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Posers and Puzzles

Posers and Puzzles

  1. 20 Feb '15 18:22 / 2 edits
    Recall that "unit circle" means a circle with a radius of one.

    Two unit circles can be positioned so that they touch in tangent fashion at just a single point. You can place a third unit circle so that it touches the first two circles in tangent fashion.

    Doing this creates a "pocket" region in between the three unit circles.

    Q1: What is the area of the pocket?

    Next we can find a small circle (radius < 1) that will fit inside the pocket so as to touch the three unit circles in tangent fashion. This creates three small regions that are between the four circles, not interior to any of them.

    Q2: What is the combined area of those three small regions?

    My source derives the answers by using good old schoolboy geometry-- no calculus needed. It gives not just a decimal approximation (such as you could work up by drawing this in CADD) but the precise answers. (Hint: irrational numbers come into play.)

    Q3: My source stops there, but you could continue this process by adding three circles smaller yet, and then three more even smaller ones, and so on. In the limit of adding an infinite number of ever-smaller circles, what is the area trapped between all the circles?

    {I suspect Q3 is very time consuming to answer, and I may not be competent to decide if an answer you give to Q3 is correct. I just throw Q3 out there in case anybody wants to really strain the brain. Don't beat your head against the wall on this one if you think it is equivalent to writing a master's thesis or something. }
  2. Subscriber joe shmo
    Strange Egg
    22 Feb '15 14:28 / 2 edits
    Originally posted by Paul Dirac II
    Recall that "unit circle" means a circle with a radius of one.

    Two unit circles can be positioned so that they touch in tangent fashion at just a single point. You can place a third unit circle so that it touches the first two circles in tangent fashion.

    Doing this creates a "pocket" region in between the three unit circles.

    Q1: What i ...[text shortened]... he wall on this one if you think it is equivalent to writing a master's thesis or something. }
    I'll kick it off with a solution for Q1:

    Joining the centers of the 3 tangent unit circles you have an Equilateral of side length 2 units.

    Within that triangle are 3 circular sectors of equal area, and equal subtending angle (the interior angles of the equilateral joining the centers)

    the "area of the pocket" is the area of the Equilateral - 3*Sector Area

    Area Eq. = ½*2*2*sin( 60° ) = 2*√3/2 = √3

    Area Sec. = ½*1²*pi/3

    Area pocket = √3 - 3*½*pi/3

    = √3 - ½*pi
  3. Subscriber joe shmo
    Strange Egg
    22 Feb '15 16:29
    Originally posted by joe shmo
    I'll kick it off with a solution for Q1:

    Joining the centers of the 3 tangent unit circles you have an Equilateral of side length 2 units.

    Within that triangle are 3 circular sectors of equal area, and equal subtending angle (the interior angles of the equilateral joining the centers)

    the "area of the pocket" is the area of the Equilateral - 3*Sect ...[text shortened]... ec. = ½*1²*pi/3

    Area pocket = √3 - 3*½*pi/3

    = √3 - ½*pi
    Q2:

    Find the center of the equilateral described in Q1's solution ( the intersection of the three line segments that bisect the equilaterals interior angles)

    Let the distance from that center to any vertex of the equilateral be:

    r + x

    From the law of sines

    sin ( 120° )/2 = sin ( 30° )/(r+x)

    x = 2*√3/3 - 1

    The remaining area is:

    A = √3 - ½*pi - pi(2*√3/3 - 1)²

    A= √3 +pi*[(8*√3 -17)/6]
  4. 22 Feb '15 17:34
    Joe has correctly answered Q1 and Q2. Nicely done.

    I am not really expecting anyone to tackle Q3.
  5. Subscriber joe shmo
    Strange Egg
    22 Feb '15 18:03
    Originally posted by Paul Dirac II
    Joe has correctly answered Q1 and Q2. Nicely done.

    I am not really expecting anyone to tackle Q3.
    Yeah, I'm thinking about it... but within the very next step it becomes quite challenging. After that step, it "appears" as though there will be 2 circles, of one radius, and 1 circle of another radius to fill the space... and the symmetries will get progressively more convoluted as it continues. I'm not very hopeful.
  6. 22 Feb '15 18:15 / 1 edit
    Originally posted by joe shmo
    After that step, it "appears" as though there will be 2 circles, of one radius, and 1 circle of another radius to fill the space...
    What I am picturing for Q3 is a sequence of ever-shrinking circles centered along each of three axes, and nowhere else. I am not sure if that is the same thing that you are picturing. (There are other choices I could have made for loading the pocket with circles, of course.)
  7. Subscriber joe shmo
    Strange Egg
    22 Feb '15 18:23 / 2 edits
    Originally posted by Paul Dirac II
    What I am picturing for Q3 is a sequence of ever-shrinking circles centered along each of three axes, and nowhere else. I am not sure if that is the same thing that you are picturing. (There are other choices I could have made for loading the pocket with circles, of course.)
    I'm thinking that the axis are changing in a fractal like manner.

    Edit: I see now that you said "add three" new circles each time.
    I'm thinking of adding 3ⁿ circles each time.
  8. 22 Feb '15 18:51 / 1 edit
    Originally posted by joe shmo
    I'm thinking that the axis are changing in a fractal like manner.
    Either way of loading the circles in is worth considering.
  9. 22 Feb '15 20:06
    The keyword "Apollonian gasket" might help.
  10. 22 Feb '15 21:06
    If the total area left decreases by more than half each go the final area after infinite turns will be zero
  11. 22 Feb '15 21:07
    Originally posted by WanderingKing
    The keyword "Apollonian gasket" might help.
    I just did a search for Apollonian gasket. I see that "If we define the curvatures (1/r) of any three circles as a, b, and c, respectively, the Theorem states that the curvature of the circle (or circles) tangent to all three, which we will define as d, is: d = a + b + c ± 2 (sqrt (a × b + b × c + c × a ))."

