AMC 10 Step by Step
Log inSign up
Printable

2010 AMC 10A Problem 20

Problem onlyBack to the problem
2010 AMC 10A Problem 20AMC 10 Step by Step
Name:                             Date: September 13, 2026
Geometric Optimization · difficulty 4 of 5 · about 6 min · position P16-20

A fly trapped inside a cubical box with side length 11 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?

  1. A)

    4+424+4\sqrt{2}

  2. B)

    2+42+232+4\sqrt{2}+2\sqrt{3}

  3. C)

    2+32+332+3\sqrt{2}+3\sqrt{3}

  4. D)

    42+434\sqrt{2}+4\sqrt{3}

  5. E)

    32+533\sqrt{2}+5\sqrt{3}

Answer

D

Key insight

Each corner has one space diagonal and the tour uses two legs per corner, so at most 4 legs are space diagonals; the rest can be face diagonals.

Solution

The path is a closed tour through the 88 corners, so it has 88 legs. A leg between two corners is an edge (length 11), a face diagonal (2\sqrt2), or a space diagonal (3\sqrt3).

Upper bound. Each corner is the endpoint of exactly one space diagonal (the one to the opposite corner). The tour passes through each corner exactly once, using two legs there, so at most one of those two legs is a space diagonal. Counting each space diagonal at both of its endpoints, the tour contains at most 82=4\dfrac{8}{2} = 4 space diagonals. The remaining legs are at most 2\sqrt2 each, so the length is at most 43+424\sqrt3 + 4\sqrt2.

Construction. Label corners by coordinates. The tour
000→111→001→110→011→100→010→101→000000 \to 111 \to 001 \to 110 \to 011 \to 100 \to 010 \to 101 \to 000
alternates: the odd legs flip all three coordinates (space diagonals) and the even legs flip exactly two (face diagonals). It visits every corner once and has length exactly 43+424\sqrt3 + 4\sqrt2.

The answer is (D) 42+43\boxed{\textbf{(D)}\ 4\sqrt{2}+4\sqrt{3}}.

Why this works

A maximum needs two halves: a bound and an example that meets it. The bound comes from a local resource count (one space diagonal per corner, two legs per corner), and the example is found by alternating "opposite corner" jumps with face-diagonal jumps. The alternating pattern is natural once you notice that after a space diagonal you must leave by something shorter.

The trap

Picking choice (E), which needs 5 space diagonals, without checking that a corner can be the endpoint of only one space diagonal.

Common mistakes

  • Picking choice (E), which needs 5 space diagonals, without checking that a corner can be the endpoint of only one space diagonal.
  • Assuming the fly must crawl along edges or faces, which caps the answer at (A) or (B).

Techniques

Bound the quantity above/below or estimate to pin it down · Consider the largest/smallest element or boundary case