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:

d**y**/d**a** = **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

d**O**/d**a** = **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.

**Q.E.D.**