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Sir Col
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Three Touching Circles
« on: Mar 8th, 2004, 4:34pm » |
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Three touching circles have centres, A, B, and C.
ABC are joined to form a triangle with sides 3, 4, 5.
Points are created where the circles meet and are joined to form another triangle, with the side opposite A labelled a, opposite B is b, and the side opposite C is labelled c.
Find the value of abc.
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Noke Lieu
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Re: Three Touching Circles
« Reply #1 on: Mar 8th, 2004, 9:51pm » |
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Aw, pants.
For that piece of scrap paper I had on the bus up to Sydney last weekend. Was trying to do the same thing, but with a square (much easier to start with, then got stuck- from www.freepuzzles.com geometry 20). Lets see, where did I get to?
area bounded by the chord (at 90) (pi-2r2)/2
thats right... I think.
Assuming the circles are 1,2,3 in radius
1
The lengths of the triangle abc are 2.5, 1.788 and 1.9 (roughly)
2 That means you can use Heron's to figure out the area. Or the area of ABC, and subtract Abc, aBc and Cba...
3 then use a tweaked version of (pi-2r2)/2
to take into account that you're not dealing with quarter arcs in two of the cases. This gives you the areas of the arcs over lapping abc. So subtract them.
Unfortunately, I am just about to be called away, and by the time I get back, some big brainy type will have given a fuller answer. Damn.
oh, wanted to add that you'd have to get up pretty early to better me. But its only the time zone thing....
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| « Last Edit: Mar 8th, 2004, 9:53pm by Noke Lieu » |
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Sir Col
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Re: Three Touching Circles
« Reply #2 on: Mar 9th, 2004, 12:28am » |
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Nice work, Noke Lieu. How did you obtained the lengths? If you've derived the exact values... ?
As an extension, if the lengths of the triangle ABC are rational, prove that abc will also be rational.
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Barukh
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Re: Three Touching Circles
« Reply #3 on: Mar 9th, 2004, 1:32am » |
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If I didn't mess up the calculations, the answer is abc = 24/5. Note that the lengths a, b, c are not needed to calculate this.
on Mar 9th, 2004, 12:28am, Sir Col wrote:| As an extension, if the lengths of the triangle ABC are rational, prove that abc will also be rational. |
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Hmm, are you sure  ?
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| « Last Edit: Mar 9th, 2004, 1:33am by Barukh » |
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towr
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Re: Three Touching Circles
« Reply #4 on: Mar 9th, 2004, 5:39am » |
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I also get ::2*2*sqrt(5)/5 * 2*3*sqrt(10)/10 * 2*1* sqrt(2)/2 = 48/10 = 24/5:: I suppose I did calculate a,b and c to do that though..
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towr
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Re: Three Touching Circles
« Reply #5 on: Mar 9th, 2004, 5:53am » |
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on Mar 9th, 2004, 12:28am, Sir Col wrote:| As an extension, if the lengths of the triangle ABC are rational, prove that abc will also be rational. |
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on Mar 9th, 2004, 1:32am, Barukh wrote:| Hmm, are you sure ? |
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Aren't you?
In general, for sidelengths x,y,z it's ::(x + y - z)2(x - y - z)2(x - y + z)2/(8xyz):: which for rational x,y,z is trivially rational as well
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Sir Col
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Re: Three Touching Circles
« Reply #6 on: Mar 9th, 2004, 8:37am » |
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Bravo! I worked out abc in terms of the three radii and got ::8(r1r2r3)2/((r1+r2)(r1+r3)(r2+r3 )):: but your version is much better in that you can explicitly determine the product from the side lenghts.
I was initially trying to write a problem where the lengths of triangle ABC are integer and the perimeter of the other triangle was also integer, but it seems that this is impossible. Someone may like to try and prove this?
Isn't it fascinating that although the perimeter of the inner triangle can never be integer, the product of its lengths is always integer!
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towr
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Re: Three Touching Circles
« Reply #7 on: Mar 9th, 2004, 9:14am » |
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on Mar 9th, 2004, 8:37am, Sir Col wrote:| I was initially trying to write a problem where the lengths of triangle ABC are integer and the perimeter of the other triangle was also integer, but it seems that this is impossible. Someone may like to try and prove this? |
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try x=y=z=2k
a rather trivial solution for
(x + y - z)·[sqrt](- (x2 - 2·x·y + y2 - z2)/(x·y))/2 + (x - y + z)·[sqrt](- (x2 - 2·x·z - y2 + z2)/(x·z))/2 - (x - y - z)·[sqrt]((x2- y2+ 2·z·y - z2)/(y·z))/2
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Barukh
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Re: Three Touching Circles
« Reply #8 on: Mar 9th, 2004, 9:42am » |
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Here's how I did it.
[smiley=blacksquare.gif]
Let A(*) be the area of the corresponding triangle. I will also use the same letters to represent both the side lengths and angles because it won't lead to ambiguity.
It is not difficult to see that the vertices of the triangle (abc) are the points of contact of the triangle (ABC) with its incircle. Therefore, the circumcircle of (abc) is the incircle of (ABC). Let its radius be r. Then r = 2A(ABC)/(A+B+C) = 1.
