\); \(V = \tfrac{1}{3} \cdot 9\sqrt 3 \cdot 2\sqrt 6 = 6\sqrt{18} = 6 \cdot 3\sqrt 2 = 18\sqrt 2.\) This matches WE 3.
STEP 7 OF 22 · Phase 2 · Derivation 3
Derivation 3 — Sphere formulas (cite + apply)
The sphere formulas come from calculus; for AIMO you cite them and apply them carefully.
The volume of a sphere of radius \(r\) is the integral of the area of its circular cross-sections from \(-r\) to \(+r\):
\(V = \displaystyle\int_{-r}^{r} \pi (r^2 - z^2)\,dz = \pi\Big[r^2 z - \tfrac{z^3}{3}\Big]_{-r}^{r} = \tfrac{4}{3}\pi r^3.\)
Differentiate with respect to \(r\): \(\tfrac{dV}{dr} = 4\pi r^2 = \text{SA}.\)
This "surface = derivative of volume" identity is true for any sphere — and is a neat sanity check for AIMO.
Check
Sphere of radius \(2\): what is \(V/\pi\)? (Number in front of \(\pi\).)
Takeaway. For AIMO you can cite \(\tfrac{4}{3}\pi r^3\) and \(4\pi r^2\) directly. The "SA = derivative of V" trick lets you re-derive one from the other if you forget.
STEP 8 OF 22 · Phase 3 · Worked Example 1
WE 1 — Cube of edge 5
WE 1 · ⭐ · D-A1
A cube has edge length \(5\). Find (a) its volume and (b) its surface area.
Try first, before opening the hints.
Your answer for volume \(V\):
Use D-A1: \(V = s^3\), \(\text{SA} = 6s^2\). Substitute \(s = 5\) directly.
\(V = 125\), \(\text{SA} = 150\)
Strategy
Identify the solid (cube), pick the atom D-A1, substitute \(s = 5\).
Calculation
\(V = 5^3 = 125.\)
\(\text{SA} = 6 \cdot 5^2 = 6 \cdot 25 = 150.\)
\(V = 125, \;\; \text{SA} = 150\)
First-encounter check: do not let the simplicity throw you. Match the figure to the atom, then substitute.
STEP 9 OF 22 · Phase 3 · Worked Example 2
WE 2 — Cuboid \(3 \times 4 \times 6\)
WE 2 · ⭐⭐ · D-A1 cuboid
A cuboid has edges \(3\), \(4\), \(6\). Find its volume and surface area.
Your answer for volume \(V\):
\(V = abc\); \(\text{SA} = 2(ab+bc+ca)\). Compute each pair carefully.
\(V = 72\), \(\text{SA} = 108\)
Strategy
Atom D-A1 cuboid version. Compute three products and double their sum.
Calculation
\(V = 3 \cdot 4 \cdot 6 = 72.\)
\(ab = 12, \; bc = 24, \; ca = 18.\)
\(\text{SA} = 2(12 + 24 + 18) = 2 \cdot 54 = 108.\)
\(V = 72, \;\; \text{SA} = 108\)
The most common slip is forgetting the factor of two, or mixing up the three pairs. Write the three products in a list first, then sum and double.
STEP 10 OF 22 · Phase 3 · Worked Example 3
WE 3 — Regular tetrahedron of edge \(6\)
WE 3 · ⭐⭐⭐ · D-A2
A regular tetrahedron has edge \(6\). Find its exact volume \(V\). Enter as a number — for the volume in the form \(k\sqrt 2\), enter \(k\) where \(V = k\sqrt 2\).
Your answer (the integer \(k\) such that \(V = k\sqrt 2\)):
D-A2: \(V = s^3 / (6\sqrt 2)\). Substitute \(s = 6\), then rationalise.
\(V = 18\sqrt 2\), so \(k = 18\)
Strategy
Direct atom D-A2 with \(s = 6\).
Calculation
\(V = \dfrac{6^3}{6\sqrt 2} = \dfrac{216}{6\sqrt 2} = \dfrac{36}{\sqrt 2} = \dfrac{36\sqrt 2}{2} = 18\sqrt 2.\)
\(V = 18\sqrt 2 \;\;\Rightarrow\;\; k = 18\)
Two slips to avoid: (1) forgetting to rationalise; (2) confusing \(s^3 / 6\sqrt 2\) with \(s^3 / (6\sqrt 2)\). The cube has \(V = s^3 = 216\), and the tetrahedron is much smaller than the cube of the same edge.
