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Math Help - 3 questions

  1. #1
    Junior Member Freaky-Person's Avatar
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    Smile 3 questions

    I got 3 questions to solve here.

    1) if: x^2 + x + 1 = 0 then calculate: x^3333 + x^333+ x^33 + x^3 + 3 = ?


    2) a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da

    Prove: a=b=c=d


    3) You are given a circles diameter and a point that lies above the diameter. without using a protractor how would one find the EXACT perpendicular line (EXACTLY 90 degrees) from the point to the diameter.

    Thanksies
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post
    I got 3 questions to solve here.

    1) if: x^2 + x + 1 = 0 then calculate: x^3333 + x^333+ x^33 + x^3 + 3 = ?
    Solve,
    x^2+x+1=0
    x=\frac{-1\pm \sqrt{1-4} }{2}
    x=-1/2+i/2\cdot \sqrt{3}, -1/2-i/2\cdot \sqrt{3}
    ---
    There are 2 ways. But I found a more elegant approach.
    The expected way is through de Moiver's theorem.
    ---
    We know that,
    x^2=-x-1 (1)
    Multiply by x,
    x^3=-x^2-x
    Substitute equation (1),
    x^3=-(-x-1)-x=1.
    Thus,
    x^3=1
    Raise by power of: 11,111,1111 respectively to get,
    x^{33}=1
    x^{333}=1
    x^{3333}=1
    Thus,
    1+1+1+1+3=7
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  3. #3
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    Quote Originally Posted by Freaky-Person View Post


    2) a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da

    Prove: a=b=c=d
    This is the Cauchy-Swartzh inequality.
    Consider,
    \bold{x}=<a,b,c,d>
    \bold{y}=<b,c,d,a>
    Then,
    |\bold{x}\cdot \bold{y}|\geq ||\bold{x} ||\cdot ||\bold{y}||
    Furthermore, we only have equality when,
    \bold{x}=\bold{y}
    Thus,
    a=b=c=d
    Last edited by ThePerfectHacker; January 2nd 2007 at 07:15 AM.
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  4. #4
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    Quote Originally Posted by Freaky-Person View Post
    3) You are given a circles diameter and a point that lies above the diameter. without using a protractor how would one find the EXACT perpendicular line (EXACTLY 90 degrees) from the point to the diameter.
    I got a problem with this question. I think in projective geometry it is shown that any construction with a straightedge and compass can be done with the compass alone. But they are certainly talking about something else there. For example, you cannot draw a line with a compass! That is what you are asking, no?
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  5. #5
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    This is the Cauchy-Swartzh inequality.
    Consider,
    \bold{x}=<a,b,c,d>
    \bold{y}=<b,c,d,a>
    Then,
    \bold{x}\cdot \bold{y}\geq ||\bold{x} ||\cdot ||\bold{y}||
    Furthermore, we only have equality when,
    \bold{x}=\bold{y}
    Thus,
    a=b=c=d
    ....

    what?
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  6. #6
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I got a problem with this question. I think in projective geometry it is shown that any construction with a straightedge and compass can be done with the compass alone. But they are certainly talking about something else there. For example, you cannot draw a line with a compass! That is what you are asking, no?
    no no, what I mean, is I need to find how to make a perfect 90 degrees angle from the diameter. The line doesn't matter really, I just need exactly 90 degrees without a compass. Just how it could be done.
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  7. #7
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    Quote Originally Posted by Freaky-Person View Post
    ....

    what?
    The Cauchy-Swartzh Inequality says,
    (a\cdot b+b\cdot c+c\cdot d+d\cdot a)\geq  \sqrt{a^2+b^2+c^2+d^2}\cdot \sqrt{b^2+c^2+d^2+a^2}
    Thus,
    ab+bc+cd+ad\geq a^2+b^2+c^2+d^2
    And the only way we have equality (like here).
    Is when, (all are equal)
    a=b=c=d
    Last edited by ThePerfectHacker; January 1st 2007 at 02:21 PM.
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  8. #8
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    Quote Originally Posted by Freaky-Person View Post
    no no, what I mean, is I need to find how to make a perfect 90 degrees angle from the diameter. The line doesn't matter really, I just need exactly 90 degrees without a compass. Just how it could be done.
    Let me butt in.

