Can you see out of the forest?
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You are standing at the centre of a circular forest of radius 500 metres. The trees of this very regularly planted forest stand in a precise rectangular lattice on the plane, each 10 metres from the next: the points $(m,n)$ within the disc-shaped forest with $m,ninmathbb{Z}$ not both zero (since $(0,0)$ is the point where you're standing). Each tree is a perfect cylinder with radius at least 20 centimetres.
Can you see out of the forest?
Source: Leith Hathout, Crimes and Mathdemeanors.
mathematics geometry
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
$endgroup$
|
show 12 more comments
$begingroup$
You are standing at the centre of a circular forest of radius 500 metres. The trees of this very regularly planted forest stand in a precise rectangular lattice on the plane, each 10 metres from the next: the points $(m,n)$ within the disc-shaped forest with $m,ninmathbb{Z}$ not both zero (since $(0,0)$ is the point where you're standing). Each tree is a perfect cylinder with radius at least 20 centimetres.
Can you see out of the forest?
Source: Leith Hathout, Crimes and Mathdemeanors.
mathematics geometry
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
$endgroup$
1
$begingroup$
@tmpearce Because you're also on the same rectangular grid, so there's no two rows you can look directly between. And it's not straightforward :-)
$endgroup$
– Rand al'Thor
Sep 13 '16 at 0:09
2
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IMO the trees should be given a radius of exactly 20cm. The phrase at least implies it doesnt matter if they are bigger in which case i would say you cant see out. And if you are making them perfectly cylindrical on a perfect grid they might as well be an exact size as well.
$endgroup$
– gtwebb
Sep 13 '16 at 0:10
1
$begingroup$
To start with, suppose the grid is aligned NS-EW. You only have to consider the angle from North through to North-East (or any of the equivalent sectors between adjacent primary and secondary cardinal directions) because of the eight-fold symmetry. Either it can be done in one and thence all of the sectors, or it can't be done in any of them...
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– Nij
Sep 13 '16 at 0:28
12
$begingroup$
lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/)
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:17
2
$begingroup$
...hence why it's a comment not an answer
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:34
|
show 12 more comments
$begingroup$
You are standing at the centre of a circular forest of radius 500 metres. The trees of this very regularly planted forest stand in a precise rectangular lattice on the plane, each 10 metres from the next: the points $(m,n)$ within the disc-shaped forest with $m,ninmathbb{Z}$ not both zero (since $(0,0)$ is the point where you're standing). Each tree is a perfect cylinder with radius at least 20 centimetres.
Can you see out of the forest?
Source: Leith Hathout, Crimes and Mathdemeanors.
mathematics geometry
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
$endgroup$
You are standing at the centre of a circular forest of radius 500 metres. The trees of this very regularly planted forest stand in a precise rectangular lattice on the plane, each 10 metres from the next: the points $(m,n)$ within the disc-shaped forest with $m,ninmathbb{Z}$ not both zero (since $(0,0)$ is the point where you're standing). Each tree is a perfect cylinder with radius at least 20 centimetres.
Can you see out of the forest?
Source: Leith Hathout, Crimes and Mathdemeanors.
mathematics geometry
mathematics geometry
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
edited Sep 13 '16 at 0:19
Rand al'Thor
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
asked Sep 12 '16 at 23:59
Rand al'ThorRand al'Thor
70.7k14235471
70.7k14235471
1
$begingroup$
@tmpearce Because you're also on the same rectangular grid, so there's no two rows you can look directly between. And it's not straightforward :-)
$endgroup$
– Rand al'Thor
Sep 13 '16 at 0:09
2
$begingroup$
IMO the trees should be given a radius of exactly 20cm. The phrase at least implies it doesnt matter if they are bigger in which case i would say you cant see out. And if you are making them perfectly cylindrical on a perfect grid they might as well be an exact size as well.
