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

*of the rower's distance to the blue dot instead of the red dot.*

**y**

The red dot, blue dot and intersection point form a right triangle whose hypotenuse is * h* = R.DELTA

*and short side is DELTA*

**a***and included angle is*

**y***.*

**a**

Thus cos(* a*) = DELTA

*/*

**y***DELTA*

**R**

**a**

Letting DELTA * a* go to zero, with have the rate of increase path for the rower:

d* y*/d

*=*

**a***cos(*

**R***)*

**a**

Call the ogre's path * O*. His path increases by

*, the shore length between the red and blue dots: DELTA*

**h***=*

**O***DELTA*

**R***.*

**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/

*= the speed ratio of ogre and rower.*

**f**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

*increases linearly with DELTA*

**O***.*

**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.**