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Team Project

  1. Apr 18, 2004 #1
    Trig Question

    Any ideas on solving this problem? The question reads...
    Four mutually tangent circles are shown. Find the radius of the shaded circle. See attachment.
    Any help will be appreciated. :confused:
     

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    Last edited: Apr 19, 2004
  2. jcsd
  3. Apr 20, 2004 #2

    honestrosewater

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    Not to be a buzzkill, but there are two shaded circles. You mean the smallest circle?
    Happy thoughts
    Rachel
     
  4. Apr 20, 2004 #3

    honestrosewater

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    Trig never held my attention for very long, but perhaps your answer will come from looking at the triangle formed by the origins of the three smallest circles?
    Happy thoughts
    Rachel
     
  5. Apr 20, 2004 #4

    honestrosewater

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    Label the circles in increasing size A, B, C, so that circle C has radius=42.6.
    Denote the origin of circle N by O_N. Denote the point where two circles N and M intersect by I_NM. Denote the radius of circle N by r_N.
    Point O_A and I_BC *look* like they lie on the same line, which would be tangent to circles B and C. Assuming you can prove this, you can consider triangle O_A O_B O_C as two right triangles;
    Triangle O_A I_BC O_B and triangle O_A I_BC O_C.
    Then note that line O_A O_B, a hypotenuse, is the sum of r_A and r_B. Similarly line O_A O_C = r_A + r_C.
    Plug all this info into the two equations for the right triangles and you may get somehwere.
    Hope this helps.
    Happy thoughts
    Rachel
     
  6. Apr 20, 2004 #5

    honestrosewater

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    Looks can be deceiving... you cannot form the right triangles as I thought. I also misspoke, as O_A and I_BC lie on some line, of course, but this line is not tangent to circles B and C.
    Still happy thoughts
    Rachel
     
  7. Apr 20, 2004 #6

    matt grime

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    As it stands there isn't a solution to this question. You need to provide more information. Proof: just draw the two circles whose radii are given. pick one point on each circle. draw in the normal lines at thos points. where they intersect is the centre of a larger circle tangential to these two. draw it. you are then claiming that the circle you can inscribe that is tangential to all three is uniquely determined which is not the case, I believe.
     
  8. Apr 20, 2004 #7
    well if you are talking about the outer circle, combine the two innner blue circles diameter using the radius to find the radius of the large blue circle...
     
  9. Apr 20, 2004 #8

    matt grime

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    but you are assuming that the centres of the smaller lie on diameter of the larger, which we don't know to be true, indeed we know it might not be true.
     
  10. Apr 20, 2004 #9

    honestrosewater

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    If all 3 inner circles are identical, i.e., if they have equal radii, then the problem can be solved, yes?
    It becomes three identical isosceles triangles whose bases are connected to form an equilateral triangle. The radius of the outer circle is then the sum of the height of one isosceles triangle and the radius of the circle inscribed in the equilateral triangle.? If you have a way to find the radius of said inscribed circle, the problem is solved?
    Then does the original problem not become a matter of proportion or ratio? Sorry, do you understand what I'm trying to say?
    Rachel
     
  11. Apr 20, 2004 #10

    matt grime

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    "If all 3 inner circles are identical, i.e., if they have equal radii, then the problem can be solved, yes?" trivially, since we know what the radius is.

    It isn't a matter of proportion or ratio in any obvious sense as i believe i just proved that there is no answer to the question as stated.
     
  12. Apr 20, 2004 #11

    honestrosewater

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    Because you can inscribe two such circles, one on each side of the point of tangency of the two given inner circles? If this is why, I don't see why it matters- both of these new circles are still uniquely determined? So there are two solutions. Or did I miss something?
    Rachel
     
  13. Apr 20, 2004 #12

    matt grime

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    the point is that we do not know anything about the larger circle that determines the smaller one, there are an infinite number of diagrams that look like the one given with those two radii that all yield different answers for the radius of the smaller one. Or did I miss something myself about the diagram? I don't believe I did. If there were one mor bit of information, such as the larger circles radius then it would be solvable, but we aren't told what that is nor are we given any information which allows us to decide what it is. there are an infinite number of larger circles one could draw, one for each pair of points on the two given sized ones with the obvious exceptions. Don't believe me? draw to little ones and draw the larger one so it touches at the oppsite sides to where the two small ones touch. ie its centre lies on the same line as the centres of the smaller one. see what happens there? No imagine that if the circles look like this: oo that you draw the degenerate case where the large circle has infinite radius, that is it is a straight line across the top. see how big the third cricle you inscribe is? there is no way of answering the question explicitly. you can form a relationship between the radii of the large circle and the third one inside and one more bit of information would provide an answer but currently it cannot be done.
     