$endgroup$
– gtwebb
Sep 13 '16 at 0:10
1
$begingroup$
To start with, suppose the grid is aligned NS-EW. You only have to consider the angle from North through to North-East (or any of the equivalent sectors between adjacent primary and secondary cardinal directions) because of the eight-fold symmetry. Either it can be done in one and thence all of the sectors, or it can't be done in any of them...
$endgroup$
– Nij
Sep 13 '16 at 0:28
12
$begingroup$
lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/)
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:17
2
$begingroup$
...hence why it's a comment not an answer
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:34
|
show 12 more comments
1
$begingroup$
@tmpearce Because you're also on the same rectangular grid, so there's no two rows you can look directly between. And it's not straightforward :-)
$endgroup$
– Rand al'Thor
Sep 13 '16 at 0:09
2
$begingroup$
IMO the trees should be given a radius of exactly 20cm. The phrase at least implies it doesnt matter if they are bigger in which case i would say you cant see out. And if you are making them perfectly cylindrical on a perfect grid they might as well be an exact size as well.
$endgroup$
– gtwebb
Sep 13 '16 at 0:10
1
$begingroup$
To start with, suppose the grid is aligned NS-EW. You only have to consider the angle from North through to North-East (or any of the equivalent sectors between adjacent primary and secondary cardinal directions) because of the eight-fold symmetry. Either it can be done in one and thence all of the sectors, or it can't be done in any of them...
$endgroup$
– Nij
Sep 13 '16 at 0:28
12
$begingroup$
lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/)
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:17
2
$begingroup$
...hence why it's a comment not an answer
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:34
1
1
$begingroup$
@tmpearce Because you're also on the same rectangular grid, so there's no two rows you can look directly between. And it's not straightforward :-)
$endgroup$
– Rand al'Thor
Sep 13 '16 at 0:09
$begingroup$
@tmpearce Because you're also on the same rectangular grid, so there's no two rows you can look directly between. And it's not straightforward :-)
$endgroup$
– Rand al'Thor
Sep 13 '16 at 0:09
2
2
$begingroup$
IMO the trees should be given a radius of exactly 20cm. The phrase at least implies it doesnt matter if they are bigger in which case i would say you cant see out. And if you are making them perfectly cylindrical on a perfect grid they might as well be an exact size as well.
$endgroup$
– gtwebb
Sep 13 '16 at 0:10
$begingroup$
IMO the trees should be given a radius of exactly 20cm. The phrase at least implies it doesnt matter if they are bigger in which case i would say you cant see out. And if you are making them perfectly cylindrical on a perfect grid they might as well be an exact size as well.
$endgroup$
– gtwebb
Sep 13 '16 at 0:10
1
1
$begingroup$
To start with, suppose the grid is aligned NS-EW. You only have to consider the angle from North through to North-East (or any of the equivalent sectors between adjacent primary and secondary cardinal directions) because of the eight-fold symmetry. Either it can be done in one and thence all of the sectors, or it can't be done in any of them...
$endgroup$
– Nij
Sep 13 '16 at 0:28
$begingroup$
To start with, suppose the grid is aligned NS-EW. You only have to consider the angle from North through to North-East (or any of the equivalent sectors between adjacent primary and secondary cardinal directions) because of the eight-fold symmetry. Either it can be done in one and thence all of the sectors, or it can't be done in any of them...
$endgroup$
– Nij
Sep 13 '16 at 0:28
12
12
$begingroup$
lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/)
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:17
$begingroup$
lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/)
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:17
2
2
$begingroup$
...hence why it's a comment not an answer
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:34
$begingroup$
...hence why it's a comment not an answer
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:34
|
show 12 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Let's call a tree front-row iff there are no other trees directly between you and it. Clearly any non-front-row trees are covered by front-row trees.
We call a pair of front-row trees adjacent iff, from your perspective, no other tree's midpoint appears between the midpoints of those two trees.
Claim: You cannot see through the gap between any two adjacent front-row trees.
