AMC 10 · 2010 · #7

Grade 8 geometry-2d
pythagorean-theoremcoordinate-geometryangle-sum-triangle identify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 2 insights
Problem
Crystal runs a four-leg loop: 1 mile due north, then 1 mile northeast, then 1 mile southeast, then a straight line back to the start. Find the length of that final straight leg.

Pick an answer.

(A)
$\ 1$
(B)
$\ \sqrt{2}$
(C)
$\ \sqrt{3}$
(D)
$\ 2$
(E)
$\ 2\sqrt{2}$

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

How to solve
Strategy Draw a Diagram

The problem is all positions and paths, so the natural tool is #1 (Draw a Diagram): drop everything onto a coordinate grid with east as +x and north as +y. The slanted miles are hard to add directly, so use Tool #7 (Identify Subproblems) to break each 45° mile into an east piece and a north piece using a 45-45-90 triangle. Add the pieces to find the endpoint, then read off a right triangle and use the Pythagorean theorem for the straight distance home.

1STEP 1

Set up coordinates

Put the start at the origin, with east as +x and north as +y. Then every leg is just a change in x and y that you can add.

Start = (0,0), east = +x, north = +y.
2STEP 2

The north leg

The first mile goes due north, so only the y-coordinate changes and she stands at (0,1).

(0,0) + (0,1) = (0,1).
3STEP 3

Split the diagonal legs

A 1-mile step at 45° is the hypotenuse of a 45-45-90 triangle, so its east and vertical pieces are each √(2)/2.

NE = (√(2)/2, √(2)/2), SE = (√(2)/2, -√(2)/2).
4STEP 4

Add the pieces

The two diagonals' vertical parts cancel, the east parts add to √(2), and y stays 1 — so she ends at (√(2), 1).

x = √(2)/2 + √(2)/2 = √(2), y = 1 + √(2)/2 - √(2)/2 = 1.
5STEP 5

Distance back to start

The way home is the hypotenuse of a right triangle with legs √(2) and 1, so it measures √(3) miles — choice (C).

d = √((√(2))² + 1²) = √(2 + 1) = √(3).
Answer
√(3)
Check the size: √(3) ≈ 1.73 miles, which sits between 1 and 2 — sensible for a loop built from 1-mile legs, not tiny and not huge. Re-trace the path: north to (0,1); northeast adds (√(2)/2, √(2)/2) giving (√(2)/2, 1+√(2)/2); southeast adds (√(2)/2, -√(2)/2) giving (√(2), 1). Distance √(2+1) = √(3). Rule out neighbors: √(2) would need the endpoint's squared distance to be 2, but (√(2))² + 1² = 3 ≠ 2; the value 2 would need the squared legs to sum to 4. So (C) is locked in.
💡Key takeaway

Put the run on a grid and split each slanted mile into an east part and a north part — the two diagonal north parts cancel, leaving a plain right triangle with legs √(2) and 1, so the walk home is √(3) miles.

  • Set up coordinates
  • The north leg
  • Split the diagonal legs
  • Add the pieces
  • Distance back to start