Consider my diagram for straight line tangential escape.
From the red dot [landing point] move clockwise a small angle DELTA a and draw a blue dot.
Call the shore length between the dots h.
Draw an arc centered on the rower's starting point clockwise from the red dot [radius is y, the rower's distance to the red dot.]
Call this arc s.
Now draw a line from the blue dot back to the rower's starting position.
The distance from the blue dot on the shore to the intersection with arc s is the change DELTA y of the rower's distance to the blue dot instead of the red dot.
The red dot, blue dot and intersection point form a right triangle whose hypotenuse is h = R.DELTA a and short side is DELTA y and included angle is a.
Thus cos(a) = DELTA y / R DELTA a
Letting DELTA a go to zero, with have the rate of increase path for the rower:
dy/da = R cos(a)
Call the ogre's path O. His path increases by h, the shore length between the red and blue dots: DELTA O = R DELTA a.
Again, in the limit we have
dO/da = R
So with respect to the angle a, the rower's and ogre's paths increase in a ratio of cos(a).
Recall that cos(a) = 1/f = the speed ratio of ogre and rower.
Thus, the difference in arrival time for the two racers with respect to angle is zero at the red dot.
This means the rower's advantage over the ogre is either a maximum or a minimum at a.
Note that the triangle we drew actually has curved sides, so the equality is only strict as DELTA a goes to zero.
They curve away from each other, so that DELTA y increases super linearly, while DELTA O increases linearly with DELTA a.
Thus, for angles greater than a, the ogre gains an advantage.
The rower's advantage is thus a maximum.
Conclusion: the red dot is the optimal landing point, and a straight line is the shortest distance to it.