Surely the shortest path for both has to be a straight line, else a straight line connecting the endpoints would lead to a shorter path. But the shortest path back to the road is too long, if they arrived at the road after making 40/41 meters the bus would have passed as it would have traveled 80/82meters. So maybe it works if they head towards the road at an angle with the shortest path like this:
Code:
B-
|
|
|b
|
| l |
P............S-
.alpha |
. |
L . |x
.|
X-
The Bus is at B when Camille/Melissa is at P, b is 70meter, l is 40 resp. 41meter, and they walk at an angle alpha they need to travel the greater distance L>l, but the bus needs to travel x meters more and there is hope that they can make the bus travel just as much as to arrive at X at the nick of time.
Now a little meditation is neccessary to get to the right inequation of the problem (it took me quite a bit of meditation as i tried to transform the problem by assuming the sisters are as fast as the bus but have to travel twice as long -- somehow it didn't work out but i spare you the messy details ... the fun of puzzling
): In order to make it before the bus, the sisters have to travel a distance of more than 2*L while the bus travels a distance of b+x:
2*L > b+x
Now get the variables on one side:
(b+x)/L > 2
So let's work out the details, looking up the sin/cos/tan formulas for a right triangle like that one made by points P,S,X:
L = l / cos alpha
x = l * tan alpha = l * sin alpha / cos alpha
and let's put that into our inequation and simplify:
b*cos alpha/l + x*cos alpha/l > 2
b/l * cos alpha + sin alpha > 2
Looks like a close call as the cos and sin functions are smaller than one most of the time and not both equal to one at the same angle. Plotting the left hand side over the angle alpha had me guessing already that Camille might make it but Melissa might have a hard time.
So it looks like we need to maximize the left hand side to see if we can get past 2. If you want to maximize a function, the differential has to be zero, so let's differentiate:
f(x) = C*cos(x) + sin(x) : cos'(x) = - sin(x), sin'(x) = cos(x)
f'(x) = cos(x) - C*sin(x)
f'(x) = 0
cos(x) - C*sin(x) = 0
cos(x) = C*sin(X)
cos(x)/sin(x) = C
1/tan(x) = C
tan(x) = 1/C
x = atan(1/C)
So we have worked out the best angle for
alpha = atan(l/b). Now all that's left to be done is plug in 40 resp 41 for l and 70 for b and work out if the right hand side of our inequation is more that 2. And this verifies that Camille arrives just in the nick of time but Melissa fails to make it. The bus travels about 93 meters before stopping for Camille, i guess she can stop the bus to let Melissa get in too.
It looks like the best angle makes the triangle B,P,X a right triangle and it seems that would be a good way to head for the bus in general (regardless of velocities), but i haven't found a reason why that should be that way.
Maybe there is a simpler approach hidden somewhere. 💡
According to my maths (which I hope are correct)
, there is a single point where one of the sisters can just exactly reach the bus.
)
But the click in your head after you solved one makes up for it. 
The Cars and Buses grabbed my attention for several hours, but it was surely worth it. 👍
