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Wizard
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Wizard
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Thanks for this post from:
MrCasinoGames
April 4th, 2020 at 10:20:30 AM permalink
This one I got from 538.com. In fact, it is their current Riddler Express of the week.

Quote: 538.com

Youíre walking along the middle of a wide sidewalk when you see someone walking toward you from the other direction, also down the middle of the sidewalk, 12 feet away. Being responsible citizens, you pass each other while maintaining a distance of at least 6 feet at all times. By the time you reach each otherís original positions, you should be back in the middle of the sidewalk again.

You should assume that the other person follows the same path you do, but flipped around (since theyíre walking in the opposite direction). For example, you could both walk 3 feet to the left, 12 feet forward and finally 3 feet back to the right, walking a total of 18 feet before swapping positions.

Being lazy (I mean, efficient), youíd like to know the shortest distance you and the other person could walk so that you can switch positions, all while staying at least 6 feet apart at all times. What is this distance?

Extra credit: Now suppose the person walking toward you has no intention of straying from the center of the sidewalk (sigh), and itís entirely up to you to maintain a distance of at least 6 feet. In this case, what is the shortest distance you would have to walk to reach the other personís original position?



538 doesn't mention the speed of the walkers, but for purposes of discussion here, let's assume both walk at the same speed.

So far, I have an answer to part 1 only. I'll put my answer in spoiler tags below.

I think this problem is too hard for the "easy puzzle" thread. However, I'm going to not apply the usual beer club rules. We're all in the same boat on this one.

6*sqrt(3) + pi = apx. 13.5339
Last edited by: Wizard on Apr 4, 2020
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ChesterDog
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April 4th, 2020 at 11:52:56 AM permalink

Pi + 6 * (square root of 3), or about 13.53 feet.
ThatDonGuy
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April 4th, 2020 at 11:59:06 AM permalink
I get the same answer as you two for #1.

The problem with #2 is, it's a "moving target."

My first thought was, find the distance d that the other person will walk so you will be on the tangent of the 6-foot circle with the other person as the center when you also walk distance d - but you can't then simply "walk around the circle" as the circle is moving.

Ace2
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April 4th, 2020 at 12:45:23 PM permalink

(45)^.5 * 2 = 13.416 feet . Which is wrong
Last edited by: Ace2 on Apr 4, 2020
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charliepatrick
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April 4th, 2020 at 1:31:47 PM permalink
Imagine a circle at the half way point radius 3'. The aim is to get to that circle, arrive tangentially, both go round the circle, and leave tangentially. It transpires if you imagine an equaliterial trangle with one side along the centre line, then chop it in half, the chop line meets the circle (as half the base is 3'). You start along the chop line and when you get to the base of the triangle are 3' from the half way point. The distance along this is SQRT (6^2-3^3 =) SQRT(27). Since the angle of the triangle to the path is 60o, you go round 60o of the circle before getting back to the triangle on the other side. Thus the distance on the circle is 2 Pi r / 6 = Pi. Then the walk back is SQRT(27) = 3 SQRT(3) again. Sum = 6 SQRT(3)+Pi.

If you don't have to go the same speed you'd rush to a half way point 6' from the path, wait for the other person to go by, and continue the second half. This gives 2 * SQRT(6^6+6^6)= 12 SQRT(2) = 16.971.

If you have to go the same speed then you could aim for a point 6' away from the path. At the same time the other person walks along the path. It transpires (think 3-4-5 triangle) if the other person walks 7.5', you can walk (4.5^2+6^6)=7.5'. However the issue is whether the two of you ever get closer than 6' (which you do!). Shame really as it was nice to have a 3-4-5 triangle in it.

My feeling is you aim along a line then an arc of some kind to a point 6' from the path. While you do that, the other person has got beyond the half way point, so the answer isn't symetrical. Also as the other person will have done your distance, this is greater than 6' are also the sum of a line and the length of an arc.
Wizard
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April 4th, 2020 at 5:23:47 PM permalink
The puzzle doesn't say, but we all assumed the walkers go at the same speed and are thus in agreement on problem 1.

I've been thinking about the second part for a couple hours now. It is going to be tricky.


