AMC 10 · 2010 · #20
Grade 8 geometry-3dA fly trapped inside a cubical box with side length 1 meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A fly starts at one corner of a $1\times1\times1$ 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.
Givens: The box is a cube with side length $1$ meter; The fly begins and ends at the same corner; It visits each of the other $7$ corners exactly once; Each move is a straight line from one corner to another; Answer choices: (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) $3\sqrt{2}+5\sqrt{3}$
Unknowns: The maximum possible total length of the closed trip
Understand
Restated: A fly starts at one corner of a $1\times1\times1$ 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.
Givens: The box is a cube with side length $1$ meter; The fly begins and ends at the same corner; It visits each of the other $7$ corners exactly once; Each move is a straight line from one corner to another; Answer choices: (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) $3\sqrt{2}+5\sqrt{3}$
Plan
Primary tool: #14 Extreme Principle
Secondary: #17 Visualize Spatial Relationships, #1 Draw a Diagram, #7 Identify Subproblems
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 $0$s and $1$s; 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.
Execute — Answer: D
8.G.B.7 Step 1 List the three possible move lengths
- Put the cube in space so its corners are the points whose coordinates are each $0$ or $1$.
- Two corners differ in $1$, $2$, or $3$ of their coordinates.
- The straight distance is the square root of the sum of the squared coordinate changes, and each change is $0$ or $1$, so the distance is just $\sqrt{\text{(number of coordinates that differ)}}$: an edge ($\sqrt{1}=1$), a face diagonal ($\sqrt{2}$), or an interior (space) diagonal ($\sqrt{3}$).
💡 On a unit cube every hop is $\sqrt{2}$ or $\sqrt{3}$-sized by the Pythagorean theorem, and $\sqrt{3}$ is the longest hop possible.
8.G.B.8 Step 2 Cap how many longest hops fit
- The trip uses exactly $8$ moves.
- To make the total as big as possible we want as many $\sqrt{3}$ moves as we can.
- A $\sqrt{3}$ move joins two opposite corners (they differ in all three coordinates).
- Each corner has exactly one opposite, so the $8$ corners split into just $4$ opposite pairs, meaning the cube has only $4$ space diagonals.
- Therefore at most $4$ of the $8$ moves can have length $\sqrt{3}$; the remaining moves can be at most $\sqrt{2}$ each.
💡 There are only four interior diagonals in the whole cube, so no trip can lean on more than four of them.
8.G.B.8 Step 3 Show the cap is reachable
- Build a tour that actually uses $4$ space diagonals and $4$ face diagonals, alternating between them.
- Labeling corners by their $0/1$ triples, follow $000\to111\to001\to110\to011\to100\to010\to101\to000$.
- The moves marked with all three coordinates flipping are the four $\sqrt{3}$ space diagonals; the moves with exactly two coordinates flipping are the four $\sqrt{2}$ face diagonals.
- Every corner appears exactly once and the fly returns home, so this is a legal trip.
💡 You can weave the four space diagonals together with four face diagonals into one unbroken loop, so the upper limit is really achievable.
7.NS.A.3 Step 4 Add up the maximum length
- The best trip has $4$ moves of length $\sqrt{3}$ and $4$ moves of length $\sqrt{2}$.
- Add these up.
- This total is larger than any option that mixes in shorter edges of length $1$, so it is the maximum.
💡 Four longest hops plus four next-longest hops give the biggest reachable sum, which is answer (D).
8.G.B.7 Put the cube in space so its corners are the points whose coordinates are each $ 8.G.B.8 The trip uses exactly $8$ moves. To make the total as big as possible we want as 8.G.B.8 Build a tour that actually uses $4$ space diagonals and $4$ face diagonals, alte 7.NS.A.3 The best trip has $4$ moves of length $\sqrt{3}$ and $4$ moves of length $\sqrt{ Review
Reasonableness: Every one of the $8$ moves is at most $\sqrt{3}\approx1.73$, so no trip can exceed $8\sqrt{3}\approx13.86$. Our answer $4\sqrt{2}+4\sqrt{3}\approx12.59$ sits just under that ceiling, exactly as expected when four of the eight moves are forced down to $\sqrt{2}$. Choices that include a plain edge of length $1$ (like (A) $4+4\sqrt{2}\approx9.66$) are smaller because they waste a move on the shortest hop, so (D) beating them all makes sense.
Alternative: Think of it as a graph without coordinates. The trip is a cycle on the $8$ corners, so it has $8$ edges. Sort every possible corner-to-corner segment by length; the only $\sqrt{3}$ segments are the $4$ interior diagonals, so a cycle can contain at most $4$ of them, and the other $4$ edges are each at most $\sqrt{2}$. Hence the length is at most $4\sqrt{3}+4\sqrt{2}$, and since a concrete cycle attains it, that bound is the answer.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the three possible move lengths: face diagonal $\sqrt{2}$ and space diagonal $\sqrt{3}$ of the unit cube.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Using $0/1$ coordinates to see which corners are opposite, count the $4$ space diagonals, and build an explicit best tour.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Adding the four $\sqrt{3}$ moves and four $\sqrt{2}$ moves to get the total $4\sqrt{2}+4\sqrt{3}$.)
⭐ 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 $4\sqrt{2}+4\sqrt{3}$.
⭐ 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 $4\sqrt{2}+4\sqrt{3}$.
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