Ring-Shaped Lakes

[leong]

This trip looks fascinating. Lake Manicouagan is the remnant of an ancient eroded impact crater and it has an unusual shape. According to this the lake is ring-shaped, bounded on the outside by an outer circle and on the inside by a concentric circular island. The area of the lake is 1492 km².

This shape begs for an interesting brainteaser. Here goes: Say you launch your kayak from the outer shore of the lake and want to paddle at a constant heading angle (a straight line). What is the distance of the longest possible trip you can take before you run aground?

Notes: Assume the lake is a perfect ring shape and neglect earth curvature. The only information needed to solve the problem is the area of the lake. All you need to solve the problem is some simple first-year high school algebra using the formula for the area of a circle and Pythagoras's theorem.

-Leon
PS

Amazingly, you don't need the diameter's of the circles to solve the problem, just the area between the two circles.

Edited February 6, 2015 by leong

[josko]

2*Sqrt(R2^2-R1^2) Manicouguan was fascinatingly gorgeous when we drove by it last summer.

OK, OK, area = pi*(R2^2-R1^2) so dist = (2/pi)*sqrt(area)

Edited February 6, 2015 by josko

[leong]

2*Sqrt(R2^2-R1^2) Manicouguan was fascinatingly gorgeous when we drove by it last summer.

OK, OK, area = pi*(R2^2-R1^2) so dist = (2/pi)*sqrt(area)

Almost right. But a small error in the algebra (a misplaced pi).

Gotta go now, my kayak is calling.

-Leon

[josko]

Do you give partial credit?

(Yeah, a sqrt(pi))

[Dan Foster]

With the right combination of slip strokes, draws, and forward and reverse strokes (or merely a circular wind pattern), you should be able to paddle an infinite distance around the torus while maintaining a constant heading.

Given a choice between doing high-school algebra or looking for loopholes, I'd rather look for loopholes.

[leong]

With the right combination of slip strokes, draws, and forward and reverse strokes (or merely a circular wind pattern), you should be able to paddle an infinite distance around the torus while maintaining a constant heading.

Given a choice between doing high-school algebra or looking for loopholes, I'd rather look for loopholes.

Dan, true, but I did emphasize a straight line. Nevertheless, I’ll give you a C+ for ingenuity.

Josko, you got full credit for your answer. I consider your error just like a typo.

The interesting thing about this problem is that the answer is the same regardless of the diameter of the island or lake.

There’s a cool similar problem in three dimensions. If you drill a cylindrical hole in a sphere, the volume remaining in the sphere depends only on the height of the hole - it is independent of the diameter of the hole or diameter of the sphere.

Perhaps only nerdy kayakers think it’s cool.

-Leon

PS

The numerical answer to the problem is 49.7 Km (30.9 statute miles). Note, in the statement of the problem I reversed a couple of digits (the area of the lake is 1942 Km 2, not 1492 Km 2). I must have been thinking of Columbus. So, Josko, each day you could start paddling from the outer shore of the lake to the tangent point of the island (about 15.5 miles) and have lunch there and then paddle along the same straight line (chord) to the outer shore of the lake to camp for the night. Then repeat going all around the lake.

PPS

Extra credit bonus question: How many days of paddling would it take to completely circumnavigate the island (the lake’s outer shore length is 821 miles) if everyday you paddled along each successive chord (30.9 miles long)?

[Dan Foster]

26.5 days.

2 pi r = 821 mi

r = 130.6 mi (radius of lake)

let two_theta be the angle around the center of the lake traversed in a day. let theta be half of that.

sin theta is half the chord distance (15.5 mi) / r

theta = 6.79 degrees

two_theta = 13.58 degrees per day

360 degrees / 13.58 = 26.5 days to complete the circle.

extra extra credit: how many days would you have to paddle to complete the "spirograph" and end up exactly back at your starting location (within 1 mile, let's say)

Bonus Manicougan reading:

http://www.myccr.com/phpbbforum/viewtopic.php?t=7780

[leong]

26.5 days.

2 pi r = 821 mi

r = 130.6 mi (radius of lake)

let two_theta be the angle around the center of the lake traversed in a day. let theta be half of that.

sin theta is half the chord distance (15.5 mi) / r

theta = 6.79 degrees

two_theta = 13.58 degrees per day

360 degrees / 13.58 = 26.5 days to complete the circle.

extra extra credit: how many days would you have to paddle to complete the "spirograph" and end up exactly back at your starting location (within 1 mile, let's say)

Bonus Manicougan reading:

http://www.myccr.com/phpbbforum/viewtopic.php?t=7780

Dan,

You did the math right, the same as I did except I used the law of cosines to get the double angle. However, you end up about 1.5 miles from shore on the last partial chord. That’s not a good answer for a paddling club (ending a circumnavigation so far from shore). I guess the solution is to paddle a beeline towards your starting point after the last full-length chord.

On the extra extra credit problem I need to familiarize myself with the parametric equations of the hypotrochoid and epitrochoid before I attempt to answer the question. So I’m requesting a time extension to hand in my solution for the extra extra credit problem. Probably offline because this is getting too far afield from kayaking topics.

-Leon