AMC 10 · 2020 · #2

Grade 8 geometry-2d
Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum
coordinate-geometrypythagorean-theoremsystematic-enumerationspatial-visualization identify-subproblemssystematic-enumeration ↑ Prerequisites: pythagorean-theoremcoordinate-geometry
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
Three letters are drawn on a grid of unit squares, using only heavy strokes running between grid points. Every stroke is either along a grid line or along a diagonal of grid squares. Add up the lengths of all the strokes that make the three letters.

Pick an answer.

(A)
17
(B)
$15 + 2\sqrt{2}$
(C)
$13 + 4\sqrt{2}$
(D)
$11 + 6\sqrt{2}$
(E)
21
How to solve
Strategy Identify Subproblems

A picture with ten separate strokes is not one length to compute — it is ten. Tool #7 (Identify Subproblems) splits the figure into single strokes so each one has an obvious length. Tool #2 (Make a Systematic List) walks the letters left to right so no stroke is skipped and none is double-counted at a junction. Tool #15 (Organize Information in More Ways) then re-sorts that list into two piles, grid-aligned and slanted, because every stroke in the first pile is a whole number of units and every stroke in the second is a multiple of √(2) — two piles that add cleanly on their own but not with each other. Tool #3 (Eliminate Possibilities) is kept in reserve: the answer choices all differ in how much √(2) they carry, so the slanted pile alone can decide the answer.

1STEP 1

List every stroke by its endpoints

Write each stroke's endpoints.

A: (0,0) - (2,2), (2,0) - (2,2), (1,1) - (2,1) M: (3,0) - (3,2), (5,0) - (5,2), (3,2) - (4,1), (4,1) - (5,2) C: (6,0) - (8,0), (6,2) - (8,2), (6,0) - (6,2)
2STEP 2

Sort into grid-aligned and slanted

Sort into grid-aligned and slanted.

grid-aligned: 2, 1, 2, 2, 2, 2, 2 slanted: (0,0) - (2,2), (3,2) - (4,1), (4,1) - (5,2)
3STEP 3

Add the grid-aligned lengths

The grid-aligned lengths simply add.

(2+1)+(2+2)+(2+2+2)=3+4+6=13
4STEP 4

Measure a slant with the Pythagorean theorem

The slanted ones need Pythagoras.

√(1²+1²)=√(2), √(2²+2²)=√(8)=2√(2)
5STEP 5

Combine the two piles

Combining gives thirteen plus four root two.

2√(2)+√(2)+√(2)=4√(2) → 13+4√(2) → (C)
Answer
13 + 4√(2)
The size is believable: 4√(2)≈5.66, so the total is about 18.7 units of ink for three letters that each stand 2 units tall inside an 8×2 region. It must also exceed 13 (the straight part alone) and stay under 13+4·2=21 (the crude bound that pretends each unit diagonal is 2), and 18.7 sits inside. Cross-checking the count another way: the figure holds four unit-diagonals in total, two hidden inside the A's long slant and one in each slant of the M, which is exactly the 4√(2) found. The wrong choices are the shapes of the common miscounts — (A) 17 and (E) 21 come from treating each diagonal as 1 or 2 units, (B) counts the A's long slant as a single √(2), and (D) miscounts which strokes are slanted.
💡Key takeaway

Break a drawing into single strokes, then split them into two piles — straight ones you add as whole numbers, slanted ones you measure with the Pythagorean theorem and count as multiples of √(2).

  • List every stroke by its endpoints
  • Sort into grid-aligned and slanted
  • Add the grid-aligned lengths
  • Measure a slant with the Pythagorean theorem
  • Combine the two piles

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