Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #13

Grade 8 geometry-3d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

pythagorean-theoremspatial-visualizationarea-triangles identify-subproblemspattern-recognition ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 2 insights

Problem

A tetrahedron ABCD has six edge lengths: AB = 2, AC = 3, AD = 4, BC = √(13), BD = 2√(5), CD = 5. Find its volume.

Pick an answer.

(A)

(B)

$2\sqrt{3}$

(C)

(D)

$3\sqrt{3}$

(E)

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch vertex A with the three edges AB, AC, AD going to B, C, D. The trick is to test whether each triangular face containing A is right-angled at A — use the Pythagorean converse (Tool #6, Guess and Check on the equation a² + b² = c²). Indeed: 2² + 3² = 13 = BC², 2² + 4² = 20 = BD², 3² + 4² = 25 = CD². All three! So the three edges from A are pairwise perpendicular — the tetrahedron is a 'corner of a box' and its volume is 1/6 · AB · AC · AD. Tool #7 (Subproblems) splits the work into: detect right angles, then plug into the corner-of-box formula. Tool #3 (Eliminate) confirms the final number against the choices.

1STEP 1

List the squared edge lengths

Square every edge: AB² = 4, AC² = 9, AD² = 16, and BC² = 13, BD² = 20, CD² = 25.

AB² = 4, AC² = 9, AD² = 16, BC² = 13, BD² = 20, CD² = 25

2STEP 2

Test the Pythagorean relation

In face ABC, AB² + AC² = 4 + 9 = 13 = BC², so by the Pythagorean converse ∠BAC = 90°.

AB² + AC² = 4 + 9 = 13 = BC² → ∠ BAC = 90°

3STEP 3

Repeat for the other two faces

Likewise 4 + 16 = 20 = BD² and 9 + 16 = 25 = CD², so ∠BAD = ∠CAD = 90° as well.

AB² + AD² = 20 = BD², AC² + AD² = 25 = CD²

4STEP 4

Place the vertices on axes

Three right angles at A mean AB, AC, AD are pairwise perpendicular — the corner of a 2 × 3 × 4 box.

A = (0,0,0), B = (2,0,0), C = (0,3,0), D = (0,0,4)

5STEP 5

Find the base area

Take right triangle ABC as the base: base area 3 (½ · 2 · 3), with height AD = 4 since AD ⊥ that plane.

[ABC] = 1/2 · 2 · 3 = 3, height = AD = 4

6STEP 6

Apply the pyramid volume formula

Pyramid volume = ⅓ · base · height = ⅓ · 3 · 4 (also ⅙ · 2 · 3 · 4), so V = 4.

V = 1/3 · 3 · 4 = 4 = 1/6 · 2 · 3 · 4

7STEP 7

Match against the choices

Only choice (C) equals 4, so the volume is (C); the others don't fit the box-corner value.

V = 4 → (C)

Answer

The 'box-corner' shape is a sanity check by itself: the full 2 × 3 × 4 box has volume 24, and a corner tetrahedron is exactly

\frac{1}{6}

of the box (because slicing a box along the three diagonal planes through one corner produces six congruent tetrahedra). So V =

\frac{24}{6}

= 4 ✓. The magnitude is reasonable for edge lengths of order 2–5: a 'typical' tetrahedron with those edges should have a volume around a handful of cubic units, and 4 fits.

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know! Square every edge: 2²+3² = 13, 2²+4² = 20, 3²+4² = 25 — these are exactly BC², BD², CD². So AB, AC, AD are all perpendicular to each other, like the three edges at the corner of a 2 × 3 × 4 box. The tetrahedron is $\frac{1}{6}$ of that box, so its volume is $\frac{24}{6}$ = 4, answer (C).

List the squared edge lengths
Test the Pythagorean relation
Repeat for the other two faces
Place the vertices on axes
Find the base area
Apply the pyramid volume formula
Match against the choices

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