AMC 10 · 2010 · #20

Grade 8 geometry-3d

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

space-diagonal-formulapythagorean-theoremoptimizationspatial-visualization extreme-principleextremal-construction ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 3 insights

Problem

A fly starts at one corner of a 1×1×1 cube, visits every other corner exactly once, and returns to the start, moving in a straight line from corner to corner. Find the largest possible total length of this trip.

Pick an answer.

(A)

$4+4\sqrt{2}$

(B)

$2+4\sqrt{2}+2\sqrt{3}$

(C)

$2+3\sqrt{2}+3\sqrt{3}$

(D)

$4\sqrt{2}+4\sqrt{3}$

(E)

$3\sqrt{2}+5\sqrt{3}$

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

Tool #14 (Extreme Principle): the question asks for a maximum, so use the longest possible moves as often as the rules allow, then prove you cannot do better. Tool #17 (Visualize Spatial Relationships): place the cube on coordinates so every corner is a triple of 0s and 1s; then the length of a move depends only on how many coordinates change. Tool #1 (Draw a Diagram): build an explicit tour to show the best bound is actually reachable. Tool #7 (Identify Subproblems): split the work into (i) list the move lengths, (ii) cap how many longest moves fit, (iii) construct a matching tour, (iv) add up.

1STEP 1

List the three possible move lengths

Put corners at 0/1 triples: a hop's length is the root of how many coordinates change — edge 1, face diagonal √(2), space diagonal √(3).

edge=√(1²)=1, face diagonal=√(1²+1²)=√(2), space diagonal=√(1²+1²+1²)=√(3)

2STEP 2

Cap how many longest hops fit

The trip takes 8 moves, but each corner has one opposite, so only 4 can be √(3) space diagonals; the rest are at most √(2).

8 moves = (at most 4)×√(3) + (rest)×( ≤ √(2))

3STEP 3

Show the cap is reachable

Tour 000→111→001→110→011→100→010→101→000 alternates space and face diagonals, hits every corner once, and closes — so the cap is reached.

000 → 111_√(3)→001_√(2)→110_√(3)→011_√(2)→100_√(3)→010_√(2)→101_√(3)→000_√(2)

4STEP 4

Add up the maximum length

Four √(3) hops plus four √(2) hops total 4√(2)+4√(3), beating every choice that wastes a move on a length-1 edge.

4√(3)+4√(2)=4√(2)+4√(3)≈ 12.59; this is choice (D).

Answer

4√(2)+4√(3)

Every one of the 8 moves is at most √(3)≈1.73, so no trip can exceed 8√(3)≈13.86. Our answer 4√(2)+4√(3)≈12.59 sits just under that ceiling, exactly as expected when four of the eight moves are forced down to √(2). Choices that include a plain edge of length 1 (like (A) 4+4√(2)≈9.66) are smaller because they waste a move on the shortest hop, so (D) beating them all makes sense.

💡Key takeaway

To make a corner-to-corner loop as long as possible, use every long diagonal you can: the cube only has four space diagonals, so pair them with four face diagonals for 4√(2)+4√(3).

List the three possible move lengths
Cap how many longest hops fit
Show the cap is reachable
Add up the maximum length

List the three possible move lengths

Cap how many longest hops fit

Show the cap is reachable

Add up the maximum length

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