[Yoruichi-san]

Since there seems to be a revived interest in math problems, here's one I got from a friend a while ago (which I couldn't resist adding my own spin on ;P):

Bonanova, Prime, Chuck Rampart, and Prof. Templeton are each standing at one of the corners of a square room of length 20, puzzling over the latest probability problem. At the exact same moment, they each come up with the solution and, simultaneously, each person begins to run towards the person adjacent to them in the counterclockwise direction to share their ideas. If they all run at the same constant speed and always directly towards their respective targets, what is the total distance they travel before meeting in the center of the square?

Please show your proof!

[Prime]

Differential equations and vector analysis again?

But if each stands in their corner, until they come up with the solution, this room is going to be very quite for a while. How do you gauge the interest then?

[andromeda]

I'm not sure

This is the picture

They kinda run one quarter of circumference (the picture is bit off)

And then we'll have for one person

4 times

So in the end the total (all four of them together will run the distance of the circles circumference) is = 62.8

But it's probably more complicated

[Guest]

10pi

my theory for this is that this is essentially a arc problem, since everyone will be circling towards each other.

so, using pi/2 for the degrees (90), multiplying that by the radius, 20, you get 10pi...

[Brandonb]

31.4 per person

Edited October 2, 2008 by Brandonb

[Yoruichi-san]

Differential equations and vector analysis again?

But if each stands in their corner, until they come up with the solution, this room is going to be very quite for a while. How do you gauge the interest then?

What do you mean by "again"? This is my first real math puzzle...

Lol...and my opinion of you four is that it won't take that long to figure out the solution...

[Yoruichi-san]

There's a reason I asked for the proof...a lot of ppl come under the impression that they run in circular arcs...but trace out their trajectories on those arcs...are they always running directly towards each other?

[unreality]

Good one It has to do a lot with logarithmic/equiangular spirals, right?

[Prime]

There's a reason I asked for the proof...a lot of ppl come under the impression that they run in circular arcs...but trace out their trajectories on those arcs...are they always running directly towards each other?

No hints, please. You already know the solution, give us a chance.

[bonanova]

Since there seems to be a revived interest in math problems, here's one I got from a friend a while ago (which I couldn't resist adding my own spin on ;P):

Bonanova, Prime, Chuck Rampart, and Prof. Templeton are each standing at one of the corners of a square room of length 20, puzzling over the latest probability problem. At the exact same moment, they each come up with the solution and, simultaneously, each person begins to run towards the person adjacent to them in the counterclockwise direction to share their ideas. If they all run at the same constant speed and always directly towards their respective targets, what is the total distance they travel before meeting in the center of the square?

Please show your proof!

we each run a distance of 20

our paths [more precisely our velocity vectors] are initially, and always remain, mutually perpendicular.

Thus the person towards whom I am running never gets closer to me nor farther from me because of his motion.

It's as if he stood still.

Thus we each run a distance equal to the side of the square room.

Nice problem.

Here is a related question.

Suppose Prime and I hold opposite ends of a taut string.

What is the area swept out by the string?

As Y-san would say, and so would I, please show your proof.

[Guest]

our paths [more precisely our velocity vectors] are initially, and always remain, mutually perpendicular.

Thus the person towards whom I am running never gets closer to me nor farther from me because of his motion.

It's as if he stood still.

Thus we each run a distance equal to the side of the square room.

Really? I would agree that the velocities are always mutually perpendicular, but not that I never get closer to my target. If my velocity is towards my target, then, unless their (equal) velocity is in the same direction I must get closer to them.

I would expect to see the velocities forming a series of inscribing squares closing in to the centre (and a diagram much like Andromeda's)

edit: BBCode

Edited October 2, 2008 by foolonthehill

[Guest]

Suppose Prime and I hold opposite ends of a taut string.

What is the area swept out by the string?

As Y-san would say, and so would I, please show your proof.

we each run a distance of 20

our paths [more precisely our velocity vectors] are initially, and always remain, mutually perpendicular.

Thus the person towards whom I am running never gets closer to me nor farther from me because of his motion.

It's as if he stood still.

Thus we each run a distance equal to the side of the square room.

Nice problem.

Here is a related question.

Brilliant, as always.

1/4 the area of the square (so 100 ft²)

This is because, between the four equivalent strings, the entire area of the square must be swept out, and all four segments must be congruent.

[Guest]

Actually, I understand bonanova's point now - he doesn't get any further/closer to me as a result of his own motion . Therefore, it is only my motion that gets me closer, and as I'm always travelling towards him, I only have to move 20m before I catch him (which also happens to be the centre because everyone else has similarly caught their targets).

Apologies bonanova, I misread your solution. As CR said, it's brilliant !

And, for what it's worth, I agree with CR for the area. (easy to see on my previous diagram)

Edited October 2, 2008 by foolonthehill

[bonanova]

Actually, I understand bonanova's point now - he doesn't get any further/closer to me as a result of his own motion . Therefore, it is only my motion that gets me closer, and as I'm always travelling towards him, I only have to move 20m before I catch him (which also happens to be the centre because everyone else has similarly caught their targets).

Apologies bonanova, I misread your solution. As CR said, it's brilliant !

And, for what it's worth, I agree with CR for the area. (easy to see on my previous diagram)

Thanks, both.

