AMC 10 · 2023 · #15

Start with scale. Walking 120 meters to travel 100 meters of field is only a 20% increase, so the legs should be gently tilted, not steeply slanted — and arccos5/6≈ 33.56° is indeed a shallow lean. A steep zigzag such as arccos3/10≈ 72.54° would have him crossing the field over and over and racking up more than three times the field's length, which the list confirms at 333.33 meters. Next check the boundary behaviour of the formula L=100/cosθ: at θ→ 0 it gives L→ 100, the straight walk down the long side, and as θ→ 90° it blows up, matching a walker who only goes up and down and never reaches BC. The answer sits sensibly between those extremes. Now stress the load-bearing assumption. The whole solution rests on the horizontal displacement being exactly 100, which comes from the phrase "ending on the segment BC" and nothing else. Step 6 tested that claim the hard way by building the path: three legs, ending at height 20√(11)-60≈ 6.33 meters up BC, with legs measuring about 54.27, 54.27, and 11.46 meters and summing to 120.00. That endpoint is inside [0,30], so the constructed walk is legal and the answer is not just algebraically forced but actually realizable. It is also worth noticing what the solution never needed: the field's 30-meter width. Changing the width changes the number of legs and where they land, but not θ — which is exactly why the problem could afford to say the leg count might differ from the figure. Finally, the distractors all decode: arccos4/5 is the answer to the same question with 125 meters, arcsin4/5 swaps sine for cosine on that same wrong pair, and arcsin5/6 is the complement of the right answer, the classic penalty for matching 5/6 to the opposite side instead of the adjacent one. Since the answer choices are arccos5/6 and arcsin5/6 both, the problem is quietly testing whether the 100 is adjacent to θ or opposite it, and Step 1 settled that: θ is measured off the long side, so the 100 runs alongside it.