IBM Research | Ponder This | May 2001 solutions
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May 2001

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PART A:

The threshold is T=4.6033388, gotten as the solution to the pair of equations
cos(B) = 1/T,
sin(B) = (1/T)*(pi + B).

For any value of k, we define the angle B=arccos(1/k), that is, cos(B)=1/k. We measure B in radians, so that a central angle of B corresponds to a circumferential distance of R*B, where R is the radius of the lake.

Assume first that the goblin starts out at the northern edge of the lake and moves counterclockwise throughout the chase. We start at the center.

Our path has two phases.

In the first phase, we trace out a semicircular arc, whose radius is (1/(2k)) times the radius of the lake. We start out heading due south, away from the goblin. We are moving at (1/k) times the speed of the goblin. When he reaches the point due west of center, we will be due east of center, (R/k) away from the center. Moreover, we will have kept diametrically opposite the goblin at all times: the center of the circle lies directly between me and the goblin.

In the second phase, we continue in a straight line, due north, towards shore. (Notice that when we ended our semicircular arc, we were heading north; we just continue in that direction.) We aim to hit the shore at an angle B away from the normal.

Still assuming that the goblin never changes direction, from the time that we finish the semicircular arc, we will travel distance R*sin(B), while the goblin will travel distance R*( pi + B). Namely, he travels R*pi to get halfway around the lake, and another R*B to cover the extra distance to our landing point.

The goblin will exactly catch up to us if R*sin(B) = (1/k)*R*(pi+B). Together with cos(B)=1/k, we solve for B=1.3518168 radians (or 77.453398 degrees) and k=T=4.6033388.

If the goblin never changes direction, then from the time we leave the semicircle, he is always slowly catching up to us in the angular sense. If he changes direction, he is only loses some of this advantage. We wait until he is once again diametrically opposite us, and then we zig back left on another chord, which will also get us to shore at an angle B.


To show that this is optimal (for both parties), consider this function of our two positions. For each of the two chords through our point which hit the shore at angle B, consider the end of the chord closer to us. Of those two ends, select the one further from the goblin's current position. Call that point P.

The function "GA" or "goblin's advantage", is the time it will take us to reach P, minus the time it will take the goblin to reach P. If we head towards P, no matter what the goblin does, the GA does not increase.

If the goblin heads towards our current position, no matter what we do, his GA does not decrease. (This is a special property of angle B, and takes a bit of calculus to verify.)

So if we both play optimally, GA does not change. For the special value k=4.6033388, the GA starts out at 0 (for our initial position (R/k,0) and his position (-R,0)), and so ends up at 0: dead heat racing for P. If k<4.6..., the GA starts out negative, and is still negative when we reach the shore, so we escape.

Al Zimmermann gives us the following answer for Part A:
I am reminded of the old physics problem in which you are asked to use a
barometer to determine the height of an apartment building. My favorite
solution is to offer the barometer to the building's superintendent in
exchange for his telling you how tall the building is. And so I wonder if
the goblin would grant us safe passage in exchange for our boat. We could
pursue this strategy by reminding the goblin of the old saying, "Give a
goblin a human and he'll eat for a day. Give a goblin a boat and he'll eat
forever."


PART B:
Why is the goblin called "Jimmy"?
The rowboat traces out a letter J (small semicircle, leading smoothly into a chord).

Those of you who answered "Because that's his name" have been with us too long.



If you have any problems you think we might enjoy, please send them in. All replies should be sent to: webmster@us.ibm.com