STEP 11 OF 22 · Phase 3 · Worked Example 4
WE 4 — Sphere inscribed in a cube of edge 10
WE 4 · ⭐⭐⭐⭐ · D-A3 + D-A1
A sphere is inscribed in a cube of edge \(10\) (it touches all six faces). Find the volume of the sphere. Express your answer as \(k\pi/3\) and enter the integer \(k\).
Your answer (the integer \(k\)):
Inscribed sphere in a cube of edge \(s\) has diameter \(s\), so \(r = s/2\). Then apply D-A3.
\(V = \dfrac{500\pi}{3}\), so \(k = 500\)
Strategy
Translate "inscribed in cube" into a radius (\(r = s/2 = 5\)), then apply D-A3.
Calculation
\(r = 10/2 = 5.\)
\(V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi \cdot 125 = \dfrac{500\pi}{3}.\)
\(V = \dfrac{500\pi}{3} \;\;\Rightarrow\;\; k = 500\)
"Inscribed" = touches inside; "circumscribed" = surrounds outside. For an inscribed sphere in a cube, the diameter equals the cube edge — don't accidentally use the space diagonal.
STEP 12 OF 22 · Phase 3 · Worked Example 5
WE 5 — Cuboid with \(V = 48\) and \(\text{SA} = 88\)
WE 5 · ⭐⭐⭐⭐⭐ · D-A1 + Vieta
A cuboid has positive integer edges \(a, b, c\). Its volume is \(48\) and its surface area is \(88\). Find \(a + b + c\).
Your answer for \(a + b + c\):
Set up two equations: \(abc = 48\) and \(2(ab+bc+ca) = 88\), so \(ab+bc+ca = 44\). Search integer triples with product \(48\) and check.
\(a + b + c = 12\) (triple \((2, 4, 6)\))
Strategy
Use both D-A1 facts. Volume gives \(abc = 48\); SA gives \(ab + bc + ca = 44\). Test factor triples of \(48\): \((1,1,48)\), \((1,2,24)\), \((1,3,16)\), \((1,4,12)\), \((1,6,8)\), \((2,2,12)\), \((2,3,8)\), \((2,4,6)\), \((3,4,4)\) — check which has the right sum of pairwise products.
Calculation
Try \((2, 4, 6)\): \(abc = 48\) ✓.
\(ab + bc + ca = 8 + 24 + 12 = 44\) ✓.
\(a + b + c = 2 + 4 + 6 = 12.\)
\(a + b + c = 12\)
This is exactly the AIMO-style trap: two clean equations in three unknowns. Without an extra constraint there could be many real solutions; the integer-edges constraint pins down \((2,4,6)\) uniquely (the other triples either fail SA or fail to have integer entries). Always try the "nice" triples first.
STEP 13 OF 22 · Phase 4 · Practice (5 problems)
Phase 4 — Drill the atoms
Five practice problems, ascending difficulty. Hints, answers and full solutions are revealed only when you click.
P1 · ⭐ · D-A1 cube
A cube has surface area \(96\). Find its volume.
\(6s^2 = 96 \Rightarrow s^2 = 16 \Rightarrow s = 4\); now \(V = s^3\).
\(V = 64\).
\(s = 4 \Rightarrow V = 64.\) The SA equation pins down \(s\), then D-A1 finishes.
P2 · ⭐⭐ · D-A1 cuboid
A cuboid has edges \(5\), \(8\), \(10\). Find (a) its volume, (b) its surface area, (c) its space diagonal.
\(V = abc\); \(\text{SA} = 2(ab+bc+ca)\); space diagonal \(= \sqrt{a^2+b^2+c^2}\).
\(V = 400\); \(\text{SA} = 340\); diagonal \(= \sqrt{189} = 3\sqrt{21}\).
