Running Paths
Alice and Bob are standing at the point marked Start on the perimeter of a park (the large, outer square). The park also has a smaller, inner square formed by joining the midpoints of its sides.
Bob starts running around the inner square and, at the same time, Alice starts running around the outer square. Both run at the same speed.
If they keep running, will they ever be in the exact same location again at the same time?
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11 solutions
I think the question is poorly worded or at best the parameters are missing or not clear. In a real world situation if they both kept moving at the same speed along their routes and they never vary they would eventually wind up in the same location at the same time as they both are in there own never ending loop. The only way for them to never meet would be for the loops and or routes to vary or to limit the number of laps/loops. This works in your math but only because you are not accounting that they are in a loop. Unless I missed something it was not accounted for and would make this problem a bit more complicated. Unless I am missing something in my thinking....please correct me if I am wrong.
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So first, if you think they will end up in the same place, you should prove how. Sentences does not help, otherwise one can just say "I can fly" and then fly away.
Second, it is not possible to find an integer that multiplied by square root of 2 gives an integer back, so the two person won't meet. It is not difficult to imagine two loops that, even with the same speed (within the loop), will keep missing each other.
The question IS poorly worded, and ignores mass. Assuming that two masses could actually overlap at the start, then following the rules of the puzzle, the masses will eventually overlap again, occupying the same space.
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One cannot assume that "the masses will overlap again eventually" just from the fact that they were overlapping once at the start. In this case, they will never overlap again because the ratio of the inner and outer perimeters is irrational.
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@Pranshu Gaba – It's not an assumption. The formulas that have been provided prove it. The formulas people are using only work for an infinitely small point - which Alice and Bob are not.
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@Heidi Cortelyou – Here's proof. Let's assume that that Alicia and Bob are 2 feet wide, and the path they are running is 1" wide. Alice and Bob will constantly overlap.
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@Heidi Cortelyou – Most of the time in Mathematics (or Physics), we simplify the problem.
For example in Physics, a ball is 10 meters away from the wall and is heading towards it in 1m/s. When will the ball hit the wall? Normally the problem assumes the ball is infinitely small. Even the formula we used made that assumption! We have v = t s instead of v = t s − r where r is the radius of the ball. This same scenario can be applied to this problem.
Hope this helps!
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@Christopher Boo – False. S does not vary with the radius of the ball. If the ball travels 10 meters, it doesn't matter what the radius is. However, collision simulations DO rely on the size of the objects. That is the nature of this problem. Size matters.
Please explain how assuming that everything is infinitely small would make this simulation work: https://www.mathworks.com/videos/modeling-contact-forces-in-a-geneva-drive-94292.html
I agree I might be wrong with my math but if they travel as the same speed across the distance shown. Say each side of the large square it 100'-0". Midpoint to midpoint equal 50'-0". So the large square would have a perimeter of 400'-0" and the smaller rotate square would have a perimeter of 200'-0". If Bob were to run two laps at the end of his second lap Bob and Alice would be at the same location when Alice completed her first lap. So Bob and Alice will ALWAYS be at the exact same location every time Alice finishes her lap so long as Bob does keep running.
Question should be reworded to state "If they both run a single lap around the squares...." or " How often would the Bob and Alice arrive in the exact same place if they both run....."
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This is not correct. If the side of the square is 100, then by pythagorean theorem , the distance from midpoint to midpoint would be 5 0 2 , which is about 70.71 and is an irrational number. Since the ratio of number of laps of Alice to number of laps of Bob is never an integer, they will never meet.
I looked at this a different way to avoid the issue of triangles. If their paths were nesting circles of the same circumferences as the two squares, they would be like two planets orbiting the sun at different distances, with points every 90 degrees corresponding to the four intersections of the squares. With the same velocities, the inner circle will be completed faster, as it is the shorter path, and with unlimited repetitions it will catch up to the longer path every so often. If at any time the two runners are at the same 90 degree points in their "orbits"mentioned earlier, then they will coincide in the same way as the two square diagram.
In an infinite loop each party continues around and around. Is there not a common denominator that allows them to be at the same point at the same time eventually. If the numbers equal out to a decimal eventually the Paths have to cross
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No, as the solution shows, there is no common denominator because 2 is an irrational number. This means that they can never be at the exact same point at the same time.
@Marta Reece - I don't think they have to be whole numbers ... they could meet up at any of four points around the lap.
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Sorry... Ignore this comment! I see now that each "1" is a corner of the field.
The only (mathematical) solution for them to meet would be a=0, so if the size of the park was indefinitely small. On the other hand, running on such a park was no fun... ;)
Where is the root 2 coming from?
