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anonymous

  • 4 years ago

"A plane contains points A and B with AB=1. Let S be the union of all disks of radius 1 in the plane that cover AB. What is the area of S?" As taken from the 2004 AMC. Here's a math challenge for those of you who like math. ;)

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  1. bahrom7893
    • 4 years ago
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    oh dang... there are just way too many circles to graph till u get to the outermost (union of all of them)

  2. bahrom7893
    • 4 years ago
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    and then u gotta subtract some complements of the intersection, uhmm no thanks!

  3. anonymous
    • 4 years ago
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    lol

  4. anonymous
    • 4 years ago
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    Too many circles to graph? Nonsense. You have two points connected by a line, the circles just pivot on each one.

  5. anonymous
    • 4 years ago
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    And then you have the circles that must lie on both A and B.

  6. anonymous
    • 4 years ago
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    I'm not redrawing it here, but you have two sectors with area 4pi/3, two sectors with area pi/6, and the intersection has area sqrt(3)/2. Answer should be 8pi/3+pi/3-sqrt(3)/2=3pi-sqrt(3)/2

  7. bahrom7893
    • 4 years ago
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    But he's asking for a union of everything

  8. bahrom7893
    • 4 years ago
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    i see six minimum

  9. anonymous
    • 4 years ago
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    Right you are, Sir! Here, have several solutions.

  10. anonymous
    • 4 years ago
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    S is the union of all the disks, but it asks for the area of S, not the area of all the disks.

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