Puzzle... killing my brain cells one by one.

[FenriR]

Leonard left at exactly 11:45:44am on Friday and walked in a straight line without stopping.
At exactly 6:53:59pm on Saturday evening he made a 90 degree turn.
He walked this direction in a straight line before stopping his walk at exactly 11:51:30pm on Sunday night.
Where is he?

Been trying to work out the answer for a week now

Don't have any indication of speed... don't know where he's 'leaving' from... it must be something to do with the context of the name. I've tried calculating where he'd be if he was walking at about 3.5mph (average walk speed) on a flat plane, but that gives an answer of about 240km away from his original position at a bearing of about 43 degrees.

However, apparently thats not right. But the answer is not presenting itself.
Help? Please?

 

[cookiecutter]

My guess is that it has something to do with timezones

 

[FenriR]

It could be. But theres no indication of what timezone Leonard starts in, or how far he'd have to travel to change timezones (not very far at the poles... quite a distance at the equator)
I've been trying to work out the context all day. I spent an hour looking at lion migrations, thats how puzzled I was.

 

[AchrisK]

Probably something like "lost" or "home". Since there is no indication of speed or location, or even the direction of the turn (it could have been a right or left turn), there is no way to nail it down to a specific place. Assuming his speed is constant, your geometry answer is the only possible non-trick answer I can think of.

 

[The Ozzman]

He crossed the international dateline?

 

[MasterOLightning]

I'm thinking the key might be in the words "straight line." Where can you walk straight for 30 hours, turn, and then walk straight for another 30 or so hours?

 

[cookiecutter]

Pretty much any plain or desert area.

 

[waif]

My math class now seems easy...

they give you exact seconds. Since if you need to find where he is, the exact seconds can't be significant unless you have his exact speed, that indicates to me that it's a trick.

 

[Valerie]

Leonard left at exactly 11:45:44am on Friday and walked in a straight line without stopping.
At exactly 6:53:59pm on Saturday evening he made a 90 degree turn.
He walked this direction in a straight line before stopping his walk at exactly 11:51:30pm on Sunday night.
Where is he?

60 hours, 5 minutes and 46 seconds away from where he started.


edit: Fixed the time.

 

[Ender Rises]

Hey... that's right. I think that's the answer.

 

[supseudonymous]

Without actually solving the problem I would expect the time traveled before the turn and time traveled after to be part of a Pythagorean triple. The answer would be given in walking time directly from start point to end point.

 

[Belligerent]

Fuck Leonard.

 

[Ender Rises]

Without actually solving the problem I would expect the time traveled before the turn and time traveled after to be part of a Pythagorean triple. The answer would be given in walking time directly from start point to end point.

Nah. I think the trick is the time thing, since technically that is traveling, since time is a dimension.

 

[Valerie]

Without actually solving the problem I would expect the time traveled before the turn and time traveled after to be part of a Pythagorean triple. The answer would be given in walking time directly from start point to end point.

So that's basically 42 and a half hours in a direct line from start to finish. Yep, I actually took the time to do that. I hope I'm right

 

[supseudonymous]

Err...no.

The tricky part, as I see it, is whether to consider the 6:53:59 turn (that second exactly) to be part of line AB, BC, or exclusive of the two. No matter, though, it does not work out to be a Pythagorean. If you consider it exclusive the answer is AC approximately equals 34:10:10.

 

[Valerie]

Err...no.

The tricky part, as I see it, is whether to consider the 6:53:59 turn (that second exactly) to be part of line AB, BC, or exclusive of the two. No matter, though, it does not work out to be a Pythagorean. If you consider it exclusive the answer is AC approximately equals 34:10:10.

It says he starts at a point (A) then travels a certain time (~31 hours), then turns at a 90 degree angle at another point (B) and travels for about another 29 hours. I devided the minutes and seconds by 60 to get a percentage of 100 so I could work it out. A squared plus B squared equals C squared, right? Yes, so AC equals 42.5089.

 

[V.V.V.V.V.]

This gives a whole new meaning to the term "math metal."

 

[supseudonymous]

Find the total time for AB, and for BC. Convert each time into only seconds. AB^2 + BC^2 = AC^2. Convert back into h:m:s. Like I said, I didn't consider the exact second the turn occurred to be part of AB or BC.

 

[Valerie]

Find the total time for AB, and for BC. Convert each time into only seconds. AB^2 + BC^2 = AC^2. Convert back into h:m:s. Like I said, I didn't consider the exact second the turn occurred to be part of AB or BC.

I converted each to hours. AB = 31:08:15 = 31.1225 hours. Then I did do that operation and converted it back. It seems to me to be really straightforward but you're making it hard

42:31:17

 

[supseudonymous]

Well now I really feel like a fucking idiot. I started at 11:45:44pm instead of am. Let me work this out with the right information...

41:47:42



In seconds:
C^2 = 108494^2 + 104250^2
C = 150462.655

I think you're missing the mark by converting time upward rather than down.