Math Problem

[rahdam]

I have an idea about the answer to the problem below, but I can't prove my answer. I was wondering what you all could come up with (fyi: a student brought this to me, I thought it was interesting).

C1 is a circle with the equation (x-1)^2 + y^2 = 1 and C2 is a contracting circle with the equation x^2 + y^2 = r^2, where 2 > r > 0. P is the point on the y-axis where C2 intercepts the y-axis (0,r). Q is the point above the x-axis where C1 and C2 intercept. R is the point where the line PQ intercepts the x-axis (x-intercept, 0). The question: What happens to R as the radius r of C1 approaches 0 from the positive side(r -> 0+)?

edit for clarity: What is the x-coordinate of R as r approaches 0+?


Monte: I'm lookin' at you...:laugh:





rahdam added to this post, 52 minutes and 45 seconds later...

I just wrote a computational simulation in C++, the answer is 4.
This answer was achieved by calculating P and Q given r, calculating the x-intercept of line PQ, and then iterating over different ranges of r to watch trends. I can get down to r = 0.001 with Rx = 3.999... before I run into floating point errors (using doubles here).

If someone can step up with a pure theoretical answer, I'd really appreciate it.

[enWTFp]

Solution.
P(px = 0 , py = r)
Q(qx = r^2/2 , qy = r.sqrt(1-r^2/4)), by solving the system [C1;C2], sqrt() denotes square root.
R(d , 0)
Note: (py - qy)/py = qx/d, by the similarity of the right triangles with hypotenuses PQ and PR. All is valid for 2 > r > 0.

From this we calculate d = 2( 1 + sqrt( 1 - r^2/4 ) ).

Hence, when r->0+, d -> 2(1+1) = 4, R -> (4,0).

[rahdam]

Thanks. In comparing what I was doing to what you did, I saw two problems that prevented the more theoretical approach.

1) My solution for qy is quantitatively the same, but is significantly more obfuscated because I plugged qx into the equation for C1, leaving (x-1)^2 rather than x^2.

2)I lost the forest for all the trees; I never though to refer back to the geometry of the problem, and the right triangle similarity eluded me entirely.