    My source did not take advantage of that.
  12. 22 Feb '15 22:26
    Originally posted by iamatiger
    If the total area left decreases by more than half each go the final area after infinite turns will be zero
    The way I was picturing of adding circles would leave nonzero area. But there are other ways of adding circles that would soak up all the available region, driving the area to zero, if my intuition is correct.
  13. Subscriber joe shmo
    Strange Egg
    23 Feb '15 18:03 / 1 edit
    Originally posted by Paul Dirac II
    The way I was picturing of adding circles would leave nonzero area. But there are other ways of adding circles that would soak up all the available region, driving the area to zero, if my intuition is correct.
    I worked out an implicit formula for the circles radius size in the next group. The good thing is that the formula should be indefinitely repeatable ( once a solution of desired precision is obtained, just change variables and start again for the following set)

    The Area remaining in the pocket after each grouping should be simple enough to figure out ( i'm not sure if you are interested in this numerical approach, so I'm not currently going to bother finding a numerical limit) If someone can verify the solution in CAD, that would be good.

    Let x be the previous circles radius
    Let y be the radii of the next circle grouping

    Bare with me, it might be messy ( I'm not a mathematician):

    Start with a right triangle OAB

    O is the center of any one of the three original circles of radius "r" (1 in this case)

    Line segment OA connects O with the point of tangency between any two original circles of radius "r"

    Line segment OB connects O with the center of the circle found in Q2 of the original question.

    The center of all sequential tangent circles to be found lies on the line segment AB

    Pick a random point "C" on AB, draw a line from "C" to "O"

    This divides the triangle OAB into two distinct regions.

    Triangle OAB has:
    OB = r+x
    OA = r
    AB = √(x²+ 2*r*x)
    Angle = δ

    Triangle OCB has:
    OC = r+y
    OB = r +x
    Angle OC/OB = ß

    Triangle OAC has:
    OA = r
    OC = r+y
    Angle OA/OC = δ - ß

    Such that:

    cos( δ ) = r/(r+x) (1)
    cos( δ-ß ) = r/(r+y) (2)

    From the Sum and Difference formulas

    cos( δ-ß ) = cos( δ )*cos( ß ) + sin( δ )*sin( ß ) (3)

    Sub (1)&(2) into (3)

    r/(r+y) = r/(r+x)*cos( ß ) + √(x²+ 2*r*x)/(r+x)*sin( ß ) (4)

    Then looking at triangle OCB and using the Law of Cosines

    (x+y)² = (r+x)² + (r+y)² - 2*(r+x)*(r+y)*cos( ß ) (5)

    From (5)

    cos(ß(r,x,y)) = {(x+y)² - [(r+x)² + (r+y)²]}/(- 2*(r+x)*(r+y)) (6)

    From Pyhtagorean Identity:

    sin(ß(r,x,y)) =√(1 - cos²(ß(r,x,y))) (7)

    Sub (6) & (7) into (4)

    r/(r+y) = r/(r+x)*cos(ß(r,x,y)) + √(x²+ 2*r*x)/(r+x)*sin(ß(r,x,y)) (8)

    from (8)

    0 = r/(r+x)*cos(ß(r,x,y)) + √(x²+ 2*r*x)/(r+x)*sin(ß(r,x,y)) - r/(r+y) (9)

    Use (9) compared with zero for a fixed "r & x" , and variable "y"

    Once "y" is obtained to a desired precision, interchange "y" with "x" and repeat for next set.

    My solution for "y" using a spreadsheet ( or the next set after Q2 ) is between 0.0628 - 0.0627 units
  14. 23 Feb '15 20:24
    Originally posted by joe shmo
    ... My solution for "y" using a spreadsheet ( or the next set after Q2 ) is between 0.0628 - 0.0627 units
    Lots of work!

    I found a source that takes it to one level past Q2. It derives that the radii of the three smallest circles (in what I will call my "three-axis construction," not some of the other constructions touched upon above) is:
    (9 -4*√3)/33
    which is about 0.06278. (Very close to your spreadsheet number.)

    Squaring that and multiplying it by pi and then multiplying it by 3 (due to 3 of the smallest circles) gives 0.03715. This is to be subtracted from your Q2 result which was about 0.08607, leaving the area of 0.04892.
  15. Subscriber joe shmo
    Strange Egg
    23 Feb '15 20:52
    Originally posted by Paul Dirac II
    Lots of work!

    I found a source that takes it to one level past Q2. It derives that the radii of the three smallest circles (in what I will call my "three-axis construction," not some of the other constructions touched upon above) is:
    (9 -4*√3)/33
    which is about 0.06278. (Very close to your spreadsheet number.)

    Squaring that and multiplying it b ...[text shortened]... is is to be subtracted from your Q2 result which was about 0.08607, leaving the area of 0.04892.
    Yeah, that is good! I took the solution further out. 0.0627818 - 0.0627817

    Seems to be a spot-on match. However, not as spot-on as (9 -4*√3)/33.

    I can't imagine rearranging the equation I derived for y, (but admittedly didn't even try to do so). if there is another way to get a nice tidy answer, I'm missing it.