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mr_homm
Member since May2205

Oct3105, 09:59 PM (EST) 

"A little circle geometry puzzle"

A few days ago, my brother brought me a problem that had been passed on to him by a student in a calculus course. The problem was supposed to be solved using the methods of differential calculus, but upon looking at it, I found an almost purely geometric solution. I say "almost" because one still needs to take a limit. Have a try at solving it; it's not too hard: On an xy coordinate system, let C1 be a circle of radius 1 centered at x=1, y=0, so that it is tangent to the y axis at the origin. Let C2 be a circle of variable radius centered at the origin. Then C2 meets C1 in two points, and C2 meets the y axis in two points. Consider the upper (positive ycoordinate) pair of intersection points, and draw the line through them, producing it until it meets the x axis at A. Let x be the xcoordinate of A. What is the limiting value of x as the radius of C2 shrinks to zero? Happy Halloween! Stuart Anderson 

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Owen
guest

Nov0105, 07:12 AM (EST) 

1. "RE: A little circle geometry puzzle"
In response to message #0

I didn't find your geometric solution, but here goes an attempt at a "standard" solution (leaving out some of the algebra): C2 meets the positive yaxis at (0,r), where r is the radius of C2. By considering the equations of the two circles, one can find that they intersect at (r^2/2, r(1r^2)^(1/2) ). From here we can find the slope of the line between these points to be (1/r)( (4  2r^2)^(1/2)  2). The line with this slope and yintercept r passes through the xaxis at x = (2 + (4  2r^2)^(1/2))/2. As r approaches 0, x approaches 2. Of course, I didn't really use the methods of Differential Calculus, either. The answer of 2 (if I'm right) does suggest that perhaps there is a geometric solution that involves triangles inscribed in circles. I'm curious about your solution... 

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Owen
guest

Nov0205, 07:09 AM (EST) 

3. "RE: A little circle geometry puzzle"
In response to message #2

Thanks for the correction. I had a little alebra error in there. The circles intersect at (r^2/2, r(1(r/2)^2)^(1/2), which leads to a slope of (1/r)((4r^2)^(1/2)  2) for the desired line. I think this will give an xintercept of x = 2 + (4r^2)^(1/2), as you said. In this case, the limit is 4, not 2. I also think it is counterintuitive that the limit exists. Now, how about an ellipse centered at (a,o) with axes lengths 2a and 2b and and another centered at the origin with corresponding axes lengths ka and kb? What happens to the xintercept of the corresponding line as k approaches 0? (I haven't worked on it.) 

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Owen
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Nov0605, 10:55 AM (EST) 

8. "RE: A little circle geometry puzzle"
In response to message #6

That is a nice explanation. I wish I would have seen it before posting my solution, because of course it is much more intuitive than my analytic solution. (I forgot that new posts get inserted in the exchange depending on which post they replied to, rather than simply being appended on the end!) 

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Owen
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Nov0505, 05:49 AM (EST) 

7. "RE: A little circle geometry puzzle"
In response to message #3

I worked on the problem with the ellipse centered at the origin with equation x^2/a^2 + y^2/b^2 = k^2. Assuming I haven't made any algebra errors this time, the line through the intersection points of this ellipse with the ellipse centered at (a,0) having axes lengths 2a and 2b will intersect the xaxis at the point x = a*k^2/(2+sqrt(k^2+4)). (The value is independent of b!)As k approaches 0, the intercept approaches 4a, which agrees with the solution to Stuart's problem (for a = b = 1). 

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mr_homm
Member since May2205

Nov0205, 11:59 AM (EST) 

