AMC 10 · 2009 · #13

Grade 10 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
law-of-cosinestrigonometric-ratiosinterval-arithmetic bound-inequality-then-enumerateextreme-principle ↑ Prerequisites: law-of-cosines
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A ship runs one leg, turns by an angle somewhere in a given range, then runs a second leg. Find which interval contains every possible squared distance back to the start.

Pick an answer.

(A)
[400,500]
(B)
[500,600]
(C)
[600,700]
(D)
[700,800]
(E)
[800,900]
How to solve
Strategy Extreme Principle

The question asks about a whole family of triangles at once, so testing the two extreme turns is the natural move. But testing two cases only proves the two endpoint values; by itself it says nothing about the turns in between, and an interval has to contain all of them. So the plan has two halves that must both be done. First, prove that AC moves in only one direction as the turn opens up, which is what makes the two extreme cases genuine bounds instead of just two data points. Second, compute those two bounds exactly. Coordinates plus the Pythagorean theorem do the computing, and they also reveal something the raw formula hides: AC² depends on the position of C only through how far east C ends up. The last step is a size comparison, because the upper bound is irrational and has to be shown to stay below 800.

1STEP 1

A turn is a deflection, not the angle at B

The turn is a deflection, so the interior angle is its supplement.

θ = turn angle, 45° ＜ θ ＜ 60°, ∠ ABC = 180° - θ ∈ (120°, 135°)
2STEP 2

AC squared depends only on how far east C is

Coordinates make the squared distance depend on one coordinate.

B=(0,0), A=(-10,0), C=(x,y), x²+y²=400 ⟹ AC²=(x+10)²+y²=500+20x
3STEP 3

Show AC only shrinks as the turn opens

A larger turn always shortens the closing distance.

AB=10, BC=20 fixed, θ₁ ＜ θ₂ → ∠ AB C₁ = 180°-θ₁ ＞ 180°-θ₂ = ∠ ABC₂ SAS ineq.{⟹} AC₁ ＞ AC₂
4STEP 4

Evaluate the two extreme turns

Evaluating the two extreme turns brackets the range.

x(45°)=20·√(2)/2=10√(2) → AC²=500+200√(2); x(60°)=20·1/2=10 → AC²=700; 700 ＜ AC² ＜ 500+200√(2)
5STEP 5

Trap the range inside one interval

That whole range fits inside [700,800], choice (B).

500+200√(2) ＜ 800 ⇔ √(2) ＜ 3/2 ⇔ 2 ＜ 9/4; 700 ＜ AC² ＜ 500+200√(2) ≈ 782.8 ⊂ [700,800]
Answer
[700,800]
Two independent checks agree. First, a crude bound: the triangle inequality gives 10 ＜ AC ＜ 30, so AC² sits between 100 and 900. The extremes of that band correspond to the two degenerate turns, a 180-degree turn that doubles the ship back to AC = 10 and a zero turn that runs it straight out to AC = 30. A turn of only 45 to 60 degrees is much nearer the straight-line case than the doubling-back case, so AC² should sit in the upper part of the band but well short of 900. The bracket 700 to 783 fits that expectation. Second, a spot check strictly inside the range using the 50-degree turn drawn in the figure: cos 50° ≈ 0.643, so AC² ≈ 500 + 400(0.643) ≈ 757, comfortably inside [700,800] and between the two computed bounds, exactly as the monotonicity claim predicts. Choice (D) survives both checks.
💡Key takeaway

When an angle is free to slide over a range, checking the two ends only counts as proof after you show the answer moves in one direction the whole way between them.

  • A turn is a deflection, not the angle at B
  • AC squared depends only on how far east C is
  • Show AC only shrinks as the turn opens
  • Evaluate the two extreme turns
  • Trap the range inside one interval

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