Friday Puzzler -- Halftime Show

Re: Friday Puzzler -- Halftime Show

Larry Barrett

Square root of 2 is part of the answer.

Re: Friday Puzzler -- Halftime Show

Alex Y

Sounds like you may have it!

A new formation

Alex Y

Larry was too quick on that one, so the band formed a new formation.

This time the band forms a square 50 yards on a side. The mascot is faster this time, and runs around the perimeter of the square as it is marching, again returning to its original position w.r.t. the square, when the square has moved exactly 50 yards.

How far did the mascot run this time?

Re: A new formation

Ed in Leaside

I would have thought it was just once around.

Re: A new formation

Larry Barrett

Seems like it must be more than once around. If the dog starts out on the goal line, runs around the band as it marches, the dog will end up on the 50 yd line. So a possible answer would be 200 + 50.

Solution?

Alex Y

Larry, why don't you post your answer and its derivation for the first version, and that may help others attacking the additional formation. The methodology of the two is the same.

Not there yet

Alex Y

The answer is between Ed's and Larry's first guesses.

Re: Not there yet

Larry Barrett

Assume the dog starts on the goal line, and runs toward the 50 yd line for the first leg. On this leg the dog runs with the band, so will run more than 50 yds before it turns to run across the field. Once the dog reaches the other side of the formation it will turn and run against the direction the band runs, so on that leg it will run less than 50 yds. Given that the dog runs at a constant rate and the band marches at a constant rate, these two legs will average 50 yds each. And the band will have covered half the distance from the goal to the 50 yd line (25 yds). So on the other two legs, where the dog runs across the field, the dog will run 50 yds + 50 yds + the remaining 25 yds. The total will be 200 + 25 yds.

Re: Solution?

Larry Barrett

Martin Gardner was well know for his 'Aha' puzzles, but no aha for me for this one.

Assume Band 1 (B1) is on the goal line and Band 2 (B2) is on the 50 yard line. Both are marching toward the other goal line, at the same rate (Rb). The mascot runs at a faster rate (Rm).

When the mascot reaches B2, B2 will be the same distance from the 50 yard line as B1 will be from its starting goal line; say this point is P yards from B1's starting point.

Using the familiar D=RxT formula, or T=D/R, we can write:

P/Rb = (50+P)/Rm

since the time for B1 to reach P is the same time for the mascot to reach 50+P.

Then the mascot turns and runs back toward B1, reaching B1 at the 50 yard line.

Since these times are equal, we can write a similar equation:

(50-P)/Rb = P/Rm

Solving equation 1 for Rb and then substituting this into equation 2, we get

(50-P)(50+P)Rm = (P^2)Rm

50^2 - P^2 = P^2

50^2 = 2P^2

so P = 50/(sqrt 2).

Since the mascot runs a total of 50 + 2P, the mascot runs 50(1 + sqrt2) yards.

Re: Not there yet

Larry Barrett

Replying to my own post to say bad assumptions, more work to do.

Re: Not there yet

Ed in Leaside

Yes, I think there's a hypotenuse or 2 involved in this.

Re: Solution?

Alex Y

Excellent. And no, this does not have a typical Gardner "Aha".

He does use a clever trick for setting up the algebra, though. He redefines the length and time units. He didn't name them, but say that 50' = one length unit and the time it takes the band to march that far is a single time unit. Then the band's speed and distance traveled, in time 1 are both 1. The mascot's velocity V is also his distance traveled in these units.

Considering the mascot's movement relative to the band, he travels one unit at speed (v-1) and one unit at speed (v+1). So his time going forward is 1/(v-1) and his time returning is 1/(v+1), and those two have to add to one. Since that totals to one time unit, with a little manipulation, we get v^2-2v-1=0, which has only one positive root, v = 1+sqrt(2). Then multiply by 50 to express in feet.

Nothing but 8th-grade algebra but HARD, IMHO.