AMC 10 · 2023 · #17

Grade 8 geometry-3d

space-diagonal-formulasystems-of-equationssurface-areavolume-rectangular-prismvieta-formulas identify-subproblemsconvert-to-algebra ↑ Prerequisites: systems-of-equationssurface-area

📏 Medium solution 💡 2 insights

Problem

A rectangular box $\mathcal{P}$ has distinct edge lengths $a$ , $b$ , and $c$ . The sum of the lengths of all $12$ edges of $\mathcal{P}$ is $13$ , the areas of all $6$ faces of $\mathcal{P}$ is $\frac{11}{2}$ , and the volume of $\mathcal{P}$ is $\frac{1}{2}$ . What is the length of the longest interior diagonal connecting two vertices of $\mathcal{P}$ ?

Pick an answer.

(A)

(B)

$~\frac{3}{8}$

(C)

$~\frac{9}{8}$

(D)

$~\frac{9}{4}$

(E)

$~\frac{3}{2}$

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Toolkit + CCSS Solution

Understand

Restated: A rectangular box has distinct edge lengths $a, b, c$. The total edge length is $13$, the total surface area is $\tfrac{11}{2}$, and the volume is $\tfrac{1}{2}$. Find the length of the longest interior (space) diagonal.

Givens: Sum of all $12$ edges: $4(a+b+c) = 13$; Sum of all $6$ face areas: $2(ab + bc + ca) = \tfrac{11}{2}$; Volume: $abc = \tfrac{1}{2}$; Answer choices: (A) $2$, (B) $\tfrac{3}{8}$, (C) $\tfrac{9}{8}$, (D) $\tfrac{9}{4}$, (E) $\tfrac{3}{2}$

Unknowns: $d = \sqrt{a^2 + b^2 + c^2}$, the space diagonal

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #13 Convert to Algebra, #8 Analyze the Units

We never need the individual edges — only the symmetric sums $s_1=a+b+c$ and $s_2=ab+bc+ca$. Tool #7 (Identify Subproblems) chops the job into three rungs: (a) read off $s_1$ and $s_2$ from the given totals, (b) plug into the identity $(s_1)^2 = a^2+b^2+c^2 + 2s_2$ to get $a^2+b^2+c^2$, (c) take the square root via the Pythagorean-theorem extension to find the space diagonal. Tool #13 (Convert to Algebra) handles the identity; Tool #8 (Analyze the Units) confirms the answer is a length (not an area or volume).

Execute — Answer: D

#7 Identify Subproblems 6.G.A.4 Step 1

Subproblem A: extract the two symmetric sums.
The box has $4$ edges of each length, so its $12$ edges sum to $4(a+b+c)=13$, giving $a+b+c=\tfrac{13}{4}$.
The box has $2$ faces of each type ($ab, bc, ca$), so surface area $2(ab+bc+ca)=\tfrac{11}{2}$.

$$a+b+c=\dfrac{13}{4},\quad 2(ab+bc+ca)=\dfrac{11}{2}$$

💡 A rectangular box's surface area and edge total are themselves "sum of edges" and "sum of face products" — exactly the two symmetric sums of $a,b,c$ we need.

#13 Convert to Algebra 8.EE.A.2 Step 2

Subproblem B: use the square-of-sum identity to get $a^2+b^2+c^2$.
The identity $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ lets us solve for the sum of squares directly.
The volume $abc=\tfrac{1}{2}$ is a red herring — we don't need it.

$$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = \left(\dfrac{13}{4}\right)^2 - \dfrac{11}{2} = \dfrac{169}{16} - \dfrac{88}{16} = \dfrac{81}{16}$$

💡 Squaring $a+b+c$ already contains every $a^2,b^2,c^2$ once plus every cross term $ab,bc,ca$ twice — exactly what we need to subtract off.

