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Homework Help: Trisectible angles | divisibility

  1. Mar 18, 2008 #1
    1) We know that if [itex]\theta[/itex] is trisectible (with straightedge and compass), then [itex]\theta[/itex]/3 is constructible.

    But is it also true that if [itex]\theta[/itex]/3 is constructible, then [itex]\theta[/itex] is trisectible (with straightedge and compass)?

    If so, then I can say that since 15o is constructible, we have that 45 o is trisectible, right? (because we can copy an angle of 15o three times, thus trisecting the angle 45 o)


    2) Let m,n be integers.
    Then m|3n3 => m|n
    and n|28n3 => n|m

    I spent half an hour thinking about this, but I still have no clue...
    Why are the implications (=>) true? Can someone please explain?


    3) How can I prove that the acute angle whose cosine is 1/10 is constructible?
    I know that if [itex]\theta[/itex] is constructible, then cos[itex]\theta[/itex] is constructible. But is the converse true? Why or why not?

    Any help is appreciated!:smile:
     
    Last edited: Mar 18, 2008
  2. jcsd
  3. Mar 18, 2008 #2
    For 1) your logic seems right. For 2), I don't see how the statements are true--are there any other qualifying statements? For example, in the first case, what if m = n2? Or what if m = 3 and n = 2?
     
  4. Mar 20, 2008 #3
    1) So is it true that [itex]\theta[/itex] is trisectible (with straightedge and compass) IF AND ONLY IF [itex]\theta[/itex]/3 is constructible (with straightedge and compass)?

    2) The whole situtation is this:
    http://www.geocities.com/asdfasdf23135/absmath1.jpg
    I circled the parts in red which corresopnds to what I've included in my top post.
    I don't understand why:
    m|3n3 => m|n
    and n|28n3 => n|m
    where m,n are integers.



    Can anyone help?
     
  5. Mar 22, 2008 #4
    Can someone please help me with Q3 as well?

    I am sure someone here understands it. Please help!!
     
  6. Mar 24, 2008 #5
    Still wondering...
     
  7. Mar 24, 2008 #6
    Isn't your first question essentially, "Can you construct an integer multiple of a constructable angle?" Well...can you?
     
  8. Mar 25, 2008 #7
    1) I think that if [itex]\theta[/itex]/3 is constructible, then we can trisect [itex]\theta[/itex] with straightedge and compass by copying the angle [itex]\theta[/itex]/3 two times (since we can always copy any angle with straightedge and compass)
     
  9. Mar 25, 2008 #8

    tiny-tim

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    … cos^-1 of constructible number is constructible angle …

    Hi kingwinner! :smile:

    Any angle whose cosine is a constructible number between -1 and 1 (like 4/5 or 1/√2) is constructible!

    Hint: draw a circle. Draw one radius. Mark 1/10 along that radius. And then … ? :smile:
     
  10. Mar 25, 2008 #9
    And then erect a pernpendicular at that point to consturuct the angle??? (since on the unit circle, x=cos(theta), where theta is counterclockwise from positive x-axis)
     
  11. Mar 25, 2008 #10

    tiny-tim

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    Yes!!!! :smile:

    (… why only three question marks? …)
     
  12. Mar 25, 2008 #11
    3) So we have theta constructible if and only if cos(theta) is constructible


    2) Let m,n be integers
    m|3n3 => m|n
    and n|28n3 => n|m
    Do you think these are actually wrong implications? (i.e. whoever was writing the solutions got it wrong...)
     
  13. Mar 25, 2008 #12
    Without additional assumptions on m and n, the implications aren't true...
     
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