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Math Help - area of quadrilateral

  1. #1
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    area of quadrilateral

    bk2 p105 q27
    question : the vertices of a quadrilateral are the centres of the circles:
    C_1 : x^2 +y^2+2tx=0
    C_2: x^2+y^2 +\frac {2y}t = 0
    and their intersecting points
    a) find the coordinates of the vertices of the quadrilaterl.
    b) find that the area of the quadrilateral is a constant.
    my working
    vertices :
    -t, 0
    0,- \frac 1 t
    0,0
    -\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}
    area:
    \frac 1 2 \begin{vmatrix}<br />
  0 & 0 \\<br />
  -t & 0 \\ <br />
0 & -\frac 1 t \\<br />
-\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\<br />
0 & 0<br /> <br />
\end{vmatrix}
    = \frac 1 2 (\frac {t^4-1}{1+t^4})
    cannot prove that the area is a constant
    thanks
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk2 p105 q27
    question : the vertices of a quadrilateral are the centres of the circles:
    C_1 : x^2 +y^2+2tx=0
    C_2: x^2+y^2 +\frac {2y}t = 0
    and their intersecting points
    a) find the coordinates of the vertices of the quadrilaterl.
    b) find that the area of the quadrilateral is a constant.
    my working
    vertices :
    -t, 0
    0,- \frac 1 t
    0,0
    -\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}
    Yes the co-ordinates are right.

    But the area is wrong. The area is a constant and it is 1.

    I did it the long way. First let O_1O_2 be the line joining the centers of C_1 and C_2.

    Then equation of the line O_1O_2 is \frac{x}{t} + yt + 1 = 0.

    Now the base of the triangle is |O_1O_2| = \sqrt{t^2 + \frac1{t^2}}.

    The perpendicular distance(height) from (0,0) to the line is \frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}}. So the area of the triangle is \frac12 |O_1O_2|\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |\frac{0}{t} + 0t + 1|= \frac12.

    The perpendicular distance(height) from (-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}) to the line is \frac{|\frac{-\frac {2t}{1+t^4}}{t} -\frac {2t^3 }{1+t^4}t + 1|}{\sqrt{t^2 + \frac1{t^2}}}. So the area of the triangle is \frac12 |O_1O_2|\frac{|-\frac {2}{1+t^4} -\frac {2t^4 }{1+t^4} + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |-\frac {2+2t^4 }{1+t^4} + 1| =\frac12 |-2 + 1| = \frac12.

    So the area of the quadrilateral is the sum of the area of triangles.

    And that is 1, a constant.
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  3. #3
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk2 p105 q27
    question : the vertices of a quadrilateral are the centres of the circles:
    C_1 : x^2 +y^2+2tx=0
    C_2: x^2+y^2 +\frac {2y}t = 0
    and their intersecting points
    a) find the coordinates of the vertices of the quadrilaterl.
    b) find that the area of the quadrilateral is a constant.
    my working
    vertices :
    -t, 0
    0,- \frac 1 t
    0,0
    -\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}
    area:
    \frac 1 2 \begin{vmatrix}<br />
0 & 0 \\<br />
-t & 0 \\ <br />
0 & -\frac 1 t \\<br />
-\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\<br />
0 & 0<br /> <br />
\end{vmatrix}
    = \frac 1 2 (\frac {t^4-1}{1+t^4})
    cannot prove that the area is a constant
    thanks
    When t = 1, A = 1. But your answer gives A = 0 ......
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    When t = 1, A = 1. But your answer gives A = 0 ......
    A = \frac{1}{2} \, |(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + (x_3 y_4 - x_4 y_3) + (x_4 y_1 - x_1 y_4)|

    where the order of the points is such that 1 connects to 2 connects to 3 connects to 4 connects to 1.

    So your answer is wrong because you have the order of the points wrong in the formula you used ......

    You should get A = \frac{1}{2} \left( \frac{2 t^4}{1 + t^4} + \frac{2}{1 + t^4}\right) = 1, as expected from the special case I considered in my earlier reply.
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  5. #5
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    Quote Originally Posted by afeasfaerw23231233 View Post
    bk2 p105 q27
    question : the vertices of a quadrilateral are the centres of the circles:
    C_1 : x^2 +y^2+2tx=0
    C_2: x^2+y^2 +\frac {2y}t = 0
    and their intersecting points
    a) ...
    b) find that the area of the quadrilateral is a constant.
    ...
    Here is a completely different attempt:

    1. Calculate the coordinates of the centers of the 2 circles:

    x^2+2tx+{\color{red}t^2}+y^2 = {\color{red}t^2} ~\iff~ (x+t)^2+y^2 = t^2

    x^2+y^2 +\frac {2y}t +{\color{red} \left(\frac1t \right)^2}={\color{red} \left(\frac1t \right)^2}~\iff~ x^2+\left(y+\frac1t\right)^2= \left(\frac1t \right)^2

    Then the area of the quadrilateral consists of 2 congruent right triangles. The legs of one of these right triangles have the lengthes(?) R = t and r = \frac1t.

    Therefore the area of the quadrilateral is:

    A = 2 \cdot \frac12 \cdot t \cdot \frac1t = 1
    Attached Thumbnails Attached Thumbnails area of quadrilateral-viereck_in2kreisen.gif  
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  6. #6
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    i joined the points in a wrong order.
    thanks for all your replies!
    Last edited by afeasfaerw23231233; June 14th 2008 at 07:17 AM.
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