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185,393 Views • Jan 27, 2025 • Click to toggle off description
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196 Comments
Top Comments of this video!! :3
That is math. It might not be formalized. It might not use as many numbers as you're used to, but that is math. Math is an extension of logic.
394 |
Draw two lines that cross each other. Prove that at some point in the line that they cross each other.
839 |
The intuitive solution is the same as the calculus solution. But the thing is you need calculus to know that the path is continuous, to know that somehow hiker A didn't instantaneously teleport from infront of hiker B to behind hiker B (thus somehow not crossing). Like "we know" that can't happen with a mountain in this set up but it's not obvious also.
Also p.s. if this was a real 3d mountain this only works if we they know they took the EXACT SAME PATH, if the up journey and down journey went on different paths up the mountain it becomes clearly false since they didn't even necessarily pass through the same points...
336 |
...but that doesn't give you an answer.
It just tells you that there is one.
30 |
The people in the comment section did not pass the vibe test. I loved the intuitive solution! Thanks so much for sharing, I had never thought about it that way.
192 |
It's just setting up the problem. When you go up the mountain it takes you 10 hours coming back down it only takes 8 hours. What you will be able to find out is at a certain amount of time the two hikers will meet somewhere on the hike. That is represented as X. When they do meet they would have each done enough of the hike that together it would equal the whole hike done. The setup equation would look like this:
X/8 + X/10 = 1;
5X/40 + 4X/40 =1;
9X/40 = 1;
9X = 40;
X = 40/9
Since they are both starting out at 8:00 a.m. they should meet 4h 26m 40s into the hike. At 12:26:40 p.m.
19 |
1. walk away from the mountain
2. Return at 5pm
3. Climb the mountain in an hour
4. Do not return to where you walked to in step 1 on your way down
And thus at no point were you at the same location at the same time.
7 |
That's funny, my immediate thought was the IVT, but it wasn't obvious to me how to prove it without it haha. For fun, here's a formal proof:
Let [t1, t2] be the overlapping time interval between ascent and descent. Define UP(t) and DOWN(t) as continuous functions mapping from [t1, t2] to [x0, x1], representing position during ascent and descent respectively.
Given UP(t1) < UP(t2) and DOWN(t1) > DOWN(t2), let f(t) = UP(t) - DOWN(t). Since f is continuous and f(t1) < 0 while f(t2) > 0, by the Intermediate Value Theorem, there exists t' in [t1, t2] where f(t') = 0.
Therefore, UP(t') = DOWN(t'), i.e., the hiker passes some point at exactly the same time on both days. QED.
142 |
My dumbass read : puzzle :hi'ler😂
30 |
I think your 'unmathematical' explanation and using the intermediate value theorem really is the same thing, just IVT formulates it rigorously in a mathematical context.
It's like saying a straight line on the xy plane that has a positive y value and a negative one must cross the x-axis because the line doesn't break and following along with my pen, it crosses the x-axis. Then claiming you didn't need calculus to prove that it did so.
101 |
This seem like type of question they would ask in physics
9 |
A second hiker who starts walking from the campsite at 8am will at some point encounter the first hiker who is walking to the campsite. Phrased in this equivalent way, the result becomes obvious.
1 |
A more complex version of this puzzle is one that proves that there are always two points on opposite ends of the planet that both have the exact same temperature and air pressure. The principle is the same as it is with this puzzle. The only differences are that the version here looks at one variable (the point on the mountain where the hiker is at the same spot at the same time of day) and deals with a 1-dimensional object (the path up and down the mountain) whereas the other version looks at two variables (points on opposing sides of the planet that have the same temperature and points on opposing sides of the planet that have the same air pressure) and deals with a 2-dimensional object (the Earth's surface). Vsauce explained it in his Fixed Points video.
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If f(t) is the function giving the distance travelled during the forward stretch of the hike on [0,s], and g(t) the distance travelled during the backward stretch of the hike on [0,s’], then 0=g(0)-f(s)>g(s)-f(s), and g(0)-f(0)>g(s’)-f(0)=0, and so using the IVT the function g(t)-f(t) has a zero (under continuity assumption) which implies that there is an intersection point between g and f.
1 |
Mirror the downhill graph.
Overlay the two graphs.
Look where lines cross.
|
I'll try to solve this as a high schooler with almost no knowledge of calculus:
So the 2 hikers cover the same distance but not at the same time.
Hiker a is a function that is equals to same distance as hiker b
Hiker b is another function that is equal to the same distance but different time
We both know that both start at 0 and end at a point at the top. For this example, we'll use y = 10 for the top.
Now let's calculate the slope of each hiker.
Assume that each hiker's journey is a straight line and their speed is the same, but the slope isn't which makes hiker b faster.
So when ha is plugged with the value 10, it is equals to the value: distance.
So: Ha(10) = 10
So it means that Ha = x
Now Hb(8) = 10
Let's use the slope formula on this
So x1, y1 = (0, 0)
And x2, y2 = (8, 10)
Which gives us 5/4 which is the slope.
Now the question is to find the time when they intersect each other.
To do that, all we need to do is just flip Hb and add a y-int of the starting point which is 10.
So now we have these functions:
Ha = x
Hb = -5/4x +10
Now all we need to do is just find the time they intersect when they are both at the same value.
Now we have x = (-5/4)x + 10
We just evaluate it.
(9/4)x = 10
9x = 40
x = 40/9
So now we know what time they intersect, assuming that they are both straight lines and no randomness included.
The answer with those conditions is 40/9 hours or 4.44 hours
5 |
I fell down the mountain. So i was before the start of the mountain at 8:01. It took me 6 more hours to crawl a bit further.
2 |
This is the same solution to the “theorem of meteorology” but with time and position rather than temperature and position. Very cool!
1 |
"doesn't require math to solve"
proceed to not do math yes but didn't solve the puzzle either XD
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@skdjrhaejdhst-cat
6 days ago
What it said: Hiker
What I read: 🇦🇹
2.2K |