If this is true, then clearly you cannot see out of the forest, since any sightline passes between some two adjacent front-row trees.
Proof of claim: Let's take any two adjacent front-row trees $A$ and $B$.
Suppose you can see between them. Then you can see their midpoint $C$, since the line $UC$ (pun intended) has equal distance to $A$ and $B$.
We know that the distance $|UC|$ is greater than $250m$, since otherwise there would be a tree at the integer coordinates $A + B = 2C$ in contradiction to the adjacency of $A$ and $B$.
Now let's look at the area of $UAB$. By Pick's Theorem, we know that $UAB$ has $0.5$ times the area of a grid square, so in this case the area of $UAB$ is $50m^2$.
Now consider the triangles $UCA$ and $UCB$ which make up $UAB$. They each have a base of over $250m$ and a height of at least $20cm$, so they must each be over $25m^2$ large. Together they must therefore have an area of strictly more than $50m^2$, which is clearly a contradiction to the previous paragraph. This proves the claim.
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
$endgroup$
2
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
1
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
1
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
2
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
1
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
|
show 14 more comments
$begingroup$
Case 1:
Assumptions: My eyes are in the exact center of my head; The trees are exactly 0.2m in radius.
Result: I cannot see out of the forest. There is a 2mm interference on the tree closest to the edge of the forest blocking my line of sight.
Case 2:
Assumptions: My eyes are like those of a normal human and they are each located 30mm from the center of my head; The trees are exactly 0.2m in radius.
Result: I can see out of the forest. There is a 1.438m gap (measured at the outer edge of the forest) between the edge of the tree closest to me and the edge of the tree farthest from me.
Kudos to the math proof (especially since I can't do that), but I'm an engineer, so I just drew it and measured it in CAD.
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
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You're hired! Just one more question... What's the volume of this light bulb?
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– Mazura
Sep 14 '16 at 1:48
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Archimedes can tell you..,
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– Stewbob
Sep 14 '16 at 2:30
add a comment |
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Facts:
- I am standing
- Trees are not less than 20cm radius
Assumptions:
- I am an average human, at least 140cm tall (lo-ball estimate)
- My eyes are positioned near the top of my body
- The majority of trees have a radius below 70cm
Therefore I can easily see out of the forest over the tops of the trees.
NB: This would have been a comment (I don't feel it really meets the spirit of the question) but I don't have the rep.
NNB: I don't perceive this as lateral thinking, just making reasonable assumptions from the stated question, but happy to be argued down!
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
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1
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The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
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– Rand al'Thor
Sep 13 '16 at 15:32
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Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
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– jymbob
Sep 13 '16 at 15:34
2
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I'm happy to defer to the proof offered by @McFry above
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– jymbob
Sep 13 '16 at 15:46
1
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I don't see how your height is relevant to the radius of the trees
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– njzk2
Sep 13 '16 at 17:08
1
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Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
|
show 1 more comment
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's call a tree front-row iff there are no other trees directly between you and it. Clearly any non-front-row trees are covered by front-row trees.
We call a pair of front-row trees adjacent iff, from your perspective, no other tree's midpoint appears between the midpoints of those two trees.
Claim: You cannot see through the gap between any two adjacent front-row trees.
If this is true, then clearly you cannot see out of the forest, since any sightline passes between some two adjacent front-row trees.
Proof of claim: Let's take any two adjacent front-row trees $A$ and $B$.
Suppose you can see between them. Then you can see their midpoint $C$, since the line $UC$ (pun intended) has equal distance to $A$ and $B$.
We know that the distance $|UC|$ is greater than $250m$, since otherwise there would be a tree at the integer coordinates $A + B = 2C$ in contradiction to the adjacency of $A$ and $B$.
Now let's look at the area of $UAB$. By Pick's Theorem, we know that $UAB$ has $0.5$ times the area of a grid square, so in this case the area of $UAB$ is $50m^2$.
