Can you find sum of this infinite squence derived from Geometry :)?

Apparently, the most important part is to find the ratio of the geometric sequence, which we will call p. For this task, we will take the first and second pair of circles.

I will be referring to the following sketch:

http://i25.tiny/ pic.com/rkmudt.png

(kindly remove the space from the URL, please – Yahoo error).

Here, |AO| = 1/2*x_1 and |PO| = 1/2*x_2. The point of touch T must be the intersection of each of the corresponding circles with a line going through its centre, thus we will find it at an extension of the line segment AP (connecting the centres of one large and one small circle).

I chose an algebraical way to solve this problem. Let us denote the vector AP in Cartesian coordinates as (x,y), the vector AT must be some positive multiple of it, (αx,αy), α > 1. The ratio of y/x is the ratio of the geometric progression, x_2/x_1, or p. After denoting y = px, we can find two equations homogeneous in x by a simple geometrical thought:

* |AT| = x_1 = 2x:

α|AP| = 2x

α^2 (x^2+y^2x^2) = 4x^2

α^2 (1 + p^2) = 4 (1)

* |PT| = x_2 = 2y:

(α-1)|AP| = 2y

(α-1)^2 (x^2+y^2) = 4y^2

(α-1)^2 (1 + p^2) = 4p^2 (2)

Dividing (2) by (1), we get

(α-1)^2 / α^2 = p^2, (3)

which we can plug into (1) to get a quadratic equation for α:

α^2 + (α-1)^2 = 4

2α^2 – 2α – 3 = 0

(α>0) … α = (1+√7)/2.

From here, it is easy to find using (3) that

p = (√7 – 1) / (√7 + 1).

Finally, let us find the sum of x_n = x_1*p^(n-1) = p^(n-1) as n goes from 1 to +∞. This is of course

Σ [n=0 to +∞] p^n = 1/(1-p) = (√7 + 1)/2

(and by some coincidence, equal to α).

Thanks to rozeta53 for being honest about the time 🙂 I started typing my answer when yours was not here, but it really took me 40 min to formulate everything in English, make a MetaPost image, proofread my answer… I’m rather slow at that.

actuality is a reason greater suitable than an success. Its not approximately veracity plenty because it particularly is merely the willingness to share suggestions unadorned with ulterior motives. you are able to decieve with the ‘actuality’ and you will tell an ‘honest’ lie. (ie – you additionally could make errors and nevertheless be honest). arithmetic is artwork as nicely as technology and prefer poetry, arithmetic too has infinite expressions (and expressions on infinity for that count) yet whilst compelled to choose for i could say that the ‘actuality’ is greater like poetry. jogs my memory of a asserting: in basic terms the poet can see previous the element and spot the entire image. (possibly in basic terms the poet can see previous the element and spot the actuality)

reality is a reason extra desirable than an success. Its no longer approximately veracity lots because it extremely is merely the willingness to share guidance unadorned with ulterior reasons. you may decieve with the ‘reality’ and you will tell an ‘sincere’ lie. (ie – you additionally could make errors and nonetheless be sincere). arithmetic is artwork besides as technology and prefer poetry, arithmetic too has infinite expressions (and expressions on infinity for that remember) yet whilst compelled to choose i’d say that the ‘reality’ is extra like poetry. jogs my memory of a announcing: purely the poet can see previous the component and notice the entire image. (possibly purely the poet can see previous the component and notice the reality)

(√(7)+1)/2

http://i299.photobucket.com/albums/mm286/rozeta53/…

Edit: (finished 4 min after Vašek)

Let R=1 be the radius of the greater circles and r be the radius of the smaller circles.

Let AC=FG=x

From the right triangle OAB ==> AB=2*OA

R=2(r√3/2+x)=r√3+2x ….. (1)

On the other hand

AC*AF=AD*AE ==>

x(R-x)=(R-2r)R ==>

x²-Rx+R²-2Rr=0

Solving the quadratic with respect to x ==>

x=[R-√(8Rr-3R²)]/2 ….. (2)

Substitute x from (2) in (1) ==>

R=r√3+R-√(8Rr-3R²)

√(8Rr-3R²)=r√3

Square the both sides of the above equation ==>

3r²-8Rr+3R²=0

r=R(4-√7)/3

r/R=(4-√7)/3

limS(n)=R/(1-r/R) = 1/(1-r/R) = (√(7)+1)/2

n –>∞

Nice problem, Vikram.

Looks like open kidney to me lol seriously nice question 😉