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In an isosceles triangle, if one of the legs is a, then the hypotenuse is (a)sqrt(2). (That's just applying the Pythagorean Theorem: a^2 + a^2 = c^2 => 2a^2 = c^2 => sqrt(2) * a = c.)
At some point, is there not a common denominator where Bob's distance is an exact multiple of Alice's which would then put them both at the start position at the same time?
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There is not a common denominator as it turns out - even after, say, a trillion trillion laps the runners would never be back at exactly the starting point at the same time (close though!). To understand this, notice that there are 8 segments on the outer square - to keep things simple, assume that each of those 8 segments is each just 1 meter (the outer runner will run exactly 8 meters). You might use Microsoft Excel's square root function (SQRT) to get the side lengths of the inner square: SQRT((1x1)+(1x1)) = SQRT(2) which is about 141.4214...the runner on the inner square therefore runs about 5.656854 meters. But the issue is that the square root of 2 (roughly 1.4142...) can't be expressed as a fraction as it is, like pi, an irrational number. If the square root of 2 were, say, exactly 99/70, then they would eventually meet. http://mathforum.org/dr.math/faq/faq.integers.html
This is the correct solution (only correct one I see), though a lot of the confusion in the comments could be eliminated if you expand on it a little bit. For example, explain why
xan=yam/sqrt(2)
(the
n
and
m
are needed because Bob will lap Alice) is only valid at the 4 intersecting points (thus n and m must be integers). Also, it would help to explain why irrationality is important.
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I don't know if Marta is going to do any edits, but that sounds lovely: you're welcome to write your own solution which includes that!
The starting conditions for the puzzle are impossible. Alice and Bob have mass, and can not exist in the same place at the same time. Assuming that they do anyway, their masses WILL OVERLAP eventually - occupying the same space. The impossible starting condition will recur.
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Okay... but that is not really the point of the problem. I am assuming Alice and Bob are point objects.
Wow. Everybody's really over thinking this. They can never meet at the same place at the same time because it's going to be a later time.
why does x and y have to be whole numbers?
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They are the number of times they went around, so yes, they have to be whole numbers.
It says "If they keep running, will they ever be in the exact same location again?" it does not specify "start position", and SPECIFICALLY does not limit the number of laps.
YES, they will eventually lap each other "if they keep running"
Now, if the scope of the problem includes physical location within the cosmos, then NO, there is no possible way, even standing still, because the universe is expanding and also constantly in motion.
Relevant wiki: Pythagorean Theorem
Let's say from the START point, both Alice and Bob started with a same velocity v.
Let's assume the side of outer square park is 2a. Therefore half of its length will be a.
According to Pythagoras Theorem, side of inner square park will be a√2.
Let's say Alice will take a time t1 to reach the start position again and Bob took time t2.
Since their velocities are same , therefore
8a/t1 = 4√2a/t2.
On solving we get. t1 =√2 t2.
There is no value possible for t1 and t2 (except 0) which makes the above expression true.
So, Alice and Bob will never meet each other at the START point.
Very poorly worded. Alice and Bob are not at exactly the same point when they start. If exactly means side by side, then the definition is inexact enough to accommodate the small difference necessary to overcome irrationality, which requires super-position. In fact, finding the point in time when they will be in the same relative position as they are at the start is trivial. In addition, since there is no time limit on how long they will run, they may run for an infinite period, and will be in the same relative position as they are at the start an infinite number of times. Sloppy writing and sloppy thinking.
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If you are unable to understand what the question wants, then who could be blamed??? But if you have sense enough to understand small things, then first of all read the very first line of the question and look at the directions of their velocities. You will get your answer. "I think your eyes must have been SLOPPY while reading the question." No problem. This time just read the question carefully.
they will meet at infinity
The question was:"will they ever be in the exact same location again at the same time", which could be at any of the 4 intersection points..... the question did not specify the "Start Point".
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It is not possible for them to meet at any of the other three points either. The proof of this works on similar lines as this solution because the ratio of the side lengths of the outer park to side length of inner park is still irrational.
They both will never meet again at same time because both will to travel different distances but with same velocity.
look carefully at every corners. there has been created a right triangle.
from the basic law of triangle we know that,
[ summation of any of the 2 sides of a triangle is greater than the other side .]
so, in every corner Alice has to go more distance than Bob.as their speed is same, so Bob will simply cross Alice & if they keep running they will never be in the exact same location.
