Problem 4 · AMC 8 Stretch
Stretch
Counting & Probability
Number TheoryGeometry & Measurement
pigeonholeparity
Now do the same idea in 3-D space. Mark any 9 points whose coordinates \((x,y,z)\) are all whole numbers. Show that 2 of them have a midpoint with all whole-number coordinates.
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Answer: 2 points share a parity pattern
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Hint 1 of 4
Use the same trick as the flat (2-D) version, but now there are three coordinates, each even or odd.
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Hint 2 of 4
Count the even/odd patterns for a triple \((x,y,z)\): each of the 3 spots is even or odd, so multiply \(2\times2\times2\).
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Hint 3 of 4
That's 8 patterns — your 8 boxes. You have 9 points.
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Approach: Pigeonhole on 3-D parity patterns — 9 points, \(2^3 = 8\) classes
- As in the flat version, the only thing that matters about a point is whether each coordinate is even or odd. With three coordinates each even or odd, the number of patterns is \(2\times2\times2 = 8\).
- Make these 8 patterns your boxes and drop the 9 points in. Since \(9 > 8\), two points land in the same box.
- They match in even/odd for \(x\), for \(y\), and for \(z\), so \(x_1+x_2\), \(y_1+y_2\), and \(z_1+z_2\) are all even.
- Dividing each by 2 gives whole numbers, so the midpoint has all whole-number coordinates.
Mark:
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