AMC 10 · 2021 · #13

Grade 8 geometry-3d

Archetype Compound Area by Decomposition / Difference

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

📏 Medium solution 💡 2 insights

Problem

What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ , $AC = 3$ , $AD = 4$ , $BC = \sqrt{13}$ , $BD = 2\sqrt{5}$ , and $CD = 5$ ?

Pick an answer.

(A)

(B)

$~2\sqrt{3}$

(C)

(D)

$~3\sqrt{3}$

(E)

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Toolkit + CCSS Solution

Understand

Restated: A tetrahedron $ABCD$ has six edge lengths: $AB = 2$, $AC = 3$, $AD = 4$, $BC = \sqrt{13}$, $BD = 2\sqrt{5}$, $CD = 5$. Find its volume.

Givens: Three edges meeting at $A$: $AB = 2, AC = 3, AD = 4$; The opposite three edges: $BC = \sqrt{13}, BD = 2\sqrt{5}, CD = 5$; Answer choices: $3,\;\; 2\sqrt{3},\;\; 4,\;\; 3\sqrt{3},\;\; 6$

Unknowns: Volume of tetrahedron $ABCD$

Understand

Restated: A tetrahedron $ABCD$ has six edge lengths: $AB = 2$, $AC = 3$, $AD = 4$, $BC = \sqrt{13}$, $BD = 2\sqrt{5}$, $CD = 5$. Find its volume.

Givens: Three edges meeting at $A$: $AB = 2, AC = 3, AD = 4$; The opposite three edges: $BC = \sqrt{13}, BD = 2\sqrt{5}, CD = 5$; Answer choices: $3,\;\; 2\sqrt{3},\;\; 4,\;\; 3\sqrt{3},\;\; 6$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities

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 $a^2 + b^2 = c^2$). Indeed: $2^2 + 3^2 = 13 = BC^2$, $2^2 + 4^2 = 20 = BD^2$, $3^2 + 4^2 = 25 = CD^2$. All three! So the three edges from $A$ are pairwise perpendicular — the tetrahedron is a 'corner of a box' and its volume is $\tfrac{1}{6} \cdot AB \cdot AC \cdot AD$. 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.

Execute — Answer: C

#1 Draw a Diagram 6.EE.A.1 Step 1

List the squares of the three edges at $A$: $AB^2 = 4$, $AC^2 = 9$, $AD^2 = 16$.
Then list the squares of the three opposite edges: $BC^2 = 13$, $BD^2 = 20$, $CD^2 = 25$.

$$AB^2 = 4,\; AC^2 = 9,\; AD^2 = 16,\;\; BC^2 = 13,\; BD^2 = 20,\; CD^2 = 25$$

💡 Grade 6 expressions: squaring each length surfaces sum-of-squares patterns.

#6 Guess and Check 8.G.B.6 Step 2

Check the Pythagorean relation in face $ABC$.
Does $AB^2 + AC^2 = BC^2$?
$4 + 9 = 13$ ✓.
So $\triangle ABC$ has a right angle at $A$ (converse of the Pythagorean theorem).
That means edge $AB$ is perpendicular to edge $AC$.

$$AB^2 + AC^2 = 4 + 9 = 13 = BC^2 \Rightarrow \angle BAC = 90^\circ$$

💡 Grade 8 Pythagorean converse: if sides satisfy $a^2 + b^2 = c^2$, the triangle is right-angled.

#6 Guess and Check 8.G.B.6 Step 3

Repeat for face $ABD$ and face $ACD$.
$4 + 16 = 20 = BD^2$ ✓, so $\angle BAD = 90^\circ$.
$9 + 16 = 25 = CD^2$ ✓, so $\angle CAD = 90^\circ$.

$$AB^2 + AD^2 = 20 = BD^2,\quad AC^2 + AD^2 = 25 = CD^2$$

💡 Same Pythagorean converse applied to the other two faces meeting at $A$.

#1 Draw a Diagram 8.G.B.8 Step 4

Translate the three right angles into a 3D picture.
All three edges from $A$ are pairwise perpendicular — exactly like the three edges meeting at a corner of a rectangular box.
Pin $A$ at the origin with $B$ on the $x$-axis, $C$ on the $y$-axis, $D$ on the $z$-axis.
The tetrahedron is the 'corner piece' of a $2 \times 3 \times 4$ box.

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

💡 Grade 8 coordinates in space: place perpendicular edges on the axes for a clean picture.