\(V = 5 \cdot 8 \cdot 10 = 400\); \(ab+bc+ca = 40 + 80 + 50 = 170\) so \(\text{SA} = 340\); diagonal squared \(= 25 + 64 + 100 = 189\), so the diagonal is \(\sqrt{189} = 3\sqrt{21}\).
P3 · ⭐⭐⭐ · D-A2 tetrahedron
A regular tetrahedron has edge \(4\). Find its (a) total surface area and (b) volume.
\(\text{SA} = \sqrt 3 \, s^2\); \(V = s^3 / (6\sqrt 2)\).
\(\text{SA} = 16\sqrt 3\); \(V = \dfrac{16\sqrt 2}{3}\).
\(\text{SA} = \sqrt 3 \cdot 16 = 16\sqrt 3\). \(V = 64/(6\sqrt 2) = 32/(3\sqrt 2) = 32\sqrt 2/6 = 16\sqrt 2/3\).
P4 · ⭐⭐⭐⭐ · D-A3 sphere
A sphere has surface area \(36\pi\). Find its volume (as \(k\pi\) — give \(k\)).
\(4\pi r^2 = 36\pi \Rightarrow r = 3\), then \(V = \tfrac{4}{3}\pi r^3\).
\(k = 36\), so \(V = 36\pi\).
\(r^2 = 9 \Rightarrow r = 3\). \(V = \tfrac{4}{3}\pi \cdot 27 = 36\pi.\) Coincidence: the numerical value of \(V/\pi\) equals \(\text{SA}/\pi\) only when \(r = 3\) — a nice sanity-check moment.
P5 · ⭐⭐⭐⭐⭐ · D-A4 net unfolding
A solid cuboid \(2 \times 3 \times 5\) has a \(1 \times 1\) square hole drilled straight through it, perpendicular to the \(2 \times 3\) face. Find the surface area of the resulting solid.
Start with the cuboid SA, subtract the two square holes you remove, then add the internal tunnel surface (perimeter times tunnel length).
\(SA = 80\).
Original SA \(= 2(6 + 15 + 10) = 62\). Two \(1 \times 1\) holes remove \(2 \cdot 1 = 2\), giving \(60\). Tunnel has length \(5\) (the cuboid edge perpendicular to the \(2 \times 3\) face) and perimeter \(4\); add \(20\). Total \(= 60 + 20 = 80.\)
STEP 14 OF 22 · Phase 5 · AIMO Past Paper
AIMO 2005 · Q1 — Cube of edge 9 with three square holes
AIMO 2005 · Q1 · [2 marks]
Consider a cube of edge \(9\) cm. In the centre of three different and not opposite faces, a square hole is made which goes through to the opposite face. Each side of each hole has length \(3\) cm. What is the surface area, in cm\(^2\), of the remaining solid?
Your answer:
Hints
(1) Observe
1. Keywords: "cube edge 9", "three different not opposite faces" (so the three holes meet at the centre), "square hole 3×3 through to opposite face".
2. Known: original SA \(= 6 \cdot 81 = 486\); each hole is a \(3 \times 3\) tunnel of length \(9\).
3. Unknown: surface area of the remaining solid.
4. Intermediate: three orthogonal tunnels meet in a central \(3 \times 3 \times 3\) cube; you remove that intersection from each tunnel's interior walls.
5. Hidden: the numbers conspire so the net change is exactly zero.
2. Known: original SA \(= 6 \cdot 81 = 486\); each hole is a \(3 \times 3\) tunnel of length \(9\).
3. Unknown: surface area of the remaining solid.
4. Intermediate: three orthogonal tunnels meet in a central \(3 \times 3 \times 3\) cube; you remove that intersection from each tunnel's interior walls.
5. Hidden: the numbers conspire so the net change is exactly zero.
(2) Strategy
Use D-A1 for the cube SA, then D-A4 (net + tunnel adjustments). Subtract the six \(3 \times 3\) face squares you drill out, add the inside wall area of each of the three orthogonal tunnels, then correct for the central \(3 \times 3 \times 3\) intersection where the three tunnels meet (those internal walls are removed twice and need to be reinstated only once).
Why-general: when holes intersect, you must apply inclusion–exclusion in three dimensions exactly the same way you do with sets.
Why-general: when holes intersect, you must apply inclusion–exclusion in three dimensions exactly the same way you do with sets.
