There is this maths competition geometry problem and my approach
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And this is my initial approach.
From the picture, the shaded circle looks slightly bigger.
What we are looking for is the $x$ which is half of the length of the side of the red square. But it seems that we have two unknowns, the $x$ and the hypotenuse of the triangle.
Are there any other way besides this? Or any useful theorem? But I don't think it is that complicated.
Many many thanks! Helps are greatly appreciated.
On
The radius of the shaped circle is $( \sqrt{3} + \sqrt{2} - 2 )\times 110 \approx 126.089080693617$ units.
For simplicity of derivation, let us rescale all the unshaded circles to have radius $1$. Let $r$ be the radius of the shaded circle. If we choose the coordinate system such the the shaded circle is centered at $A = (0, 0)$ and the $x$-axis passing through the nearest unshaded circles on its right, the centers of 3 of the unshaded circles on its right are given by
$$B = (r+1,0),\; C = (r+3,0)\quad\text{ and }\quad D = (x_D,y_D) = (r+2,\sqrt{3})$$
Construct a line passing through $A$ tangent to the circle centered at $D$. Let the line touch the circle at point $E = (x_E,y_E)$. By symmetry, it is clear $\angle BAE = 45^\circ$. It is easy to see
$$ \begin{cases} x_E &= x_D - \cos45^\circ = r + 2 - \frac{1}{\sqrt{2}}\\ y_E &= y_D + \sin45^\circ = \sqrt{3} + \frac{1}{\sqrt{2}} \end{cases} $$ As a result, $$\angle BAE = 45^\circ \quad\implies\quad x_E = y_E \quad\implies\quad r = \sqrt{3} + \sqrt{2} - 2$$
Following is a figure illustrating the arrangement of the circles. The unshaded/shaded circles in question have been colored in red/blue in this figure.
$\hspace1in$ 
Hint. Sorry, not good at diagrams so will answer in words, but I'm sure you can draw your own diagram.
Look at the five circles which form a "ring" in the top left of the diagram. Specifically, start with the shaded circle then go left, up/left, up/right, down/right and down to the shaded circle again. Call the centres of the circles $A,B,C,D,E$ respectively. Then you should be able to work out the lengths $BC,\,CD,\,DE$, and you can see that $AB=EA$. You should also be able to find the five angles in the pentagon, and this will be enough to solve the problem.