In the 2009 Stanford Olympics, Willy and Sammy are two bikers. The circular race track has two
lanes, the inner lane with radius 11, and the outer with radius 12. Willy will start on the inner lane,
and Sammy on the outer. They will race for one complete lap, measured by the inner track. 
What is the square of the distance between Willy and Sammy's starting positions so that they will both race
the same distance? Assume that they are of point size and ride perfectly along their respective lanes
1. **Determine the length of one lap on the inner track:**
   The inner track has a radius of 11 units. The circumference \(C\) of a circle is given by:
   \[
   C = 2\pi r
   \]
   Substituting \(r = 11\):
   \[
   C_{\text{inner}} = 2\pi \times 11 = 22\pi
   \]

2. **Determine the length of one lap on the outer track:**
   The outer track has a radius of 12 units. Using the same formula for circumference:
   \[
   C_{\text{outer}} = 2\pi \times 12 = 24\pi
   \]

3. **Calculate the unused portion of the outer track:**
   Since the race is measured by the inner track, both bikers will complete a distance of \(22\pi\). The outer track's circumference is \(24\pi\), so the unused portion of the outer track is:
   \[
   24\pi - 22\pi = 2\pi
   \]

4. **Determine the fraction of the outer track that is unused:**
   The fraction of the outer track that is unused is:
   \[
   \frac{2\pi}{24\pi} = \frac{1}{12}
   \]

5. **Calculate the central angle corresponding to the unused portion:**
   The total angle in a circle is \(360^\circ\). The central angle corresponding to the unused portion is:
   \[
   \frac{1}{12} \times 360^\circ = 30^\circ
   \]

6. **Use the law of cosines to find the square of the distance between Willy and Sammy:**
   The law of cosines states:
   \[
   c^2 = a^2 + b^2 - 2ab \cos(\theta)
   \]
   Here, \(a = 11\), \(b = 12\), and \(\theta = 30^\circ\). Substituting these values:
   \[
   c^2 = 11^2 + 12^2 - 2 \times 11 \times 12 \times \cos(30^\circ)
   \]
   We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), so:
   \[
   c^2 = 121 + 144 - 2 \times 11 \times 12 \times \frac{\sqrt{3}}{2}
   \]
   Simplifying further:
   \[
   c^2 = 121 + 144 - 132\sqrt{3}
   \]
   \[
   c^2 = 265 - 132\sqrt{3}
   \]

The final answer is \(\boxed{265 - 132\sqrt{3}}\).