#7 Identify Subproblems 8.G.B.7 Step 3

Subproblem C: take the space-diagonal length via $d=\sqrt{a^2+b^2+c^2}$ (the $3$D Pythagorean theorem).
Substitute $a^2+b^2+c^2 = \tfrac{81}{16}$.

$$d = \sqrt{\dfrac{81}{16}} = \dfrac{9}{4} \;\Rightarrow\; \textbf{(D) }\dfrac{9}{4}$$

💡 The space diagonal of a box is the hypotenuse of a right triangle whose legs are a face diagonal and the third edge — applying Pythagoras twice gives $d^2 = a^2 + b^2 + c^2$.

#8 Analyze the Units 5.MD.A.1 Step 4

Unit check: $a+b+c$ has units of length, so $(a+b+c)^2$ and $2(ab+bc+ca)$ both have units of length-squared (area).
Their difference $a^2+b^2+c^2$ also has area units, and the final square root brings us back to a length — consistent with what the diagonal must be.

$$[\text{length}]^2 - [\text{length}]^2 = [\text{length}]^2,\ \sqrt{[\text{length}]^2} = [\text{length}]$$

💡 Tracking units catches errors fast: a diagonal must come out as a length, not a number times an area.

[1] #7 6.G.A.4 Subproblem A: extract the two symmetric sums. The box has $4$ edges of each leng

[2] #13 8.EE.A.2 Subproblem B: use the square-of-sum identity to get $a^2+b^2+c^2$. The identity

[3] #7 8.G.B.7 Subproblem C: take the space-diagonal length via $d=\sqrt{a^2+b^2+c^2}$ (the $3$

[4] #8 5.MD.A.1 Unit check: $a+b+c$ has units of length, so $(a+b+c)^2$ and $2(ab+bc+ca)$ both h

Review

Reasonableness: Sanity-check the magnitudes: $a+b+c = 3.25$ and a typical edge is $\sim 1$, so the diagonal $\sqrt{a^2+b^2+c^2}$ should be a bit bigger than any single edge but smaller than $a+b+c$. The answer $\tfrac{9}{4}=2.25$ comfortably fits between $\sim 1$ and $3.25$. Also, $\left(\tfrac{9}{4}\right)^2 = \tfrac{81}{16}$ matches our intermediate value exactly. The volume $abc=\tfrac{1}{2}$ was never used — which is fine, because the diagonal is determined by $s_1$ and $s_2$ alone.

Alternative: Tool #6 (Guess and Check): each answer choice gives a candidate $a^2+b^2+c^2 = d^2$. Test (D): $d^2 = \tfrac{81}{16}$. Check whether this matches $(a+b+c)^2 - 2(ab+bc+ca) = \tfrac{169}{16} - \tfrac{88}{16} = \tfrac{81}{16}$. It does — same answer (D) without ever solving for $a$, $b$, $c$ individually.

CCSS standards used (min grade 8)

6.G.A.4 Represent three-dimensional figures using nets and find surface area (Reading the given totals as the symmetric sums of a box: $12$ edges $\to 4(a+b+c)$, $6$ faces $\to 2(ab+bc+ca)$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Applying the identity $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ to solve for $a^2+b^2+c^2$, then taking $\sqrt{\tfrac{81}{16}} = \tfrac{9}{4}$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using the $3$D form of the Pythagorean theorem, $d=\sqrt{a^2+b^2+c^2}$, to express the space diagonal in terms of the edges.)
5.MD.A.1 Convert among different-sized standard measurement units within a given system (Tracking length/area units through the identity to confirm the final answer has length units.)

⭐ You never need the individual edges $a,b,c$: the totals give $a+b+c=\tfrac{13}{4}$ and $2(ab+bc+ca)=\tfrac{11}{2}$, and the identity $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ delivers $a^2+b^2+c^2 = \tfrac{81}{16}$. The space diagonal is $\sqrt{\tfrac{81}{16}} = \textbf{(D) }\tfrac{9}{4}$ — and the volume was a red herring.