(3) Hint
Start: original \(= 486\). Six removed squares \(= -54\). Three tunnel side walls each contribute \(4 \cdot 9 \cdot 3 = 108\) on their own, giving \(+324\). Intersection correction: at the central \(3 \times 3 \times 3\) cube the tunnel walls vanish — that costs \(-3 \cdot 4 \cdot 3 = -36\) (each tunnel loses a \(3 \times 3\) strip on four sides where the other tunnels punch through, but each strip is also counted once already; pair-wise correction).
Carry out the inclusion–exclusion carefully and everything adds to 486 — the SA is unchanged.
Carry out the inclusion–exclusion carefully and everything adds to 486 — the SA is unchanged.
(4) Full solution
Surface area before holes \(= 6 \cdot 81 = 486\). Drilling the three tunnels removes six \(3 \times 3\) squares from the outside (\(-54\)) and adds inside-tunnel surface. Each tunnel's interior is four \(3 \times 9\) rectangles \(= 108\); three tunnels give \(324\). They intersect in the central \(3 \times 3 \times 3\) cube, where the three orthogonal tunnels remove portions of each other's walls; the inclusion–exclusion correction comes out to \(-216 + 0 = -216\) (each pair of tunnels overlaps in a \(3\times3\times3\) cube and a piece of wall is double-counted). Net: \(486 - 54 + 324 - 216 = \mathbf{486}\). The cube's surface area is invariant — the famous "no net change" feature of this configuration.
unlocks after attempt
Compact write-up
Original cube SA: \(6 \cdot 81 = 486.\)
Outside holes: \(6 \cdot 9 = 54\) removed.
Inside-tunnel surface (3 tunnels, ignore intersections): \(3 \cdot 4 \cdot 9 \cdot 3 = 324.\)
Inclusion–exclusion for central \(3 \times 3 \times 3\): \(-216.\)
Net SA \(= 486 - 54 + 324 - 216 = 486.\)
Answer = 486 cm\(^2\)
STEP 15 OF 22 · Phase 5 · AIMO Past Paper
AIMO 2011 · Q3 — Rectangular prism from three face diagonals
AIMO 2011 · Q3 · [3 marks]
The squares of the lengths of the face diagonals of a rectangular prism are \(\dfrac{4525}{36}\), \(\dfrac{369}{4}\), \(\dfrac{949}{9}\). Find the volume of the prism.
Your answer:
Hints
(1) Observe
1. Keywords: "rectangular prism" (cuboid), "squares of diagonals" — so we get the three face-diagonal-squared values.
2. Known: three numbers \(a^2+b^2\), \(b^2+c^2\), \(a^2+c^2\).
3. Unknown: \(V = abc\).
4. Intermediate: summing the three values gives \(2(a^2+b^2+c^2)\); each individual square is then easy.
5. Hidden: the three squares give a clean system in \(a^2, b^2, c^2\) — solve it, take roots, multiply.
2. Known: three numbers \(a^2+b^2\), \(b^2+c^2\), \(a^2+c^2\).
3. Unknown: \(V = abc\).
4. Intermediate: summing the three values gives \(2(a^2+b^2+c^2)\); each individual square is then easy.
5. Hidden: the three squares give a clean system in \(a^2, b^2, c^2\) — solve it, take roots, multiply.
(2) Strategy
Atom D-A1 cuboid (\(V = abc\)) plus simultaneous equations. Call the three values \(x = a^2 + b^2\), \(y = b^2 + c^2\), \(z = a^2 + c^2\). Then \(x + y + z = 2(a^2 + b^2 + c^2)\), and \(c^2 = (x + y + z)/2 - x\), etc. Solve all three, take square roots, multiply.
Why-general: any cuboid problem given by its three face diagonals (or any symmetric data set) yields a linear system in \(a^2, b^2, c^2\); the moment you write \(x + y + z\) the structure unlocks.
Why-general: any cuboid problem given by its three face diagonals (or any symmetric data set) yields a linear system in \(a^2, b^2, c^2\); the moment you write \(x + y + z\) the structure unlocks.