#7 Identify Subproblems 6.G.A.1 Step 5

Compute the volume using face $ABC$ (a right triangle, legs $2$ and $3$) as the base.
Its area is $\tfrac{1}{2} \cdot 2 \cdot 3 = 3$.
The height from $D$ to the plane of $ABC$ (the $xy$-plane) is simply $AD = 4$, since $AD$ is perpendicular to that plane.

$$[ABC] = \tfrac{1}{2} \cdot 2 \cdot 3 = 3, \quad \text{height} = AD = 4$$

💡 Grade 6 base-times-height area of a right triangle, plus 'perpendicular edge = height'.

#7 Identify Subproblems 7.G.B.6 Step 6

Apply the pyramid volume formula.
$V = \tfrac{1}{3} \cdot \text{base} \cdot \text{height} = \tfrac{1}{3} \cdot 3 \cdot 4 = 4$.
This is also the well-known 'corner of a box' formula $V = \tfrac{1}{6} \cdot AB \cdot AC \cdot AD = \tfrac{1}{6} \cdot 2 \cdot 3 \cdot 4 = 4$.

$$V = \tfrac{1}{3} \cdot 3 \cdot 4 = 4 = \tfrac{1}{6} \cdot 2 \cdot 3 \cdot 4$$

💡 Grade 7 volume formula: any pyramid's volume is one-third of base times height.

#3 Eliminate Possibilities 6.EE.B.5 Step 7

Match $V = 4$ to the choices.
It's (C).
The other values ($3, 2\sqrt{3}, 3\sqrt{3}, 6$) don't agree with the box-corner calculation.

$$V = 4 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 6 multiple-choice match: only one option equals $4$.

[1] #1 6.EE.A.1 List the squares of the three edges at $A$: $AB^2 = 4$, $AC^2 = 9$, $AD^2 = 16$.

[2] #6 8.G.B.6 Check the Pythagorean relation in face $ABC$. Does $AB^2 + AC^2 = BC^2$? $4 + 9

[3] #6 8.G.B.6 Repeat for face $ABD$ and face $ACD$. $4 + 16 = 20 = BD^2$ ✓, so $\angle BAD = 9

[4] #1 8.G.B.8 Translate the three right angles into a 3D picture. All three edges from $A$ are

[5] #7 6.G.A.1 Compute the volume using face $ABC$ (a right triangle, legs $2$ and $3$) as the

[6] #7 7.G.B.6 Apply the pyramid volume formula. $V = \tfrac{1}{3} \cdot \text{base} \cdot \tex

[7] #3 6.EE.B.5 Match $V = 4$ to the choices. It's (C). The other values ($3, 2\sqrt{3}, 3\sqrt{

Review

Reasonableness: The 'box-corner' shape is a sanity check by itself: the full $2 \times 3 \times 4$ box has volume $24$, and a corner tetrahedron is exactly $1/6$ of the box (because slicing a box along the three diagonal planes through one corner produces six congruent tetrahedra). So $V = 24/6 = 4$ ✓. 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.

Alternative: Tool #6 (Guess and Check) using the scalar triple product. With the coordinates $A = (0,0,0)$, $B = (2,0,0)$, $C = (0,3,0)$, $D = (0,0,4)$, the volume of the tetrahedron is $\tfrac{1}{6} |{\vec{AB} \cdot (\vec{AC} \times \vec{AD})}| = \tfrac{1}{6} |\det\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix}| = \tfrac{1}{6} \cdot 24 = 4$. Same answer (C). (This is the high-school version of the same calculation we did with the pyramid formula.)

CCSS standards used (min grade 8)

6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Computing $AB^2, AC^2, AD^2, BC^2, BD^2, CD^2$ to surface Pythagorean patterns.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Matching the computed $V = 4$ against the five answer choices to select (C).)
6.G.A.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons (Computing the area of right triangle $ABC$ as $\tfrac{1}{2} \cdot 2 \cdot 3 = 3$.)
7.G.B.6 Solve real-world and mathematical problems involving area, volume and surface area (Applying $V = \tfrac{1}{3} \cdot \text{base} \cdot \text{height}$ to the tetrahedron.)
8.G.B.6 Explain a proof of the Pythagorean Theorem and its converse (Using the converse to detect right angles at $A$ in faces $ABC, ABD, ACD$.)
8.G.B.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Placing $A, B, C, D$ on the coordinate axes once $AB, AC, AD$ are pairwise perpendicular.)

⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know! Square every edge: $2^2+3^2 = 13$, $2^2+4^2 = 20$, $3^2+4^2 = 25$ — these are exactly $BC^2, BD^2, CD^2$. So $AB, AC, AD$ are all perpendicular to each other, like the three edges at the corner of a $2 \times 3 \times 4$ box. The tetrahedron is $1/6$ of that box, so its volume is $24/6 = 4$, answer (C).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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