Let's say the horizontal movement is straight ahead along the sidewalk, as if the other guy wasn't there.

I think you will start out walking in either direction 60 degrees away from straight ahead. After walking 6 feet, you'll be 6 feet away from the rude guy walking straight ahead.

Then it gets tricky. I think you'll take some kind of an altered cycloid path -- getting pushed back horizontally and up vertically. I think at some point you'll be 6 feet vertically from the straight walker. Then continue on this curved path a little until you can leave it along its tangent line and head straight to the point 12 feet directly ahead of where you started.

I think this is going to involve parametric equations and arc length. Here is where the air starts to get thin for me.

I tend to think I won't be the first to solve this one, knowing the brain power here. Whoever solves it first, please give a good solution.


p.s. Choice 8 in the poll should have read, "I hope the quarantine is over by April 8 2024.
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
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April 4th, 2020 at 5:38:45 PM permalink
For Part 1, I got the same answer.

But I think you should mentioned that both person walk with same speed ? Am I missing something ?

For part 2, Assuming both walk with same speed, it involve arc length, min distance = 18.1633 . . .should be very close but not 100% accurate. LOL
Last edited by: ssho88 on Apr 5, 2020
charliepatrick
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CrystalMath
April 5th, 2020 at 7:13:26 AM permalink
Quote: ssho88

...For part 2, Assuming both walk with same speed, it involve arc length, min distance = 18.1633 . . .should be very close but not 100% accurate. LOL

My initial ideas, not fully proved, get an upper bound of about 18.125 using the following logic.

The Walker goes from A to B along the blue path.
You follow the red dots and at some stage have to cross 6 feet apart (shown by the right hand gold line).

Part 1: PQ PQC is an equilateral triangle. If you go to point Q, then you will have walked 6 feet and then be 6 feet from the Walker.
Part 2 : QR If you continue on that path to R, then you're going outside what you have to (Note: I haven't proved this so am happy to be corrected, in which case it's likely to be similar to the 18.163 above.)
Part 3 : RS Once you're on the six foot path, go along it until you're at the same height as the Walker (D S).
In theory I think you follow an arc (the blue one is an elipse with radii 3 and 6). Also you should have aimed for a point to meet it at right angles, therefore Q might be further up.
Parts 4 and 5 are nearly symettric since you start out by following the same path as in part 3 and part 2. (In theory you follow the arc in identical manner until you're travelling at right angles to the walker...
Part 6 : UV ...then head straight home.

Each part has a number, typically involving square roots (e.g. UV = SQRT(75) ), but as this is all an approximation.

I think the puzzle involves somehow proving an elipse is the path needed and by weird magic working out where the perpendiculars meet the curve.
ssho88
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April 5th, 2020 at 7:27:32 AM permalink
Quote: charliepatrick

My initial ideas, not fully proved, get an upper bound of about 18.125 using the following logic.

The Walker goes from A to B along the blue path.
You follow the red dots and at some stage have to cross 6 feet apart (shown by the right hand gold line).

Part 1: PQ PQC is an equilateral triangle. If you go to point Q, then you will have walked 6 feet and then be 6 feet from the Walker.
Part 2 : QR If you continue on that path to R, then you're going outside what you have to (Note: I haven't proved this so am happy to be corrected, in which case it's likely to be similar to the 18.163 above.)
Part 3 : RS Once you're on the six foot path, go along it until you're at the same height as the Walker (D S).
In theory I think you follow an arc (the blue one is an elipse with radii 3 and 6). Also you should have aimed for a point to meet it at right angles, therefore Q might be further up.
Parts 4 and 5 are nearly symettric since you start out by following the same path as in part 3 and part 2. (In theory you follow the arc in identical manner until you're travelling at right angles to the walker...
Part 6 : UV ...then head straight home.

Each part has a number, typically involving square roots (e.g. UV = SQRT(75) ), but as this is all an approximation.

I think the puzzle involves somehow proving an elipse is the path needed and by weird magic working out where the perpendiculars meet the curve.




Similar idea ! I use 3 known points to form quadratic equation . . .then use integration to calculate the arc length . . .BUT STILL AN ESTIMATION.
unJon
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April 5th, 2020 at 8:28:31 AM permalink
Nevermind
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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