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Math Help - more circle

  1. #1
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    more circle

    I am studying, and there are just 3 problems I don't quite understand. Plz help me.

    1. Let Γ be a circle with center Q, let [AB] and [CD] be chords of Γ(so that the endpoints lie on the circle), and let G and H be the midpoints of [AB] and [CD]. Prove that d(Q, G) = d(Q, H) if and only if d(A, B) = d(C, D), and d(Q, G) < d(Q,H) if and only if d(A, B) > d(C, D).

    2. Let Γ be a circle with center Q, and let L be a line containing a point X on
    Γ. Prove that X is the only common point of Γand L if and only if QX is perpendicular to L. (These are the usual synthetic descriptions for the tangent line to Γat X.) [ Hint : If L also meets Γat another point Y, explain why angleQXY is acute. ]


    3. Let Γ be a circle with center Q, let X be a point in the exterior of Γ, and let A and B be two points of Γwhich lie on opposite sides of QX such that XA and XB are tangent to Γin the sense of the preceding exercise. Prove that d(X, A) = d(X, B).
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by jenjen View Post
    I1. Let Γ be a circle with center Q, let [AB] and [CD] be chords of Γ(so that the endpoints lie on the circle), and let G and H be the midpoints of [AB] and [CD]. Prove that d(Q, G) = d(Q, H) if and only if d(A, B) = d(C, D), and d(Q, G) < d(Q,H) if and only if d(A, B) > d(C, D).
    The triangles \triangle QAG and \triangle QHC are right triangles and QG=QH, \ QA=QC=R.
    Then \triangle QAG\equiv\triangle QHC\Rightarrow AG=CH\Rightarrow AB=CD.

    We have R^2=QA^2=QG^2+AG^2
    R^2=QC^2=QH^2+CH^2.
    Then:
    QG<QH\Rightarrow QG^2<QH^2\Rightarrow QG^2-QH^2<0
    QG^2-QH^2=(R^2-AG^2)-(R^2-CH^2)=CH^2-AG^2
    Then CH^2<AG^2\Rightarrow CH<AG\Rightarrow 2CH<2AG\Rightarrow CD<AB
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by jenjen View Post
    I am studying, and there are just 3 problems I don't quite understand. Plz help me.

    2. Let Γ be a circle with center Q, and let L be a line containing a point X on
    Γ. Prove that X is the only common point of Γand L if and only if QX is perpendicular to L. (These are the usual synthetic descriptions for the tangent line to Γat X.) [ Hint : If L also meets Γat another point Y, explain why angleQXY is acute. ]
    .
    X is the only common point of \Gamma and L

    \Longleftrightarrow L is tangent to \Gamma

    \Longleftrightarrow the line connecting X and the center of \Gamma is perpendicular to L..

    if we use the hint..
    suppose L meets \Gamma at points X and Y.. then \Delta QXY is an isosceles triangle with \overline {QX}=\overline {QY}. hence, m\angle QXY = m\angle QYX
    now, if m\angle QXY \geq 90^o, then the sum of the internal angles of \Delta QXY would be > 180^o

    therefore, m\angle QXY < 90^o.. QED
    Last edited by kalagota; November 24th 2007 at 05:31 AM.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by jenjen View Post
    3. Let Γ be a circle with center Q, let X be a point in the exterior of Γ, and let A and B be two points of Γwhich lie on opposite sides of QX such that XA and XB are tangent to Γin the sense of the preceding exercise. Prove that d(X, A) = d(X, B).
    consider \triangle QAX and \triangle QBX

    XA and XB are tangent lines, \implies m\angle QAX = 90^o = m\angle QBX.. also, \overline {QA} = \overline {QB} and \overline {QX} = \overline {QX}.. hence, \triangle QAX \sim \triangle QBX. therefore, \overline {XA} = \overline {XB}. QED
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  5. #5
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    Hello, jenjen!

    Here's #3 . . .


    3. Let C be a circle with center Q.
    Let X be a point in the exterior of C,
    and let A and B be two points of C which lie on opposite sides of QX
    such that XA and XB are tangent to C.
    Prove that: . XA = XB
    Code:
                  * * *
              *           *  A
            *               *
           *              /  * *
                        /         *
          *           /       *     *
          *        Q* - - - - * - - - *X
          *           \       *     *
                        \         *
           *              \  * *
            *               *
              *           *  B
                  * * *

    Tangent XA is perpendicular to radius QA.
    Tangent XB is perpendicular to radius QB.
    . . Hence, triangles QAX and QBX are right triangles.

    Since QA and QB are radii of circle Q, QA = QB.
    . . And, of course: . QX = QX.

    Hence: . \Delta QAX \cong \Delta QBX

    Therefore: .  XA = XB

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  6. #6
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    thank you sooo much!
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