  14. Apr 20, 2004 #13

    honestrosewater

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    Ohhhh... I see. Thanks for being patient.

    Though I'm not satisfied that the problem cannot be solved if restated as three mutually tangent- um disjoint?- circles, that is, no circle contains another. I still think these must follow some proportional rule. I will work on this, noting that their origins form a triangle and so (r_1+r_2)+(r_1+r_3)>(r_2+r_3) and so on.
    Happy thoughts
    Rachel
     
    Last edited: Apr 20, 2004
  15. Apr 20, 2004 #14

    matt grime

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    sorry, but you shouldn't waste your time on that: given any two the third can take on any other radius, within reason. they can't be tangential and disjoint by the way. here's how to see it visually, perhaps.

    just fix two circles both of radius 1 touching like: oo. now imagine balancing any sized hoop on top of them. suppose also that you've got a really big lassoo that has a stiff circular end. loop it over the whole lot and jiggle it until it lightly touches all three. see, you can do that for all possible hoops you put on top whenever it's top sticks out above the oo ones. if there were some relation ship between the three interior ones then i couldn't do that. there is a relationship between the hoop and top and the size of the lasso, but exactly what i'm not sure.
     
  16. Apr 20, 2004 #15

    honestrosewater

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    (Okay, I need to say something- I kind of feel bad because I am asking questions that I could probably figure out on my own with some effort- I should be able to figure them out anyway. But I enjoy this, and think it is beneficial to all, but if this is not what the forum is intended for, then someone needs to tell me so, otherwise I'll proceed.)

    Given our 3 circles, their radii r_1, r_2, and r_3 form a triangle with sides (r_1+r_2), (r_2+r_3), and (r_1+r_3). How many solutions {r_1, r_2, r_3} are there for our mutually tangent circles?
    Ex. for a triangle with sides 4, 5, 5, one solution is {2, 2, 3} are there any others?
    Rachel
    I think this is the relationship I was trying to pinpoint.
     
    Last edited: Apr 20, 2004
  17. Apr 20, 2004 #16

    matt grime

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    I believe you're asking gieven the sides of the triangle a b and c, then letting r,s,t be the desired radii, can we solve:

    r+s=a
    s+t=b
    t+r=c

    the answer being yes. and it is unique
     
  18. Apr 20, 2004 #17

    honestrosewater

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    I think there are is only one solution.

    Given the equations
    r_1+r_2=a
    r_2+r_3=b
    r_1+r_3=c
    if we change one r, the equations change thusly
    (r_1+n)+(r_2-n)=a
    (r_2-n)+(r_3+n)=b
    (r_1+n)+(r_3+n)=c -> not true.
    If we increase(decrease) one radius by some amount, we must decrease(increase) the corresponding radius by the same amount so as not to alter the lengths of the sides of the triangle. And since we are dealing with 3 radii, this is not possible, as one radius ends up having two different lengths.
    Happy thoughts
    Rachel

    Matt, seems I was writing while you were ;)
     
  19. Apr 20, 2004 #18

    honestrosewater

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    So back to the original problem... to simplify things, note 28.4*1.5=42.6, so we have r_1=1, r_2=1.5, and r_3.
    1+r_3>1.5 and 1+1.5>r_3
    2.5>r>.5
    yes?
     
  20. Apr 20, 2004 #19

    honestrosewater

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    This is true of the other origins as well- the line from the origin to the opposite point of tangency does not form a right angle. Thus r_3 does not equal r_1 or r_2, because either would create an isosceles triangle, and opposite points would bisect opposite sides, at their points of tangency, creating right angles. Yes?
     
    Last edited: Apr 20, 2004
  21. Apr 20, 2004 #20

    mathman

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    In order for the solution to be unique, I have to assume something that is apparent from the diagram, but not stated, i.e. the center of the outer circle is on the line connecting the two given inner circles. Undet this assumption, the problem can be solved using the law of cosines.
    Let r and R be the given radii of the two inner circles (28.4 and 42.6).
    Let s be the radius of the outer circle, s=r+R.
    Let x be the radius of the unknown circle.
    Notation p=center of r circle, P=center of R circle, c=center of s circle, q=center of x circle.

    We will look at triangles (1) pPx, (2) psx. Apply law of cosines for both triangles, angle A at p.

    (1) (R+x)2=(r+x)2+(R+r)2-2(r+x)(R+r)cos(A)
    (2) (s-x)2=(r+x)2+R2-2(r+x)Rcos(A)

    Going through the algebra, cos(A) can be eliminated between the 2 equations and alll the x2 terms drop out and we get:

    x=rRs/(r2+R2+rR).

    Good luck (I hope I don't have any typos.)
     
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