Now consider the triangles $UCA$ and $UCB$ which make up $UAB$. They each have a base of over $250m$ and a height of at least $20cm$, so they must each be over $25m^2$ large. Together they must therefore have an area of strictly more than $50m^2$, which is clearly a contradiction to the previous paragraph. This proves the claim.
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
$endgroup$
2
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
1
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
1
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
2
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
1
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
|
show 14 more comments
$begingroup$
Let's call a tree front-row iff there are no other trees directly between you and it. Clearly any non-front-row trees are covered by front-row trees.
We call a pair of front-row trees adjacent iff, from your perspective, no other tree's midpoint appears between the midpoints of those two trees.
Claim: You cannot see through the gap between any two adjacent front-row trees.
If this is true, then clearly you cannot see out of the forest, since any sightline passes between some two adjacent front-row trees.
Proof of claim: Let's take any two adjacent front-row trees $A$ and $B$.
Suppose you can see between them. Then you can see their midpoint $C$, since the line $UC$ (pun intended) has equal distance to $A$ and $B$.
We know that the distance $|UC|$ is greater than $250m$, since otherwise there would be a tree at the integer coordinates $A + B = 2C$ in contradiction to the adjacency of $A$ and $B$.
Now let's look at the area of $UAB$. By Pick's Theorem, we know that $UAB$ has $0.5$ times the area of a grid square, so in this case the area of $UAB$ is $50m^2$.
Now consider the triangles $UCA$ and $UCB$ which make up $UAB$. They each have a base of over $250m$ and a height of at least $20cm$, so they must each be over $25m^2$ large. Together they must therefore have an area of strictly more than $50m^2$, which is clearly a contradiction to the previous paragraph. This proves the claim.
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
$endgroup$
2
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
1
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
1
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
2
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
1
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
|
show 14 more comments
$begingroup$
Let's call a tree front-row iff there are no other trees directly between you and it. Clearly any non-front-row trees are covered by front-row trees.
We call a pair of front-row trees adjacent iff, from your perspective, no other tree's midpoint appears between the midpoints of those two trees.
Claim: You cannot see through the gap between any two adjacent front-row trees.
If this is true, then clearly you cannot see out of the forest, since any sightline passes between some two adjacent front-row trees.
Proof of claim: Let's take any two adjacent front-row trees $A$ and $B$.
Suppose you can see between them. Then you can see their midpoint $C$, since the line $UC$ (pun intended) has equal distance to $A$ and $B$.
We know that the distance $|UC|$ is greater than $250m$, since otherwise there would be a tree at the integer coordinates $A + B = 2C$ in contradiction to the adjacency of $A$ and $B$.
Now let's look at the area of $UAB$. By Pick's Theorem, we know that $UAB$ has $0.5$ times the area of a grid square, so in this case the area of $UAB$ is $50m^2$.
Now consider the triangles $UCA$ and $UCB$ which make up $UAB$. They each have a base of over $250m$ and a height of at least $20cm$, so they must each be over $25m^2$ large. Together they must therefore have an area of strictly more than $50m^2$, which is clearly a contradiction to the previous paragraph. This proves the claim.
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
$endgroup$
Let's call a tree front-row iff there are no other trees directly between you and it. Clearly any non-front-row trees are covered by front-row trees.
We call a pair of front-row trees adjacent iff, from your perspective, no other tree's midpoint appears between the midpoints of those two trees.
Claim: You cannot see through the gap between any two adjacent front-row trees.
If this is true, then clearly you cannot see out of the forest, since any sightline passes between some two adjacent front-row trees.
Proof of claim: Let's take any two adjacent front-row trees $A$ and $B$.
Suppose you can see between them. Then you can see their midpoint $C$, since the line $UC$ (pun intended) has equal distance to $A$ and $B$.
We know that the distance $|UC|$ is greater than $250m$, since otherwise there would be a tree at the integer coordinates $A + B = 2C$ in contradiction to the adjacency of $A$ and $B$.
