🇺🇸 AMC 8 ⇄ switch contest
1996 AMC 8 Stretch

Problem 7

Problem 7 · AMC 8 Stretch Stretch
Geometry & Measurement Counting & Probability pigeonholevisual-representation
Nine points are placed anywhere inside a square whose sides are 1 unit long. Show that some 3 of these points form a triangle with area less than \(\tfrac18\). You may use this fact: any triangle that fits inside a rectangle has area less than half the rectangle's area.
Unit square partitioned into four quarter-squaresA(0,0)B(1,0)C(1,1)D(0,1)area 1/4
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Answer: a triangle of area less than 1/8
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Hint 1 of 4
The fact about triangles turns this into a 'get 3 points into a small region' problem. If 3 points sit in a region of area \(\tfrac14\), their triangle has area less than half of \(\tfrac14\).
Still stuck? Show hint 2 →
Hint 2 of 4
Cut the unit square into 4 equal small squares. What is the area of each?
Still stuck? Show hint 3 →
Hint 3 of 4
Each small square has area \(\tfrac14\) — and there are 4 of them (your boxes). You have 9 points.
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Approach: Pigeonhole — 9 points into 4 quarter-squares, then the triangle-area fact
  1. By the given fact, 3 points inside a region of area \(\tfrac14\) make a triangle of area less than \(\tfrac12 \times \tfrac14 = \tfrac18\). So we just need 3 of the 9 points in one region of area \(\tfrac14\).
  2. Cut the unit square into 4 equal small squares (slice it in half across and half down). Each small square has area \(\tfrac14\); these 4 squares are our boxes.
  3. Drop the 9 points into the 4 squares. Since \(9 = 4\times 2 + 1\), some square holds at least 3 points.
  4. Those 3 points form a triangle of area less than \(\tfrac18\).
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