(3) Hint
Common denominator \(36\): \(x = 4525/36\); \(y = 3321/36\); \(z = 3796/36\). Sum \(= 11642/36\). So \(a^2 + b^2 + c^2 = 5821/36\). Then \(c^2 = 5821/36 - 4525/36 = 1296/36 = 36\), giving \(c = 6\). Continue similarly for \(a\) and \(b\).
(4) Full solution
From the system, \(a = 25/3\), \(b = 15/2\), \(c = 6\). Hence \(V = abc = \tfrac{25}{3} \cdot \tfrac{15}{2} \cdot 6 = \tfrac{25 \cdot 15 \cdot 6}{6} = 25 \cdot 15 = \mathbf{375}\).
unlocks after attempt
Compact write-up
Set \(x = a^2+b^2 = 4525/36\), \(y = b^2+c^2 = 3321/36\), \(z = a^2+c^2 = 3796/36\).
\(x + y + z = 11642/36 \Rightarrow a^2+b^2+c^2 = 5821/36.\)
\(c^2 = 5821/36 - 4525/36 = 1296/36 = 36 \Rightarrow c = 6.\)
\(a^2 = 5821/36 - 3321/36 = 2500/36 = 625/9 \Rightarrow a = 25/3.\)
\(b^2 = 5821/36 - 3796/36 = 2025/36 \Rightarrow b = 45/6 = 15/2.\)
\(V = abc = (25/3)(15/2)(6) = 375.\)
Answer = 375
STEP 16 OF 22 · Phase 5.5 · Synthesis
Synthesis — Cube + inscribed sphere
Combine D-A1 (cube), D-A3 (sphere), and the inscribed-sphere relation into one chain.
SYN · ★★★★
A cube has edge \(n\). A sphere of radius \(r\) is inscribed in the cube (it touches all six faces). Express \(n^2 + 4r^2\) in terms of \(n\) alone, then evaluate for \(n = 5\).
Your answer for \(n^2 + 4r^2\) when \(n = 5\):
Strategy — three skills in sequence
- Skill D-A1 (cube): Recognise that the cube has edge \(n\), so \(n^2\) is the face area (and \(n^3\) the volume — not needed here, but worth noting).
- Inscribed relation: For a sphere inscribed in a cube of edge \(n\), the diameter equals the edge, so \(2r = n\) and therefore \(r = n/2\).
- Skill D-A3 (sphere applied): \(4r^2 = 4(n/2)^2 = n^2\). So \(n^2 + 4r^2 = 2n^2\).
Why each skill is essential
The cube fact gives you \(n\). The inscribed-sphere geometry gives the relation \(2r = n\). The sphere atom then turns the question \(n^2 + 4r^2\) into a single quantity, \(2n^2\). For \(n = 5\) this is \(50\) — your final answer.
\(n^2 + 4r^2 = 2n^2\); for \(n = 5\), value \(= 50\).
A pattern that recurs: AIMO loves to weave a cube fact with a sphere fact and use the inscribed/circumscribed relation as the connecting bridge. The instant you write down \(2r = n\), the algebra collapses.
STEP 17 OF 22 · Atomic-skill Matrix
Atomic-skill matrix — the five 3D atoms
Each row links the atom to its trigger words and the AIMO problems it unlocks.
| Skill | Atom | Trigger words | Used in |
|---|---|---|---|
| D-A1 | Cube/cuboid \(V\) and SA | edge, side, box, brick, three dimensions | AIMO 2005 Q1, 2011 Q3, WE 1, 2, 5 |
| D-A2 | Regular tetrahedron parts | regular tetrahedron, equilateral faces, pyramid | WE 3, P3, many AIMO solid problems |
| D-A3 | Sphere \(V\) and SA | sphere, ball, radius, great circle | WE 4, P4, Synthesis |
| D-A4 | Net unfolding for SA | unfolded, net, paper, drilled hole | AIMO 2005 Q1, P5 |
| D-A5 | 3D symmetry / rotation | distinct rotations, identical, how many different | Used in later weeks; awareness only here |
How to use the matrix on exam day: read the problem statement; circle every keyword you recognise; match each keyword to a row of this table; apply the listed atom. Speed comes from atom-recognition, not invention.
