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  #1  
Old 06-14-2008, 05:54 AM
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Default 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}  0 & 0 \\  -t & 0 \\ 0 & -\frac 1 t \\-\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\0 & 0\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  
Old 06-14-2008, 06:17 AM
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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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Old 06-14-2008, 06:22 AM
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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}0 & 0 \\-t & 0 \\ 0 & -\frac 1 t \\-\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\0 & 0 \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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Old 06-14-2008, 06:43 AM
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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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Old 06-14-2008, 07:15 AM
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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
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  #6  
Old 06-14-2008, 09:06 AM
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i joined the points in a wrong order.
thanks for all your replies!

Last edited by afeasfaerw23231233; 06-14-2008 at 09:17 AM.
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