Now let's look at the area of $UAB$. By Pick's Theorem, we know that $UAB$ has $0.5$ times the area of a grid square, so in this case the area of $UAB$ is $50m^2$.
Now consider the triangles $UCA$ and $UCB$ which make up $UAB$. They each have a base of over $250m$ and a height of at least $20cm$, so they must each be over $25m^2$ large. Together they must therefore have an area of strictly more than $50m^2$, which is clearly a contradiction to the previous paragraph. This proves the claim.
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
answered Sep 13 '16 at 3:11
AnonAnon
2,5741718
2,5741718
2
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
1
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
1
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
2
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
1
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
|
show 14 more comments
2
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
1
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
1
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
2
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
1
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
2
2
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
$begingroup$
@randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated.
$endgroup$
– Anon
Sep 13 '16 at 3:23
1
1
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
$begingroup$
@ArturKirkoryan No, it involves proving Minkowski's Theorem using Blichfeldt's Lemma, then using that on an arbitrary 80cm strip across the origin to find a tree whose center lies on that strip (and thus blocks the axis of that strip). Wait a second, that's not true at all, this means that the proof given in the book isn't even correct! And just reducing the strip width to 40cm doesn't work either, because that would break an important inequality necessary for Minkowski's Theorem to work! It appears the proof in the book is in fact irreparably broken.
$endgroup$
– Anon
Sep 13 '16 at 3:53
1
1
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
$begingroup$
@ArturKirkoryan Here you go
$endgroup$
– Anon
Sep 13 '16 at 3:57
2
2
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
$begingroup$
+1 for avoiding Minkowski's theorem. And for "the line $UC$" :)
$endgroup$
– Rosie F
Sep 13 '16 at 5:50
1
1
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
$begingroup$
Brilliant proof! And my sincere apologies for the book proof being wrong; we're lucky that there was still an elementary proof, otherwise my question might have inadvertently been unsolvable or even wrong.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 12:26
|
show 14 more comments
$begingroup$
Case 1:
Assumptions: My eyes are in the exact center of my head; The trees are exactly 0.2m in radius.
Result: I cannot see out of the forest. There is a 2mm interference on the tree closest to the edge of the forest blocking my line of sight.
Case 2:
Assumptions: My eyes are like those of a normal human and they are each located 30mm from the center of my head; The trees are exactly 0.2m in radius.
Result: I can see out of the forest. There is a 1.438m gap (measured at the outer edge of the forest) between the edge of the tree closest to me and the edge of the tree farthest from me.
Kudos to the math proof (especially since I can't do that), but I'm an engineer, so I just drew it and measured it in CAD.
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
$endgroup$
$begingroup$
You're hired! Just one more question... What's the volume of this light bulb?
$endgroup$
– Mazura
Sep 14 '16 at 1:48
$begingroup$
Archimedes can tell you..,
$endgroup$
– Stewbob
Sep 14 '16 at 2:30
add a comment |
$begingroup$
Case 1:
Assumptions: My eyes are in the exact center of my head; The trees are exactly 0.2m in radius.
Result: I cannot see out of the forest. There is a 2mm interference on the tree closest to the edge of the forest blocking my line of sight.
Case 2:
Assumptions: My eyes are like those of a normal human and they are each located 30mm from the center of my head; The trees are exactly 0.2m in radius.
Result: I can see out of the forest. There is a 1.438m gap (measured at the outer edge of the forest) between the edge of the tree closest to me and the edge of the tree farthest from me.
Kudos to the math proof (especially since I can't do that), but I'm an engineer, so I just drew it and measured it in CAD.
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
$endgroup$
$begingroup$
You're hired! Just one more question... What's the volume of this light bulb?
$endgroup$
– Mazura
Sep 14 '16 at 1:48
$begingroup$
Archimedes can tell you..,
$endgroup$
– Stewbob
Sep 14 '16 at 2:30
add a comment |
$begingroup$
Case 1:
Assumptions: My eyes are in the exact center of my head; The trees are exactly 0.2m in radius.