    Can you use a ruler or any straigth edge and a measuring device for distances if the straight edge has none?
    If yes, then draw two equal line segments from the point to the diameter. Each of these equally long line segments (rays) must be longer than the shortest distance from the point to the diameter. [You can approximate that, I'm sure.]. Then get the midpoint of the line segment on the diameter that is cut by these two rays using the measuring device. Then connect this midpoint to the given point. There, you have a 90-degree angle.

    [The perpendicular bisector of the base in an isosceles triangle passes through the apex of the triangle.]
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  9. #9
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    Hello, Freaky-Person!

    I solved #1 in a manner similar to ThePerfectHacker.


    1) If x^2 + x + 1 \:=\:0, then calculate: x^{3333} + x^{333} + x^{33} + x^3 + 3

    We are given: . x^2 + x + 1 \:=\:0

    Multiply both sides by (x - 1)\!:\;\;(x - 1)(x^2 + x + 1) \:=\:0(x-1)

    . . and we have: . x^3 - 1 \:=\:0\;\;\Rightarrow\;\;x\:=\:\sqrt[3]{1}

    Hence, x = the three cube roots of 1.
    . . They happen to be: . x_1\,=\,1,\;x_2\,=\,\frac{-1 + i\sqrt{3}}{2},\;x_3\,=\,\frac{-1-i\sqrt{3}}{2}
    But we really don't care about their exact values,
    . . we only care that: . x_i^3\,=\,1


    So we have: . x^{3333} + x^{333} + x^{33} + x^3 + 3 \;= \;\left(x^3\right)^{1111} +\left(x^3\right)^{111} + \left(x^3\right)^{11} + \left(x^3\right) + 3

    . . . . . . = \;1^{1111} + 1^{111} + 1^{11} + 1 + 3 \;=\;1 + 1 + 1 + 1 + 3 \;=\;\boxed{7}

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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Freaky-Person View Post
    3) You are given a circles diameter and a point that lies above the diameter. without using a protractor how would one find the EXACT perpendicular line (EXACTLY 90 degrees) from the point to the diameter.
    You can use neither a protractor nor a compass? (You didn't mention you couldn't use a compass in your problem statement.) I presume you can't use a ruler, either, but I'm hoping you are at least allowed a straight-edge.

    I was reviewing The Elements. I don't think this can be done without a compass and a straight-edge.

    -Dan
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  11. #11
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    Here is another way to solve #1.

    We can write,
    x=-\cos \frac{\pi}{3}\pm i\sin \frac{\pi}{3}
    When we cube it by de Moivers theorem,
    x^3=-\cos \pi \pm i\sin \pi=1
    The rest is trivial.
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  12. #12
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    Quote Originally Posted by topsquark View Post
    You can use neither a protractor nor a compass? (You didn't mention you couldn't use a compass in your problem statement.) I presume you can't use a ruler, either, but I'm hoping you are at least allowed a straight-edge.
    As I said any construction that is doable with a straightedge and compass can be done with a compass along. Thus, if you can do it with both then you can do it with a single one.
    I was reviewing The Elements. I don't think this can be done without a compass and a straight-edge.
    You know there is a way to prove impossibility.
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  13. #13
    Junior Member Freaky-Person's Avatar
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    you can use a straight edge, but the thing is, my teacher said that not all edges we are given are exactly 90 degrees and he said we're using bad Canadian rulers with all the edges rounded XD
    Last edited by ThePerfectHacker; January 2nd 2007 at 01:25 PM.
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  14. #14
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    Quote Originally Posted by Freaky-Person View Post
    you can use a straight edge, but the thing is, my teacher said that not all edges we are given are exactly 90 degrees and he said we're using bad Canadian rulers with all the edges rounded XD
    I see this was last edited by Hacker...I wonder what he changed



    Also!

    You could use LOCI construction of the right angles if you had a compass. Maybe you could fold paper
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  15. #15
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    Quote Originally Posted by anthmoo View Post
    I see this was last edited by Hacker...I wonder what he changed
    We have the most strict proper language policy than any other forum on the internet.
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