4. "RE: A little circle geometry puzzle"
In response to message #2

>>The line with >>this slope and yintercept r passes through the xaxis at x >>= (2 + (4  2r^2)^(1/2))/2. > >It's rather x = 2 + (4  r^2)^(1/2). >Owen lost a factor of 2 in the y coordinate of the intersection point of the circles. >Curious, without the calculations, I would vouch that in the >limit the xintersept goes to infinity. Yes, that was what I expected too when I first glanced at the problem. The reason this problem was in a calculus class is that the students were not expected to rationalize the denominator in the formula for x, but rather to use L'Hopital's rule to take the limit as r> 0 in its original form. Of course rationalizing the denominator is easier, but in my experience, university calculus courses do not emphasize insight, but rather applications of rules from a standard "calculus toolkit." Now to the geometric approach: Let A be the point (0,r), B be the intersection of circles C1 and C2, C be the center of C1 at (1,0), O be the origin, and X be the intersection of line AB with the x axis. Draw the chord OB of C1 and consider the triangle OCB. This is of course isoceles, so the angle bisector at C cuts the chord OB at its midpoint D, and meets OB at a right angle. Therefore angle AOB = angle DCO = 1/2 angle BCO. Next, look at triangle AOB, which is also isoceles, and repeat the same process, so that there is an angle bisector at O meeting chord AB perpendicularly at its midpoint E. Then since OE is perpendicular to BX, angle BXO = angle AOE = 1/2 angle AOB. Therefore angle BXO = 1/4 angle BCO. This is true regardless of the size of r, so in the limit as r > 0, B moves down to the origin and in the small angle limit we get OX = 4*OC. But OC = 1, so OX > 4. One could put more details into taking this limit, but it is pretty clear that the ratio approaches 4. Although the other methods also give the limit as 4, this appears somewhat mysteriors. The fact that the angles BCO and BXO are in 4:1 ratio gives the clearest most satisfying way to see why the limit is 4, at least in my opinion. Stuart Anderson 

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alexb
Charter Member
1682 posts 
Nov0205, 12:06 PM (EST) 

5. "RE: A little circle geometry puzzle"
In response to message #4

>The reason this problem was in a calculus >class is that the students were not expected to rationalize >the denominator in the formula for x, but rather to use >L'Hopital's rule to take the limit as r> 0 in its original >form. Such a disrespect towards students. I hope the professor would not disqualify against rationalizing the denominator. >The >fact that the angles BCO and BXO are in 4:1 ratio gives the >clearest most satisfying way to see why the limit is 4, at >least in my opinion. > Yes, it carries a solid grain of explanation of the phenomenon. Thank you. 

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mpdlc
guest

Nov1305, 08:12 AM (EST) 

9. "RE: A little circle geometry puzzle"
In response to message #0

A few days ago, my brother brought me a problem that had been passed on to him by a student in a calculus course. The problem was supposed to be solved using the methods of differential calculus, but upon looking at it, I found an almost purely geometric solution. I say "almost" because one still needs to take a limit. Have a try at solving it; it's not too hard: On an xy coordinate system, let C1 be a circle of radius 1 centered at x=1, y=0, so that it is tangent to the y axis at the origin. Let C2 be a circle of variable radius centered at the origin. Then C2 meets C1 in two points, and C2 meets the y axis in two points. Consider the upper (positive ycoordinate) pair of intersection points, and draw the line through them, producing it until it meets the x axis at A. Let x be the xcoordinate of A. What is the limiting value of x as the radius of C2 shrinks to zero? Happy Halloween! Stuart Anderson Back to the desktop after a few days off. I found this interesting question, which I solve initially using analytic geometry using polar coordinates for the straight line joining the two intersection points of the circunferences. I called them M1 and M2 and after I find the intersection of this line with the polar axis (OX in our case) and taking limits, since there is nothing exciting about it that being different of what everybody have been written already so I omit this part.
However dealing with the polar equation of the straight line is obvious I need to express it the variable distance of the line M1M2 to the origin O. This distance is preciselly the height of isosceles triangle M1OM2. Let us call M the intersection of this height with the base M1M2, being M the midpoint of the base obviously. We we will also call P the intersection of this height OM with C1, this point P it is also the center of the circunference which inscribes the mentioned isosceles triangle, so M1M2 is chord and an arc of this circunference centered in P. Since P belongs to C1 it will form an straight triangle with the diameter of C1 that is contained in the X axis; lets call OPD this triangle being D the intersection of C1 with X axis. Obviously the hypothenuse PD is paralell to M1M2, and the ratio between OM/OP = OA/OD is getting equal to 2 in the limit when M1, M2 and M become the same point, then OM will be twice OP . So we must conclude that OA is four times the radius of C1, what means four units in our case. 

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mpdlc
guest

Nov1305, 01:06 PM (EST) 

10. "RE: A little circle geometry puzzle"
In response to message #9

After posting in a hurry I found an error I call hypothenuse PD when I should call it leg... I think much faster than I can type in English. 

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