All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Chord Harmonics (Posted on 2005-01-11) Difficulty: 3 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Still no proof, but... | Comment 4 of 9 |

Ok, I still donít have a proof, but here are my thoughts so far.

First, in order to have a circle that contains a chord of length 9, the diameter of the circle must be at least 9 (so the radius must be at least 4.5).

Second, I made an approximation to get me in the right ballpark. Ok, imagine the perfect circle, with these 6, 7, 8, 9 chords in them making a quadrilateral. Draw a line from each of the four vertices to the center of the circle. Now we have 4 pie slices of the circle, each with a different angle at the center of the circle. My approximation is that the angles are proportional to the chord lengths. So calling the angles of the cords a, b, c and d, I should see that 6:7:8:9 and a:b:c:d are in the same proportions. Well, 6+7+8+9 = 30 and a+b+c+d = 360. So the factor difference is 360/30 = 12. So a:b:c:d = 6*12 : 7*12 : 8*12 : 9*12 = 72:84:96:108.

Using those approximated angles, and looking at right triangles made by the center of the circle, the midpoint of a chord, and an endpoint of that chord, I get the following relationships. R*sin(a/2)=6/2, R*sin(b/2)=7/2, R*sin(c/2)=8/2 and R*sin(d/2)=9/2. I get that R is about 5.1039-5.5623. So thatís my ballpark.

Third, I started to go into detail with equations. I thought they made plenty of sense, but then everything got screwy and I donít know what I did. I used those same 4 equations: R*sin(a/2)=6/2, R*sin(b/2)=7/2, R*sin(c/2)=8/2 and R*sin(d/2)=9/2. Then I solved for the angle in each case, so I had a=2*sin-1(6/(2r)), b=2*sin-1(7/(2r)), c=2*sin-1(8/(2r)) and d=2*sin-1(9/(2r)). Knowing that a+b+c+d = 360 degrees (or 2*pi), I have one equation with one unknown:

a+b+c+d = 2*sin-1(6/(2r)) + 2*sin-1(7/(2r)) + 2*sin-1(8/(2r)) + 2*sin-1(9/(2r)) = 360 = 2*pi
sin-1(3/r) + sin-1(3.5/r) + sin-1(4/r) + sin-1(4.5/r) = 180 = pi

I didnít know anything neat to do at this point, so I put the equation into excel and did my little zeroing in thing. I got a completely wrong answer though. I got R = 4.242640687. I know this is wrong on so many levels. First, it is less than 4.5 which I said was the minimum. A chord of length 9 canít fit in this circle. Second, itís not in my 5.1039-5.5623 ballpark. And third, all of the angles are 90 degrees! Thatís impossible. That means all the chords should be identical in length.  Then when I go back and plug 90 degrees in I get different radii.  So I don't know what I screwed up (and no, I didn't get my radians and degrees confused... I think).

Then I mocked it up in a CAD program, where the lengths were fixed, and the endpoints were fixed on a circle. Then I could change the radius of the circle they were on and zero in on an answer. Thatís how I got my R = 5.363825 solution.

I hope someone is more enlightened than I am =)

Oh wait... I think it might have to do with some of the angles being obtuse and messing up the inverse sine.  I'll check into that and get back to you.


  Posted by nikki on 2005-01-12 16:51:38
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information