A simple challenge?

Question

A friend of mine asked me this question and I think the answer is correct so I'm challenging you guys if you could give me the answer.

Three circles are tangent to each other. An oh so small circle can be place between the three circles making it tangent to all three of them.

The question now is... derive a formula on how to get the radius of the "oh so" small circle

let the radius of the "oh so" small circle be A

Gud luck guys!!!

btw way if your thinking that it's someone else's homework... it's not...

I've done this already and I have an answer... I just want to know if someone else can do the same or if there is another method to it

Another thing... the circles may or may not be of the same radius

C'mon guys the circles aren't of the same radius!!!

Answer 1

This is just a method of calculating A to any degree of accuracy, not an explicit formula:

Suppose the radii of the three large circles are r1,r2,r3. Use the three centers of the circles to form a triangle with sides:

r1+r2,r2+r3,r3+r1

The area of this triangle using Heron's formula is:

A = sqrt(r1*r2*r3*(r1+r2+r3))

Now divide this triangle into three triangles each with one side from the original triangle and with the center of the small circle (of radius A) as a vertex. These triangles have sides:

r1+r2,A+r1,A+r2
r2+r3,A+r2,A+r3
r3+r1,A+r3,A+r1

So the sum of these areas must be the same as the area of the original triangle using Heron's formula:

sqrt(A*r1*r2*(r1+r2+A))
+sqrt(A*r2*r3*(r2+r3+A))
+sqrt(A*r3*r1*(r3+r1+A))
=sqrt(r1*r2*r3*(r1+r2+r3))

Defining S = r1+r2+r3+A and dividing by r1*r2*r3*A, this can also be written:

sqrt(S/A-1)=sqrt(S/r1-1) + sqrt(S/r2-1) + sqrt(S/r3-1)

==>

S/A = 1 + [sqrt(S/r1-1) + sqrt(S/r2-1) + sqrt(S/r3-1)]^2

This can be solved iteratively by starting with S=r1+r2+r3, using the above equation to calculate A, then setting S = r1+r2+r3+A and doing it again.

Since the circle is indeed oh so small, the difference between S and r1+r2+r3 (i.e. A) is very small and the process converges very quickly.

Using Surveyor's numbers 30,40,50 (and his answer A=5.96032619) as an example:

S = 30+40+50=120 --> A = 6.0776728
S = 126.0776728 --> A = 5.9581718
S = 125.9581718 --> A = 5.960365

etc.

Answer 2

The radius of the larger circles will be B.

Connecting the centers of the circles forms an equilateral triangle. The height of this equaliteral triangle will be sqrt(3)*B.

The center of the smallest triangle will be where the bisectors of the big triangle meet or 2/3 of the way down the height. The line from the top circle to the center of the smallest circle equals that 2/3 of the height and it also equals A + B.

So 2/3 * sqrt(3) * B = A + B
and A = B * (2/3 * sqrt(3) - 1)

Answer 3

Wow! I'm trying this in AutoCad and it has me baffled. It may have somthing to do with the triangle formed by the three larger circles. This will probably reduce my productivity at work today unless some genius gives the correct answer before I figure it out.
Okay, using Autocad, and tangent circles with radii of 30, 40 and 50, circle "A" has a radius of 5.96032619. I solved it graphically, but I can't get the formula to come out of my brain.

Answer 4

Draw a line using the center point of each circle to create a triangle. The area of that triangle is the relationship to the area of that 'oh so' small circle from which the radius can be derived.

Answer 5

If the radius of the "oh so" small circle is A, the formula for A is A=
a((sq. rt.(3))/3 - 1), where a is radius of the big circle. This is correct if the tangent circles are equally big and form with their radiuses a triangle with all angles of 60°(i can't remember how this triangle is called in english).
Now... I would like those 10 points, please :)

Answer 6

1.The triangle formed by the centers of the major circles of radius R is an equilateral triangle of side k. Where k=2R
2.The center of this triangle is the center of the inner circle radius of r which we seek.
3.Draw three lines to bisect its 60 degree angles and apposing sides. They will intercept in the middle of the triangle and in the center of the minor circle. You should have six right and equal triangles inside the equilateral triangle. Got that?
4.The sides of these small triangles in terms of the larger equilateral triangle is a= 0.5k, b= .5k tan(30 degrees), c= 0.5k /cos(30) or since k=2R we have a=R, b=R tan(30), c=R/cos(30).

5. The curios thing is that b is the radius of a minor circle.
So r=R tan(30)= 1/sqrt(3)=0.577 R

Have fun

Answer 7

I'm too lazy to do it, but the center of the small circle must be at the intersection of three hyperbolas with the centers of the given circles as foci, defined by the differences of the radii.

Answer 8

The above answers seem to assume that the three large circles are equal. That was not stated in the problem. If so, it greatly simplifies the soluton.

Answer 9

enable me ask you the alternative. in case you have been to choose that God replaced into actual for in elementary terms one hour, could you completely replace your morality? in all threat no longer as a results of fact in that in need of a length of time you does no longer be waiting to triumph over the philosophies you have stepped forward, the morals that society has imposed on you, the values you have found out out of your loved ones, etc. it may take plenty longer than one hour for all those impacts to countered. Plus atheist do no longer bypass out and do all those issues on an hourly foundation. they could in all threat could have a reason to desire to kill somebody. (except they have so plenty greater incorrect with them that basically atheism.) Or a keen important different to commit adultery. yet whilst a individual does no longer believe that there are absolutes of suitable and incorrect - ethical regulations given by utilising God - they are greater probable to persuade themselves they have the main dazzling to kill somebody or to cheat on their important different in the event that they are "in love" with somebody else. yet they do no longer do all those issues on a hourly foundation. as a thank you to show out to a Christian that in the event that they have been to resign believing in God, they does no longer do all those issues in one hour is an illogical argument. despite the fact that in case you do away with the source of absolute values from their life, you enhance the possibilities that given time, given the inducement and given the possibility they could be greater probable to do those issues. Sorry, yet your argument fails....