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 Chord Harmonics (Posted on 2005-01-11)
Four chords of a circle are arranged in such a way that they form a quadrilateral shape, the length of the chords are as follows.

6cm, 7cm, 8cm and 9cm.

what is the area of the circle that will intersect all of the points?

 No Solution Yet Submitted by Juggler Rating: 4.0000 (3 votes)

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 re: nikki - step towards algebraic solution | Comment 6 of 9 |
(In reply to Still no proof, but... by nikki)

Here's a small step towards an algebraic solution.  I'm working off nikki's analysis in "Still no proof, but..."

We have a sum of four doubl-angle arcsin terms as equal to 360 degrees ():
2*a + 2*b + 2*c + 2*d = 360

Take the sine of each side of the equation:
sin( 2*(a + b + c + d) ) = sin(360) = 0

using the double-angle identity for sines { sin(2*a) = 2*sin(a)cos(a) } gives us:
2 * sin( a + b + c + d ) cos( a + b + c + d) = 0

now we can reduce this equation using the additive identities for sin and cos:
{ sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
and cos(a + b) = cos(a)cos(b) - sin(a)sinb) }
we get:
2 * ( sin( a+b)cos(c+d) + cos(a+b)sin(c+d) ) * (cos(a+b)cos(c+d) - sin(a+b)sin(c+d) ) ) = 0
<=> ( (sin(a)cos(b) + cos(a)sin(b))*(cos(c)cos(d) - sin(c)sin(d) ) + (cos(a)cos(b) - sin(a)sin(b))*(sin(c)cos(d) + cos(c)sin(d)) ) * ( (cos(a)cos(b) - sin(a)sin(b)) * (cos(c)cos(d) - sin(c)sin(d)) - (sin(a)cos(b) + cos(a)sin(b)) * (sin(c)cos(d) + cos(c)sin(d)) )  = 0
<=> ( sin(a)cos(b)cos(c)cos(d) - sin(a)cos(b)sin(c)sind(d) + cos(a)sin(b)cos(c)cos(d) - cos(a)sin(b)sin(c)sin(d) - cos(a)cos(b)sin(c)cos(d) + cos(a)cos(b)cos(c)sind(d) - sin(a)sin(b)sin(c)cos(d) - sin(a)sin(b)cos(c)sin(d) )  *  (
cos(a)cos(b)cos(c)cos(d) - cos(a)cos(b)sin(c)sin(d) - sin(a)sin(b)cos(c)cos(d) + sin(a)sin(b)sin(c)sin(d) - sin(a)cos(b)sin(c)cos(d) - sin(a)cos(b)cos(c)sin(d) - cos(a)sin(b)sin(c)cos(d) - cos(a)sin(b)cos(c)sin(d) ) = 0

Now all that's left is to plug in expressions for a,b,c, and d and continue to simplify.  We have
a = arcsin(6/(2r)) , b = arcsin(7/(2r)) , c = arcsin(8/(2r)) , and d = arcsin(9/(2r) so then:
sin(a) = sin(arcsin(6/(2r))) = 6/(2r)
and cos(a) = cos(arcsin(6/2r))) = sqrt( (2r)^2 - 6^2 ) / (2r)

The easiest way to convince yourself cos(a) is this value is to draw a right triangle and consider one non-right angle.  If that angle is arcsin(6/2r) then the opposite side is length 6 and the hypotenuse length 2r.  The adjacent side is then length sqrt( (2r)^2 - 6^2) so the cos of that angle is sqrt( (2r)^2 - 6^2 ) / (2r).

Actually plugging in these values and solving for r looks way more hairy than I have time to work on now, so good look if anybody tries it :)

 Posted by Jay Schamel on 2005-01-14 22:37:42

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