# Problem 256

Name Message
John Nolton
Posts: 14

Joined: 10/11/2011
 Problem 256 Flag »  Reply » The solution to this problem is hit the Enter key on the computer and you get;r= 434.605395867015829+This type of solution seems to be OK with Shelly. I sent her an Email and told her I object to using a computer as Lindell suggest in his solution. In his final equation (which he does not number) he says, "This can be set up in "Goal Seek" in Excell or solved by any good on-line equation solver.If you know a little Calculus you can solve his final equation with ease.Lets look at the problem; There were 2 typo errors which I reported. One was corrected in 3 days but the other is still there. In equation [2] in the denominator it should read (s-a)(s-b)(s-c)(s-d).There is no s in front of this part of the equation for the cyclic quadrilateral. In other words it should NOT READ s(s-a)(s-b)(s-c)(s-d).Now by inspection of of the 4th equation down you can see that no s was used.I guess a direct solution has it place but I always try to look for the simple solution and move on.I set this problem up in one minute and it took about 20 minutes to solve. I would speculate it took much longer than that if you try to use the way it is presented.We can solve this problem by simple iteration and we will use a simple equation that every surveyor should remember; C=2r sin 1/2 delta           C=chord and delta is an angle. (2r=d)From the radius point draw a line to each point around the circle, A,B,C,D and ENow we are going to solve for the Diameter of this problem by the following equation;[2*ArcSin[300/d]+2*ArcSin[400/d]+2*ArcSin[500/d]+2*ArcSin[600/d]+2*ArcSin[700/d]]-360 degree=0What we are doing is solving for each  angle at the radius point using the Chord formula then adding up the angles to see if we get 360 degrees. But we are really solving for d.Graph this equation, I used 800,850,900 for d and found that the solution fell between 850 and 900.I then graphed it again using 860 and 870 and scaled a value of d= 869+.I then plugged this value of 869 into my equation and used linear interpolation to solve for the final answer. It took 5 iterations to have the answer to 8 decimal places past 869, or my final answer wasd=869.21079173+         or r=434.60539586+I checked my answer using  a math program Mathematica and it took 0.015 seconds for the solution.John NoltonTombstone, AZ
Monday, November 26, 2012 at 2:57:51 PM

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