@Mohammad Khaza - that doesn't hold. E.g. if the outer path was length 10 and inner was length 9 then after Bob had done 10x laps he would be in the same place as Alice (she would have done 9 longer laps, but also be on the starting point)
your solution is wrong --
the answer is yes they will both be at the start at the same time --
if i use 2.8 as the square root to 8, --
they will be both at the start after ---
bob has run 160 laps and --
alice has run 112 laps --
to the objection that i have rounded off square root of two, --
the logic is the same for any number of decimal places --
of course what this means is that bob and alice will have to run an infinite ---number of laps to avoid the rounding error --
that means they will meet at the end of time --
kind of romantic
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You assume the square root of 8 is 2.8 which is a rational number, in reality square root of 8 is irrational. So one can say that they will kinda be in the same spot at the same time again, but not precisely at the same spot at the same time.
Besides comments below, question is not clear. Will they ever be at the same point is not the same as will the both ever be at their starting point
Would you say the same thing if they were on crossing circular paths? Of radius, say, 10 and 20. Without the irrationality of root 2, as in Marta's solution, I don't think it's enough to say one path is longer than another.
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i just thought with the given problem. but if there is circular paths or any other paths the answer will be same because if the inner part is half inch less than[suppose] the outer part,then with the same speed the person inside is always a bit ahead than the person outside. Marta just calculated the hypotenuse of the triangle.
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Let's take your circle example. The outer circle is X inches long, the inner circle is X-1 inches long. If both runners run at the same speed then it will be the case that: After the outer runner has run 1 lap the inner runner will be 1 inch ahead,
After the outer runner has run 2 laps the inner runner will be 2 inches ahead,
.......................................
After the outer runner has run X-1 laps the inner runner will be X-1 inches ahead.
The inner circle was only X-1 inches long, so being X-1 inches ahead just means he is exactly one lap ahead. No more no less. The inner and outer runner is at the same position.
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@Kim Rylund – that would have been a super argument but unfortunately not.did you notice what is asked from the question?will they meet at the starting point again?
however ,if you have proper logic now,please tell me as i have also a little doubt with this question.
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@Mohammad Khaza – The question was actually if they ever would be in the exact same location again -> will they ever meet :P
Actually, it would be enough to figure out if they ever both for a time t is at a point where their 2 paths meet since that automatically would make them be in the exact same position at time = 4t.
The reason they never meet is that you can't choose an integer x for the formula x sqrt(2) such that the formula returns another integer. x sqrt(2) would be all the moments where Bob is at one of the meeting points (where the 2 paths meet), but Alice would only ever be at one of these points at an even time. (Hope this makes sense, my English is a little rusty :P)
But if instead of running the hypotenuse of a (1,1,sqrt(2)) triangle Bob were running the hypotenuse of a (3,4,5) triangle while Alice were running the outside boundary, there would be a point at which they would be in the same position. This is clear because Bob would run 7 circuits of the inner square while Alice ran 5 circuits of the other square. (Diagram left to the reader.) You have to use the geometry of this figure and the fact that sqrt(2) is an irrational number.
If they can make more than ONE lap. they WILL meet at a common point.
I heard they wound up in bed together with Ted and Carol. seriously, if they run a large number of laps Bob will "lap" or overtake Alice numerous times but for him to be passing her at exact the point where they both started would take an inordinate amount of time and unless Bob and Alice are from the Hopi Indian tribe they will be exhausted long before that would happen.
Bob's distance around the box differs by Alice's by a factor of sqrt(2), which is an irrational number and therefore, while they will get close to contacting each other if they keep running for infinity, they will never overlap exactly.
To solve this first arbitrarily assign the distance Alice traverses from Start to the first corner a value of 2 miles (it's a big park). And then from the first corner to the first midpoint is another 2 miles. The distance Bob traverses from Start to the first midpoint is then the hypotenuse of a right triangle described by the equation 2^2 + 2^2 = C^2, which, solving for C, i.e. Bob's distance from Start to first midpoint, gives a value of sqrt(8), or 2sqrt(2), since he travels this distance 4 times to get back where he started, the total distance he travels is 8sqrt(2) or 11.31 miles, whereas Alice has to travel the 2 miles 8 times, which is 16 miles total (the only difference between these numbers is the factor of sqrt(2)). Bob travels less distance than Alice and since they are traveling at the same speed, it takes less time for Bob to complete one lap around his square.
