Diameter of a circle touching three inner circles of diameter 1 If the diameters of three three inner circle are $1$ meter, what is the radius of the big circle?

(Note: the OP provided own answer below, after getting a hint).
 A: The radii of the inner circles are $r=1/2$. Therefore you have an equilateral triangle with sides $2r=1$.
You can draw lines from every corner of the triangle to the opposite edges. You then get smaller triangles that are scaled (smaller) versions of half the original triangle. By this argument (I don't know the correct term for this by heart) you can find that the distance from the center of the outer circle to the corners of the triangle is $1/\sqrt{3}$. From that point you are only one radius of an inner circle away from the edge of the outer circle.
Thus, the radius of the outer circle is:
$$R = \frac{1}{\sqrt{3}}+\frac{1}{2}$$
A: Descartes' circle theorem is a fast shortcut for such problems. Since for the three equal circles we have $\kappa_i = \frac{1}{R_i} = 2$, the curvatures of the outer and inner tangent circles are given by the solutions of:
$$2(\kappa_4^2+12) = (6+\kappa_4)^2 $$
hence:
$$\kappa_4 = 3\pm 2\sqrt{3} $$
and the inner and outer radii are:
$$\frac{1}{\sqrt{3}}\pm \frac{1}{2}.$$
A: Thanks alot SPK,for such a concise information.
Detailed Answer:
Draw Three Triangles

Now as every side of triangle is $(1/2+1/2)=1$
Hence it forms Equilateral Triangle

Now Use Cosine Formula
$$b^2 = x^2 + x^2 - 2x \cdot x \cos(120^{\circ})$$
Hence 
$$1=x^2+x^2-2 \cdot x \cdot x \cdot 0.5$$
$$\implies 1=3x^2;$$
$$\implies x=1/\sqrt{3};$$
And Radius of Big Circle: 
$$R=\frac{1}{2}+x= \frac{1}{2}+\frac{1}{\sqrt{3}} $$

