Mike and Ike’s Trike

[Smith]

Mike and Ike buy a trike and ride in circles around a spike. They each ride at a different time of day. They both pedal at the same rate. They both make perfect circles with the spike at the exact center. Mike rides twice as long (time) as Ike, but they cover the exact same distance. How is this possible?

(Edit - minor change in terms)

Edited January 24, 2012 by Smith

[Guest]

Ike travels Pi*r1^2 * X (m/s) * Y (seconds). Mike travels Pi*r2^2 * X(m/s) * 2Y (seconds). These values are equal because they travel the same distance.

Pi * r1^2 * X * Y = Pi * r2^2 * X * 2Y. Divide both sides by Pi, X and 2Y r1^2 / 2 = r2^2. Take the square root of both sides sqrt (r1^2 / 2) = r2. r2 = r1 / sqrt(2). 1/sqrt(2) = .707ish. Thus, they travel the same distance because Mike's radius is .707 times Ike's radius and must travel longer to cover the same distance.

[Smith]

How do you define distance? I remember my physics teacher talking about displacement- the distance between the start and end point. If Ike completed a whole number of circles (1, 2, 3, etc), the displacement would be 0. If Mike rides in a circle with the same radius, going around for twice as long, he would also end at the start point and the displacement would alsi be 0.

Probably not what you are looking for but might as well try.

In the spoiler, Thalia asks "How do you define distance?" then discusses the concept of displacement and puts this into the perspective of start and end points. Complete circles could then be considered as a distance of 0. Excellent thought, displacement, but not what I had in mind at all. Read on.

The spike is in the middle of a playground carousel. One of them rides against the spin of the carousel, one of them rides with the spin. To an outside observer on solid ground, one has covered more distance than the other without pedaling any faster..

Another amazing idea brought to bear on the riddle! If the riding surface is moving relative to "solid ground" then the distance covered by the rider relative to solid ground can be equalized. I am constantly amazed at the ingenuity of the Brain Denizens! However, my trickery was much simpler than this. Read on.

It is possible if you define "distance" in terms of angles instead of the usual metric.

I feel that Superprismatic has touched so closely upon my solution that I can no longer maintain the charade. Below you will find my intention and the solution.

Simon Legree first mentions the concept of distance from the spike, but reversed the distance (Mike would have to be farther from the spike, not closer). flamebirde agrees, but neither of them explains their reasoning. Given the "twice as long" wording, it could be a shot in the dark to suppose a "half the distance" solution, so I had to pan those responses.

opnurisyder then gives the correct relationship between the riders and the spike, but likewise does not give any more details by which to discern whether this is a guess or a well-considered solution. I left the door open for more details to be provided but none followed.

Superprismatic makes a statement regarding the definition of "distance" in terms of angles. It might have been more satisfying had he gone on to mention the required relationship in terms of their distance from the spike, but knowing the caliber of answers usually coming from Superprismatic, I must assume he had this detail in mind.

So, here is the key... If the distance covered by each rider is stated in terms of "angular distance" as is used in certain circles (astronomy comes to mind), then if Mike rides in a circle with twice the radius of Ike's circle, and travels twice the linear distance that Ike does, their ANGULAR DISTANCE will be exactly the same. That was it. Bravo to those who conceived so many alternate solutions (the viable ones, anyway).

Kudos to Thalia for raising the question of the definition of distance, a very salient point, to opnurisyder for giving the correct relationship of distances from the spike, and to Superprismatic for identifying the exact nature of the key - the use of the word "distance" "in terms of angles".