Problem 1 · AMC 8 Stretch
Stretch
Geometry & Measurement
symmetryvisual-representation
Two flagpoles stand on flat ground, 8 meters apart. The left pole is 6 m tall and the right pole is 4 m tall. A rope 10 m long is tied to the top of each pole. A ring slides freely on the rope and a weight hangs from it, so the two parts of the rope make equal angles with the vertical (a perfectly balanced clothesline). How high above the ground does the weight hang (in meters)?
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Answer: 2 meters
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Hint 1 of 4
Draw the picture: two poles, the rope, the weight. The rope makes equal angles with the vertical on both sides. Keep that word 'equal' in mind.
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Hint 2 of 4
Equal angles is exactly what a mirror does. Reflect the right pole (and its half of the rope) straight down across the horizontal line through the weight. The whole rope becomes one straight line of length 10 — the hypotenuse of a right triangle.
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Hint 3 of 4
The horizontal leg is the distance between the poles, 8 m. The vertical leg is how far the two tops are apart once one is flipped down: \((6 - h) + (4 - h)\), where \(h\) is the weight's height.
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Approach: Reflect to straighten the rope, then Pythagorean theorem
- Let \(h\) be the weight's height. The rope makes equal angles with the vertical, which is the signature of a mirror reflection, so reflect the right pole and its rope half straight down across the horizontal line through the weight \(W\).
- After reflecting, the right rope segment lines up with the left one and the whole 10 m rope becomes a single straight line — the hypotenuse of a right triangle.
- The horizontal leg is the 8 m between the poles. The vertical leg: the left top is \((6 - h)\) above \(W\) and the reflected right top is \((4 - h)\) below \(W\), so the legs differ by \((6 - h) + (4 - h) = 10 - 2h\).
- Pythagoras: \(10^2 = 8^2 + (10 - 2h)^2\), so \((10 - 2h)^2 = 36\), giving \(10 - 2h = 6\), so \(h = 2\). Check: \(6^2 + 8^2 = 100 = 10^2\). The weight hangs 2 meters above the ground.
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