AMC 10 · 2021 · #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 $\frac{1}{6}$ · 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

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

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

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

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

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

[ABC] =

\frac{1}{2}

· 2 · 3 = 3, height = AD = 4

6STEP 6

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

V =

\frac{1}{3}

· 3 · 4 = 4 =

\frac{1}{6}

· 2 · 3 · 4

7STEP 7

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).

More like this