However, the question asks will they ever be in the exact location again, which I'm taking to mean, will they ever overlap exactly if they keep going forever? If they were on a circle, yes you could figure out where they would overlap, but there are only 4 overlap points in this problem which are whole number fractions of Alice's total distance, 16 miles. They are the 4 mile point (one quarter point), the 8 mile point (halfway point) and the 12 mile point (three quarter point), and the 16 mile point. Yet, Bob is traveling a distance that is different from Alice by a factor of sqrt(2) which is an irrational number and therefore there will never by an exact location where they will meet up. Yes, he laps her and yes, it does get close some times, for example, if they are both traveling a 1 mile per hour, Bob passes the halfway point (opposite the start) at 39.6 hours while on his 4th lap, where as Alice is in the vicinity on her 3rd lap and slightly ahead of him by 0.4 miles. There are also a few more run ins, when Bob is on his 7th lap at the 3/4 point he would be slightly ahead of Alice by 0.36 miles while she is on 5th lap. They get close but they never overlap exactly.
If you use an arbitrary number for the side of the box and an arbitrary rate for their speeds and then figure out what times they will be at each of the four meeting points using a spreadsheet you will see that they never meet up. An interesting follow up to this question is given the rate of 1 mile per hour and the fact that it takes Alice 16 hours to complete the journey what is the closest they will ever meet up? That I have no idea how to solve.
it should be precise that their track has no width otherwise it is possible. So it is only true in a pure mathematical univers :-)
And Alice and Bob have no mass...
The solution is obviously posted many times below! And has to do with the distances each travel per second. but imagine the ramifications of many nested square shapes. Perhaps train track style vehicles all traveling st the same set speed could transfer across mutual tracks seamlessly. Very cool problem.
The problem assumes that Alice and Bob are infinitely small. How many trains are built like that?
For ease of calculation, assume that the length of one side of the outer square is 2 units, and that Bob and Alice both run at a rate of 1 unit/interval. Pythagorean theorem tells us that the length of one side of the inner square must be 2 . We know that at this speed, Bob will reach any given corner of the inner square at n ∗ 2 where n is the number of corners that Bob has already visited, while Alice will reach any given corner of the inner square at n ∗ 2 . Since 2 is irrational, every time Bob reaches a corner will be an irrational number, while the times when Alice reaches an inner square corner will all be even integers, so there is no way for Alice to reach one of the four common points at exactly the same time as Bob.
A more interesting analysis might be to find the position and point in time when Bob and Alice are closest to each other. I'm not sure how to find it mathematically, but I did a quick spreadsheet model of their positions and times and discovered that if Bob and Alice continue running for 1000 laps around their respective squares each, there will come a moment when Bob has run 348.25 laps and Alice has run 246.25 laps where they are less than 0.001 units away from one another... which is pretty close.
Anyone know how to formalize my analysis to find the smallest possible distance between Bob and Alice given an infinite number of laps each?
If a is the side of the outer square, the side of inner square is a/√2. Then perimeter of outer is 4a is not equal to 4a/√2. So answer is NO
No. If they did it would be equivalent to proving that the square root of 2 was a rational number.
For A and B to meet, it has to be at one of the midpoints of the square sides.
Without loss of generality we can assume that a half square side measures 1 and that A and B move at unit speed (1 distance unit per 1 time unit).
A travels 2 between possible meeting points hitting them at times 2m. B travels √2 between possible meeting points hitting them at times n√2. Where m and n are positive integers.
So for them to meet there must be a common time when both are at a meeting point. However, it is impossible for them to even arrive at any two midpoints of the square at the same time, let alone the same one.
A common midpoint time would satisfy:
2m = n√2 or √2 = n/m. That would mean √2 is a rational number, a contradiction.
So A and B cannot be exact same point again.
Since there only 4 points where the 2 squares meet, there 4 possible encounters between Alice and Bob. However, Alice runs (\frac{2x}{\sqrt{2}}) of Bob's distance. Therefore, Alice, having to run more than Bob at the same rate, will never meet Bob.
if they meet each other in the future time t with the same speed v, the total distances of A and B will be the same (sA=sB); sA must be an integer and sB must be an irrational number， it is impossible，so they will not see each other anymore
The problem is incorrectly worded. The question was whether they would ever be in the exact same location again. The location was not specified. The wording should have been: Would they ever be at the START at the exact same time ever again.
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Good idea. I've fixed the question. You've got a good eye!
Of course they will meet each time Bob passes Alice. No where in the question is it stated that the answer must be an integer. The answer could just as well be an irrational number.
Relevant wiki: Irrational Numbers
If the side of the outer square is a , then the side of the inner one is 2 a × 2 = 2 a . There is no way x a = y 2 a for any whole numbers x and y since that would require x = 2 y and 2 is irrational.