Three circles touching each other, of radius 1, 2, 3. What is the radius of the largest circle inside these?

Question

Sorta hard to ask that in the title. This is one of those problems which just wont go away. It annoys me, and i need an answer.

The problem in more detail is as follows:
There are three circles of radius 1, 2, 3
They are all touching each other, in such a way they leave an area between them
Within this area, draw the largest possible circle. What is the radius of this circle
Its actually an easy question using trig, the catch is there are no calculators allowed.

I came across this problem in a math competition i sat a year back or so, it was the last question so expect it to be hard.

This is what ive worked out, if youd rather try solve it for yourself then i guess dont read this.

At the point were the circles meet, the slope of the tangents are equal on both circles. Therefore you can connect two radii of two circles and pass through the touching point, making a 3 4 5 triangle.

With Pythagorean triples, the radius of the circle inscribed inside the triangle is given by
r=(a+b-c)

Answer 1

The centers of three circles are located at
A = (4,0)
B = (0,3)
C = (0,0)

The center of the fourth circle is of radius r is (x,y).

x² + y² = (1 + r)²
x² + (y - 3)² = (2 + r)²
(x - 4)² + y² = (3 + r)²

(y - 3)² - y² = (2 + r)² - (1 + r)²
6 - 6y = 2r
and
(x - 4)² - x² = (3 + r)² - (1 + r)²
8 - 8x = 4r

x = 1 - r/3
y = 1 - r/2

x² + y² = (1 + r)²
(1 - r/2)² + (1 - r/3)² = (1 + r)²
2 - 2(1/2 + 1/3)r + (1/4 + 1/9)r² = 1 + 2r + r²
-36 + 132r + 23r² = 0
r = 1/46 (-132 + 144) = 12/46 = 6/23

Answer: r = 6/23

x = 21/23
y = 20/23

Answer 2

If a circle is inscribed in an equilateral triangle the formulation for radius = R = a /3/3 the region a = component of triangle 9 = a /3/3 a = 9 X3 / /3 = 9 /3 field of triangle = s^2 /3/4 the place s = area of triangle = (9/3)^2 /3/4 = 243/3/4 you have not given the parent so are actually not waiting to furnish the area of shaded parent

Answer 3

Yes, indeed,the lines of centers right a triangle (3,4,5).
But I don't understand your statement that the slope of the line where two circles meet is the same. Of course it is , since it is the same line. Nor do I see its relevance. The area you are talking about is composed of arcs of the three circles.

If you connect the points of tangency, you will form a triangle consisting of chords of the three circles. If you now draw lines parallel to each chord and tangent to the corresponding circle, you will form yet another triangle. The circle inscribed in this triangle will be the circle sought.

Answer 4

Take a bow Alexander!
Elegant!

I did an easier problem, four circles, all tangent, 3 R plus 1 r and got r = R - 2R/sqrt3.

I had doubts about a rational solution but you did it.