The figure shows a $60^o$  angle in which are placed $2007$ numbered circles (only the first three are shown in the figure). The circles are numbered according to size. The circles are tangent to the sides of the angle and to each other as shown. Circle number one has radius $1$. Determine the radius of circle number $2007$.
[img]https://1.bp.blogspot.com/-1bsLIXZpol4/Xzb-Nk6ospI/AAAAAAAAMWk/jrx1zVYKbNELTWlDQ3zL9qc_22b2IJF6QCLcBGAsYHQ/s0/2007%2BMohr%2Bp4.png[/img]
1. Let Point \( A \) be the vertex of the \( 60^\circ \) angle, \( O_i \) be the center of Circle \( i \), and \( T_i \) be the point on Circle \( i \) closest to Point \( A \). We notice that if we connect \( A \), \( O_i \), and the point where Circle \( i \) is tangent to a line, we get a \( 30^\circ \)-\( 60^\circ \)-\( 90^\circ \) triangle.

2. Consider the triangle with point \( O_1 \) and let \( r_i \) denote the radius of Circle \( i \). We see that the point \( T_1 \) is on \( AO_1 \) and we note that \( O_1T_1 = r_1 = 1 \). We also see that the length of \( O_1 \) to the point tangent to Circle \( 1 \) on the line is \( r_1 = 1 \). From the properties of a \( 30^\circ \)-\( 60^\circ \)-\( 90^\circ \) triangle, this length is \( \frac{1}{2}(AO_1) \), so \( AO_1 = 2 \).

3. We see that \( AO_1 = AT_1 + O_1T_1 = AT_1 + 1 \) and \( AT_1 = O_1T_1 = r_1 \). Continuing in this fashion, we see that \( r_2 = 3 \) since \( AT_2 = 3 \).

4. To generalize, we see that \( 2r_i = AT_i + O_iT_i = AT_i + r_i \) due to the properties of a \( 30^\circ \)-\( 60^\circ \)-\( 90^\circ \) triangle, so \( AT_i = r_i \) and \( r_{(i+1)} = AT_{(i+1)} = AT_i + T_iT_{(i+1)} = 3r_i \).

5. Therefore, \( r_{2007} = 3r_{2006} = 9r_{2005} = \cdots = 3^{2006}(r_1) = 3^{2006} \).

The final answer is \( \boxed{3^{2006}} \).