AMC 12 word problem modified to be considerably harder The original problem is stated as follows:
Rachel and Robert run on a circular track. 
Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert 
runs clockwise and completes a lap every 80 seconds. 
Both start from the same line at the same time. 
At some random time between 10 minutes and 11 minutes after they begin to run, 
a photographer standing inside the track takes a picture that shows one-fourth of 
the track, centered on the starting line. What is the probability that both 
Rachel and Robert are in the picture?

This problem is quite simple -- you just see for how long both of them are in the area in the time period and divide that by $60$ (the answer is $\frac{3}{16}$). I wanted to take this a step further and say, what if the position of the photographer is also randomized? How would one go about solving that?
A possible solution I have in mind involves is not elegant at all and I don't know if it would even work. I would consider four places equally spaced around the track and find the probability for each. Then I would add in a point in the middle of each of the four places to see what that answer would be, continuing to do that to see if there was any noticeable pattern.
Any ideas? Is there a nice solution to this problem (doesn't involve use of computer programs, etc.)?
 A: An idea: Treat the problem like calculus. Let's say the probability of photographer being centered on the starting line is $dt$. Then you have your probability, $3/16$. Then if the photographer moves $dt$ to the right, you have a new probability. 
The interval for $x$ is $[600, 660]$ seconds. If I imagine this correctly, Rachel's distance in each lap is $1/3 \to -1/3$ and Robert is $-1/2 \to 1/4 $. The photographer's area is $-1/4 + dt \le y \le 1/4 + dt$, which is the same as $-1/4 \le  y - dt \le 1/4$. The area represents the space of every possible $dt$.
Rachel's distance $y=1/3 - x/90$, Robert's distance $y = -1/2 + x/80$. 
Thus the answer is the area of $\{-1/4 \le 1/3 - x/90 - dt \le 1/4 \}\cap \{-1/4 \le -1/2 + x/80 - dt \le 1/4 \}$ divided by $1 \times 60$
I believe this probability works out to the area of a parallelogram, $\frac{\sqrt{2073601}}{8160} \approx 0.1765$
A: If L is a lap distance, then Rob runs at a speed of L/(80/60) and Rachel at a speed of L/(90/60). The time taken until Rob and Rachel pass each other is given by:
$(L/(80/60) + L/(90/60)) * t_p = L$
$\Rightarrow t_p = 12/17$ minutes
So Rob and Rachel meet when $t = n*t_p$. The distance between them increases linearly to $L/2$ at $(n+\frac{1}{2})*t_p$, then decreases linearly back to $0$ at $(n+1)*t_p$. $10$ minutes corresponds to $(14 + \frac{1}{6})* t_p$, $11$ minutes corresponds to $(15 + \frac{7}{12}) * t_p$.
The probability of a random shot covering $\frac{1}{4}$ of the track capturing both Rob and Rachel when they are $k*L$ apart is $max(\frac{1}{4} - k, 0)$. It follows the probability is given by:
$P = \int_{10}^{14.25t_p} \frac{14.25 t_p - t}{t_p} dt + \int_{14.75t_p}^{15t_p} \frac{t-14.75 t_p}{t_p} dt + \int_{15t_p}^{15.25t_p} \frac{15.25 t_p - t}{t_p} dt$
