Rethinking...
How big is this ant?
Is it a male or female ant?
Does he/she start with the tip of his nose at point A or with its back?
Do ants have noses? If not, what else is in front?
How efficient do ants walk?
What if the ant refuses to walk?
What if the ant is drunk? Which side will he/she drop off the brick?
questions, questions, questions...
P.S. Who left that brick there anyway?
A non-trivial problem
37 posts
• Page 2 of 3 • 1, 2, 3
Non trivial question
by Cec » Mon May 01, 2006 5:58 am
Ruud wrote:P.S. Who left that brick there anyway?
This has been interesting and topped up by Ruud's above quite funny comments - it's distracting me from working out my "code button" homework in MM(2)
If it reaches a stalemate I could always throw in a good golf joke!Cec
- Cec
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by udosuk » Mon May 01, 2006 7:02 am
Tarek, the answers in all your replies have left me in a headache because they're all different to mine (except for Q2)...
Let's first sum up the questions:
1. Locate the point on the brick which takes the ant the longest travelling distance from corner A (assuming the ant is intelligent enough to take the shortest possible route - I know Ruud would have some more questions to ask about this
)... And calculate this travelling distance... The ant could go under the brick if it likes...
2. Similar to (1), but the ant cannot go under the brick...
3. Locate 2 points on the brick, that would take the ant the longest travelling distance from one to another... No limitations on ant movement again...
Here are my answers (distances):
1. 11.401754250991379791360490255668
2. 12.292725943057183080003032375684
3. 12.04776943320180962759310746754
Will show the working next time...
PS: Just realised we found the same distance on Q2... But the point I found was 1.33333 units down the opposite corner, or 2.66667 up from the bottom, so what's what?
Let's first sum up the questions:
1. Locate the point on the brick which takes the ant the longest travelling distance from corner A (assuming the ant is intelligent enough to take the shortest possible route - I know Ruud would have some more questions to ask about this
)... And calculate this travelling distance... The ant could go under the brick if it likes...2. Similar to (1), but the ant cannot go under the brick...
3. Locate 2 points on the brick, that would take the ant the longest travelling distance from one to another... No limitations on ant movement again...
Here are my answers (distances):
1. 11.401754250991379791360490255668
2. 12.292725943057183080003032375684
3. 12.04776943320180962759310746754
Will show the working next time...
PS: Just realised we found the same distance on Q2... But the point I found was 1.33333 units down the opposite corner, or 2.66667 up from the bottom, so what's what?
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by tarek » Mon May 01, 2006 7:31 am
tarek wrote:The answer should be 2.33333333 units up the edge of the corner on the same level & opposite (or 1.6666666 down the edge from the corner on the top & opposite...
The distance covered is 12.292725943057183080003032375684 units
That was my final answer however I was intending to say 2.666666666 & 1.333333333 as you've said, I apologise if I made you re-solve things to double check.
previous answers were just (thinking out loud), should I delete them
I have no graphical display to show, but if you unfold the brick in 2D....
one way to unfold it is to keep both squares attched to the side face (rectangle)...the other is to keep the squares attached to the middle rectangle
the shortest way to any paoint on the edge is a straight line (arc)
1st triangle we have 12,x & distance (x distance up from opposite same lavel corner)
2nd triangle we have 12-x,4 & distance........
both distances should be equal.......
Equation........ we get x=2.66666 or precisley 8/3 (so 2.333333 is a typo mistake from my side udosuk, late night & a headache)....
The distance covered would then be 12.292725943057183080003032375684 units
tarek
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tarek - Posts: 3762
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by tarek » Mon May 01, 2006 9:22 am
I think it is slightly easier now, as the method is known....
I've got 12.5 units as the longest distance possible.
Starting point: Ground level, half-point between 2 corners of square
End Point: Opposite square .5 unit down the half point between 2 top corners.
tarek
I've got 12.5 units as the longest distance possible.
Starting point: Ground level, half-point between 2 corners of square
End Point: Opposite square .5 unit down the half point between 2 top corners.
tarek
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tarek - Posts: 3762
- Joined: 05 January 2006
by SimulatedAnnealer » Mon May 01, 2006 11:16 pm
The answer to Q1 is square root of 130.
A = starting vertex
B = farthest point
R = rectangular face including A
Q = rectangular face adjoining R not including A
S = square face not including A
V = vertex not sharing any face with A
W = vertex diagonally opposite V across S
Then arguments by symmetry and contradiction show that B must be on S on the diagonal connecting V and W.
Now the distance from A to B is the smaller of the distances using a route via R then Q then S and using a route via R and S only.
Let the co-ordinates of B on S with respect to origin V be (b,b). (0 <= b <= 4)
Consider the route via RS only and flatten the brick with S adjoining R:
The distance from A to B is the hypotenuse of a right-angled triangle with sides 4-b & 12-b.
Distance via RS = SQRT ((4-b)(4-b) + (12-b)(12-b))
= SQRT (160 - 32*b + 2*b*b)
Differentiating shows this is decreasing in b (in the range 0 <= b <= 4)
Note for b = 0 distance = SQRT(160) and for b = 2, SQRT(104)
Now consider the route via RQS and flatten the brick with S adjoining Q:
The distance from A to B is the hypotenuse of a right-angled triangle with sides 8-b & 8+b.