Also, A(abc) = abc/(4r), so it remains to calculate the area of (abc).
I will leave as an excersize for the interested to show that A(abc) = r2(sin(A) + sin(B) + sin(C))/2 = r2A(ABC)(1/AB + 1/AC + 1/BC) = r2A(ABC)(A+B+C)/ABC = 6/5, and the result follows.
[smiley=blacksquare.gif]
on Mar 9th, 2004, 12:28am, Sir Col wrote:| As an extension, if the lengths of the triangle ABC are rational, prove that abc will also be rational. |
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Oh, yes, of course, now even I am convinced... 
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Barukh
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Re: Three Touching Circles
« Reply #9 on: Mar 9th, 2004, 11:24am » |
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on Mar 9th, 2004, 8:37am, Sir Col wrote:| I was initially trying to write a problem where the lengths of triangle ABC are integer and the perimeter of the other triangle was also integer, but it seems that this is impossible. Someone may like to try and prove this? |
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I agree with towr: when all the 3 circles have equal radii, the triangle (abc) is half of (ABC), and both are equilateral.
Is it the only case possible? In fact, (abc) will have rational perimeter if cos(A/2) + cos(B/2) + cos(C/2) is rational.
[Modified to remove the erroneous statement.]
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| « Last Edit: Mar 10th, 2004, 1:05am by Barukh » |
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Sir Col
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Re: Three Touching Circles
« Reply #10 on: Mar 9th, 2004, 11:46am » |
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Yes, thank you guys.
Obviously I wasn't considering the case with perfect rotational symmetry.
I derived a=2r1[sqrt](r2r3)/[sqrt]((r1+r2)(r1+r 3)) and the b and c have similar symmetrical forms.
Hence a+b+c=2(r1[sqrt](r2r3(r2+r3))+r2[sqrt](r1r3(r1+r3))+r3[sqrt](r1r2(r1+r2))/[sqrt]((r1+r2)(r1+r3)(r2+r 3)).
However, I couldn't convince myself that this had no rational solutions. Clearly making the denominator rational, r1+r2 and other terms perfect squares is one way, but then the numerator fails on associated products r1r2. Having said that, it could be possible that the denominator is irrational, but is a rational multiple of the numerator. Hmm?
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| « Last Edit: Mar 9th, 2004, 11:50am by Sir Col » |
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Noke Lieu
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Re: Three Touching Circles
« Reply #11 on: Mar 9th, 2004, 3:30pm » |
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just to answer your question Sir Col....
to find the lengths of abc, I bisected the angles A,B,C, yeilding right triangles. sohcahtoa, Robert is you father's brother.
Struck me after I posted that I'd made my solution a bit trickier than it needed to be.
You only need to calculate the area of ABC and angles (easy). Then subtract the area of segments, (also very easy to find, but introduces pi)...
Previously, I was doing a lot of double handling.
could someone check my working on this on though:
(there has to be a prettier way.)
to get the area in any chord and minor arc-
(a.pi.r2)/360 minus rsin(a/2).rcos(a/2)
a is angle, r is radius.
Which pops some questions into my head....
Given the length of the arc(ooops, and the distance between the two ends) , could you figure out the radius of the original circle.
have a suspicion you might be able to.
ah, but you don't need to do it with area...
[note] reminded myself about definition of arc and chord...[/note]
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| « Last Edit: Mar 11th, 2004, 3:25pm by Noke Lieu » |
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Sir Col
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Re: Three Touching Circles
« Reply #12 on: Mar 9th, 2004, 4:42pm » |
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Yes, your method for finding the area of the segment is sound, but I don't think it is necessary. Here is a full method that you could use to solve the original problem, although I did it for the general case initially...
In general, you can use the cosine rule to obtain an exact value for cosA. In my particular example it was overkill, because it was a 3-4-5 right angle triangle. So assuming 3 is opposite angle A, cosA=4/5.
This can be repeated for each of the angles B and C: cosB=3/5, and cosC=0 (C is a right angle).
Then applying the cosine rule (assuming the radii have already been worked out):
a2=32+32–2*3*3*4/5=18/5. Similarly, b2=16/5, and c2=2.
Hence a2b2c2=(18/5)*(16/5)*2=576/25, and abc=24/5.
Of course, the more general problem is more difficult, but I've given expressions for a, b, and c, in terms of the three radii in my previous post.
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| « Last Edit: Mar 9th, 2004, 4:45pm by Sir Col » |
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Noke Lieu
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Re: Three Touching Circles
« Reply #13 on: Mar 9th, 2004, 6:33pm » |
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Lunch is good.
came back, a tad confused why the segments were not important. Then re-read the question.
I thought abc was area between the circles... 
Completly forgot about the triangle. Which leads me to pose the question with a square. (link provided in my original answer)
awfully sorry, slightly shamefaced...
(oh, Noke rhymes with oak)
here's the text from that puzzle. The picture helps: The length of each side of the square ABCD is 10 cm. The shaded area EFGH is surrounded by 4 curves of quarter circle. How big is the shaded area EFGH?
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| « Last Edit: Mar 9th, 2004, 6:59pm by Noke Lieu » |
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