Curious... is it the song or the book? Or are they related? [foth i mean]

[Prof. Templeton]

I appreciate being cast in your riddle with such astute thinkers. I feel the other players are in another league when it comes to math problems, but thank you anyway. I see this one was settled while I slept. I really enjoy this type of problem where the solution can be had by the easy method or the hard method. Similar to bonanova's hole in a sphere, or the fly between two moving trains. You can utilize complex equations, or you can sit and think about it for awhile. I prefer to sit and think, but I seldom have time for either.

Here's an equation for a square start

d = rSqrt2

[Prime]

Brilliant, as always.

1/4 the area of the square (so 100 ft²)

This is because, between the four equivalent strings, the entire area of the square must be swept out, and all four segments must be congruent.

That's an awesome solution by Bonanova. However, I'll hold back on praising for now, on the account of the "related" problem. To me that's an indication that Bonanova already knew the problem and/or read about it. If that's the case, he simply denied our chance of solving it. If it is not the case, I appologize. The solution is really ingeneous.

With respect to the area sweeped, I don't see how it is 1/4.

The way I see it...

The way I understood it:

The rope between B and P is always a straight line passing through the center of the square.

There are only two pairs of men which can hold the rope. And I don't see why the area sweeped (highlighted in yellow) is 1/4 of the total area.

If you can derive the function for the "h" (distance from the man's path to the square diagonal) as a function of the coordinate on the diagonal, or projected coordinate on the side, then integrating that function in corresponding boundaries would give the area.

[Guest]

That's an awesome solution by Bonanova. However, I'll hold back on praising for now, on the account of the "related" problem. To me that's an indication that Bonanova already knew the problem and/or read about it. If that's the case, he simply denied our chance of solving it. If it is not the case, I appologize. The solution is really ingeneous.

With respect to the area sweeped, I don't see how it is 1/4.

The way I see it...

The way I understood it:

The rope between B and P is always a straight line passing through the center of the square.

There are only two pairs of men which can hold the rope. And I don't see why the area sweeped (highlighted in yellow) is 1/4 of the total area.

If you can derive the function for the "h" (distance from the man's path to the square diagonal) as a function of the coordinate on the diagonal, or projected coordinate on the side, then integrating that function in corresponding boundaries would give the area.

My solution was based on the two of you starting on adjacent corners.

[Prime]

My solution was based on the two of you starting on adjacent corners.

For the adjacent corners it seems to be the case. There are four identical non-intersecting areas covering the entire square.

My initial instinct tends towards more complex. But since Y-s put Prime and Bonanova into adjacent corners, your interpretation must be the true one.

[bonanova]

Bonanova ... denied our chance of solving it.

Earth to Prime.

The solution was given 6 hours after OP, in post #10, and 2 of the previous posts were yours. .

If I felt like tweaking you, I'd claim that I just got impatient waiting for you.

But that's not my style.

btw, you don't have to posture so much. you're well recognized for your insightful analysis already.

The fact that you were included in to OP proves that.

[unreality]

Any more similar problems, Ysan? This one was great

Edited October 2, 2008 by unreality

[Prime]

Earth to Prime.

The solution was given 6 hours after OP, in post #10, and 2 of the previous posts were yours. .

If I felt like tweaking you, I'd claim that I just got impatient waiting for you.

But that's not my style.

btw, you don't have to posture so much. you're well recognized for your insightful analysis already.

The fact that you were included in to OP proves that.

I don't seek recognition. Especially, for dazzling solutions, which are not my own. I just enjoy solving problems. And for that matter, I disagree with posting answers from books and other websites.

At any rate, it should have been up to Y-s to post a book answer on this topic. Which, as I looked it up, is d_n = 1/(1 - cos(2pi/n)), where n is the number of sides of a regular polygon.

There are different schools of thought on how to ejnoy riddles.

Some people like to look at the problem for a few minutes and then head for the answers section.

Others only want to find solution on their own.

Both ways are perfectly legitimate. Unfortunately, there is a conflict of interest there at times.

Logically, I don't see any reason for posting solutions which you have found in literature. Those who want to find the answer can do the same. Also, note that people come upon posts at different times. And it would be interesting to someone who stumbled upon this problem few months from now and found it was still unsolved by the forum.

[unreality]

This may be out of place, so any moderator/admin can delete it as they please: Prime, spoiler tags are there for a reason If you don't want the answer, don't open it ;D Assuming bn used spoilers of course, if he didn't then it's his fault

[Yoruichi-san]

Thanks everyone for the interest!

And Bonanova demonstrates his brilliance once again!

Just to add another way of thinking about it:

In Bonanova's frame of reference, i.e., if I am an ant sitting on Bonanova's shoulder, what I see is Prof. Templeton approaching me directly from the front in a straight line and Prime approaching me directly from the side in a straight line, each starting from distance 20 away, so each travels a distance of 20 to get to me.

Now that poses another interesting question: What does Chuck Rampart's motion look like to me?

[Prime]

Thanks everyone for the interest!

And Bonanova demonstrates his brilliance once again!

Just to add another way of thinking about it:

In Bonanova's frame of reference, i.e., if I am an ant sitting on Bonanova's shoulder, what I see is Prof. Templeton approaching me directly from the front in a straight line and Prime approaching me directly from the side in a straight line, each starting from distance 20 away, so each travels a distance of 20 to get to me.

Now that poses another interesting question: What does Chuck Rampart's motion look like to me?

I can't resist: "Your view is blocked by Bonanova's head."

Or did you mean CR approaches you from behind?