Distance via RQS = SQRT ((8+b)(8+b) + (8-b)(8-b))
= SQRT (128 + 2*b*b) which is increasing in b
(Actually if b becomes close enough to 4 so that the hypotenuse stops crossing Q then the distance follows a dogleg which is also increasing in b)
Note for b = 0 distance = SQRT(128) and for b = 2, SQRT (136) (since the route is still the hypotenuse at this point)
So the graphs of the 2 distance functions intersect precisely once in [0,2], and this value is the farthest distance.
Solving the the 2 hypotenuse equations simultaneously yields b = 1 and thus AB = SQRT (130) = 11.40175.
A = starting vertex
B = farthest point
R = rectangular face including A
Q = rectangular face adjoining R not including A
S = square face not including A
V = vertex not sharing any face with A
W = vertex diagonally opposite V across S
Then arguments by symmetry and contradiction show that B must be on S on the diagonal connecting V and W.
Now the distance from A to B is the smaller of the distances using a route via R then Q then S and using a route via R and S only.
Let the co-ordinates of B on S with respect to origin V be (b,b). (0 <= b <= 4)
Consider the route via RS only and flatten the brick with S adjoining R:
The distance from A to B is the hypotenuse of a right-angled triangle with sides 4-b & 12-b.
Distance via RS = SQRT ((4-b)(4-b) + (12-b)(12-b))
= SQRT (160 - 32*b + 2*b*b)
Differentiating shows this is decreasing in b (in the range 0 <= b <= 4)
Note for b = 0 distance = SQRT(160) and for b = 2, SQRT(104)
Now consider the route via RQS and flatten the brick with S adjoining Q:
The distance from A to B is the hypotenuse of a right-angled triangle with sides 8-b & 8+b.
Distance via RQS = SQRT ((8+b)(8+b) + (8-b)(8-b))
= SQRT (128 + 2*b*b) which is increasing in b
(Actually if b becomes close enough to 4 so that the hypotenuse stops crossing Q then the distance follows a dogleg which is also increasing in b)
Note for b = 0 distance = SQRT(128) and for b = 2, SQRT (136) (since the route is still the hypotenuse at this point)
So the graphs of the 2 distance functions intersect precisely once in [0,2], and this value is the farthest distance.
Solving the the 2 hypotenuse equations simultaneously yields b = 1 and thus AB = SQRT (130) = 11.40175.
- SimulatedAnnealer
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A Non-trivial problem
by Cec » Tue May 02, 2006 6:53 am
If we can catch this ant I move we preserve her - it has to be a female - what male would attract this interest
Cec
Cec
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by SimulatedAnnealer » Tue May 02, 2006 10:07 am
I solved for Q2 with the brick on the ground using similar methods to Q1.
Using symmetry and proof by contradiction the farthest point, B, must be on the vertical edge through V. There are more distinct potential shortest routes to consider than in Q1 since there is less symmetry.
Given that B is on this vertical edge then there are only 2 distinct algebraic entities for the distance calculations along the various potential routes.
If B is distance b along the edge from V then the squared lengths of the 2 hypotenuses are:
(8)(8) + (8+b)(8+b) increasing in b in [0,4]
(12)(12) + (4-b)(4-b) decreasing in b in [0,4]
No issues with doglegs this time, and the end point values once again mean the graphs intersect once only at the farthest point.
b = 4/3
distance = SQRT (1360/9)
Using symmetry and proof by contradiction the farthest point, B, must be on the vertical edge through V. There are more distinct potential shortest routes to consider than in Q1 since there is less symmetry.
Given that B is on this vertical edge then there are only 2 distinct algebraic entities for the distance calculations along the various potential routes.
If B is distance b along the edge from V then the squared lengths of the 2 hypotenuses are:
(8)(8) + (8+b)(8+b) increasing in b in [0,4]
(12)(12) + (4-b)(4-b) decreasing in b in [0,4]
No issues with doglegs this time, and the end point values once again mean the graphs intersect once only at the farthest point.
b = 4/3
distance = SQRT (1360/9)
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by SimulatedAnnealer » Tue May 02, 2006 12:36 pm
And for Q3 the points are on parallel diagonals on each of the 2 square faces distance at the same distance from the face centres but in opposite directions.
With cartesian axes centred on the square face centres and parallel to the edges, the co-ordinates of the points are [4-SQRT(12), 4-SQRT(12)].
The distance is 8*SQRT(4-SQRT(3)) (as per udosuk)
(I didn't fully check the logic that the points must be on the diagonals the same distance from the face centre but in opposite directions: it looked straightforward but time-consuming.)
With cartesian axes centred on the square face centres and parallel to the edges, the co-ordinates of the points are [4-SQRT(12), 4-SQRT(12)].
The distance is 8*SQRT(4-SQRT(3)) (as per udosuk)
(I didn't fully check the logic that the points must be on the diagonals the same distance from the face centre but in opposite directions: it looked straightforward but time-consuming.)
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by SimulatedAnnealer » Tue May 02, 2006 1:21 pm
Yes, the point [2-SQRT(2), 2-SQRT(2)] (relative to the face centre) also gives a distance of 12. In this case the shorter route is via 2 rectangular faces rather than one as for the face centres.
Why?
Why?
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37 posts
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