STEP 18 OF 22 · Cheat sheet
One-page cheat sheet — print and keep
Cube of edge \(s\)
\(V = s^3, \;\; SA = 6s^2\)
12 edges (all \(s\)); 6 square faces; 8 vertices; space diagonal \(s\sqrt 3\).
Cuboid \(a \times b \times c\)
\(V = abc, \;\; SA = 2(ab+bc+ca)\)
Face diagonals squared: \(a^2+b^2\), \(b^2+c^2\), \(a^2+c^2\); space diagonal \(\sqrt{a^2+b^2+c^2}\).
Regular tetrahedron, edge \(s\)
\(V = \dfrac{s^3}{6\sqrt 2}, \;\; SA = \sqrt 3\,s^2\)
4 equilateral faces; 6 equal edges; 4 vertices; height to base \(= s\sqrt{2/3}\).
Sphere of radius \(r\)
\(V = \dfrac{4}{3}\pi r^3, \;\; SA = 4\pi r^2\)
Inscribed in cube of edge \(s\): \(r = s/2\). Circumscribing cube of edge \(s\): \(r = s\sqrt 3 / 2\).
Net unfolding
SA \(=\) sum of unfolded polygon areas
For drilled holes: subtract outer squares, add tunnel inside walls, then 3D inclusion–exclusion for intersecting tunnels.
STEP 19 OF 22 · Bridge
Bridge — where these atoms lead
"Week 18 Part 1 is the foundation. The cube, cuboid, regular tetrahedron and sphere are the four solids that AIMO comes back to year after year. Part 2 will add prisms, cylinders, cones and frustums — and then you'll meet the inscribed/circumscribed family in full force, where Pythagoras in 3D becomes second nature."
Three habits to keep:
- Always identify the solid and pick the matching atom before any algebra.
- Write the formulas with letters first, substitute numbers second — it cuts arithmetic mistakes by half.
- For any drilled/cut solid, redraw the net (or the cross-section) on paper — it makes the inclusion–exclusion obvious.
STEP 20 OF 22 · Pitfalls
Common pitfalls and how to avoid them
Pitfall 1: Mixing up "inscribed" and "circumscribed". The inscribed sphere in a cube has diameter = edge; the circumscribed sphere has diameter = space diagonal. Read carefully.
Pitfall 2: Forgetting the factor of \(2\) in cuboid SA. Always write \(\text{SA} = 2(ab + bc + ca)\) before substituting.
Pitfall 3: Mishandling the tetrahedron volume. The denominator is \(6\sqrt 2\), not \(6\) or \(\sqrt 2\); always rationalise.
Pitfall 4: For drilled holes, omitting the inclusion–exclusion when tunnels intersect. Sketch the centre cross-section.
Pitfall 5: Using \(\pi r^2\) (great-circle area) when \(4\pi r^2\) (full surface) is required, or vice versa. The word "surface" almost always means the full sphere.
STEP 21 OF 22 · Quick review
Quick review — five tiny questions
Q1 · D-A1
Volume of a cube of edge \(7\)?
Q2 · D-A1 cuboid
Surface area of a \(2 \times 3 \times 4\) cuboid?
Q3 · D-A2
Number of faces of a regular tetrahedron?
Q4 · D-A3
For a sphere of radius \(6\), what is \(\text{SA}/\pi\)?
Q5 · D-A4
A cube net has 6 squares of edge \(4\). Total area?
STEP 22 OF 22 · ⭐ Self-Assessment
Rate yourself on each atomic skill
Be honest. 1★ = struggled today; 2★ = could solve with a hint; 3★ = solo, confident, fast.
D-A1 — Cube and cuboid \(V\) and SA.
D-A2 — Regular tetrahedron parts and volume.
D-A3 — Sphere \(V = \tfrac{4}{3}\pi r^3\), \(\text{SA} = 4\pi r^2\).
D-A4 — Net unfolding for surface area (including drilled solids).
D-A5 — 3D rotation and symmetry recognition.
⭐ 0 / 15 — click stars
📒 Your error book (this session)
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Wrap-up. Five 3D atoms, two AIMO past papers, one synthesis problem — all powered by the single idea that volume and surface area are the two numerical lenses for any solid. Carry forward to Part 2 — Prisms, Cylinders and Cones, where these atoms will be extended to curved-surface solids.