Problem 16 · AMC 8 Stretch
Stretch
Geometry & Measurement
Counting & Probability
pigeonholevisual-representation
Thirteen points are placed in a rectangle with side lengths 3 and 2. Show that some 3 of them form a triangle with area at most \(\tfrac12\). You may use this fact: any triangle that fits inside a rectangle has area less than half the rectangle's area.
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Answer: a triangle of area at most 1/2
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Hint 1 of 4
Same trick as the unit-square problem: get 3 points into a small region of area 1, and the triangle is smaller than half of 1.
Still stuck? Show hint 2 →
Hint 2 of 4
The big rectangle is 3 wide and 2 tall, so its area is 6. Cut it into unit squares. How many do you get?
Still stuck? Show hint 3 →
Hint 3 of 4
You get 6 unit squares (3 across, 2 down), each of area 1 — your 6 boxes. You have 13 points.
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Approach: Pigeonhole — 13 points into 6 unit squares, then the triangle-area fact
- Cut the \(3 \times 2\) rectangle into 6 unit squares (3 across, 2 down), each of area 1. These 6 squares are the boxes.
- Drop the 13 points into the 6 squares. Since \(13 = 6\times2 + 1\), one square must hold at least 3 points.
- By the given fact, 3 points inside a region of area 1 form a triangle of area less than \(\tfrac12 \times 1 = \tfrac12\).
- So some 3 of the points form a triangle of area at most \(\tfrac12\).
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