Result: I cannot see out of the forest. There is a 2mm interference on the tree closest to the edge of the forest blocking my line of sight.
Case 2:
Assumptions: My eyes are like those of a normal human and they are each located 30mm from the center of my head; The trees are exactly 0.2m in radius.
Result: I can see out of the forest. There is a 1.438m gap (measured at the outer edge of the forest) between the edge of the tree closest to me and the edge of the tree farthest from me.
Kudos to the math proof (especially since I can't do that), but I'm an engineer, so I just drew it and measured it in CAD.
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
$endgroup$
Case 1:
Assumptions: My eyes are in the exact center of my head; The trees are exactly 0.2m in radius.
Result: I cannot see out of the forest. There is a 2mm interference on the tree closest to the edge of the forest blocking my line of sight.
Case 2:
Assumptions: My eyes are like those of a normal human and they are each located 30mm from the center of my head; The trees are exactly 0.2m in radius.
Result: I can see out of the forest. There is a 1.438m gap (measured at the outer edge of the forest) between the edge of the tree closest to me and the edge of the tree farthest from me.
Kudos to the math proof (especially since I can't do that), but I'm an engineer, so I just drew it and measured it in CAD.
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
answered Sep 13 '16 at 17:07
StewbobStewbob
1514
1514
$begingroup$
You're hired! Just one more question... What's the volume of this light bulb?
$endgroup$
– Mazura
Sep 14 '16 at 1:48
$begingroup$
Archimedes can tell you..,
$endgroup$
– Stewbob
Sep 14 '16 at 2:30
add a comment |
$begingroup$
You're hired! Just one more question... What's the volume of this light bulb?
$endgroup$
– Mazura
Sep 14 '16 at 1:48
$begingroup$
Archimedes can tell you..,
$endgroup$
– Stewbob
Sep 14 '16 at 2:30
$begingroup$
You're hired! Just one more question... What's the volume of this light bulb?
$endgroup$
– Mazura
Sep 14 '16 at 1:48
$begingroup$
You're hired! Just one more question... What's the volume of this light bulb?
$endgroup$
– Mazura
Sep 14 '16 at 1:48
$begingroup$
Archimedes can tell you..,
$endgroup$
– Stewbob
Sep 14 '16 at 2:30
$begingroup$
Archimedes can tell you..,
$endgroup$
– Stewbob
Sep 14 '16 at 2:30
add a comment |
$begingroup$
Facts:
- I am standing
- Trees are not less than 20cm radius
Assumptions:
- I am an average human, at least 140cm tall (lo-ball estimate)
- My eyes are positioned near the top of my body
- The majority of trees have a radius below 70cm
Therefore I can easily see out of the forest over the tops of the trees.
NB: This would have been a comment (I don't feel it really meets the spirit of the question) but I don't have the rep.
NNB: I don't perceive this as lateral thinking, just making reasonable assumptions from the stated question, but happy to be argued down!
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
$endgroup$
1
$begingroup$
The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 15:32
$begingroup$
Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
$endgroup$
– jymbob
Sep 13 '16 at 15:34
2
$begingroup$
I'm happy to defer to the proof offered by @McFry above
$endgroup$
– jymbob
Sep 13 '16 at 15:46
1
$begingroup$
I don't see how your height is relevant to the radius of the trees
$endgroup$
– njzk2
Sep 13 '16 at 17:08
1
$begingroup$
Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
|
show 1 more comment
$begingroup$
Facts:
- I am standing
- Trees are not less than 20cm radius
Assumptions:
- I am an average human, at least 140cm tall (lo-ball estimate)
- My eyes are positioned near the top of my body
- The majority of trees have a radius below 70cm
Therefore I can easily see out of the forest over the tops of the trees.
NB: This would have been a comment (I don't feel it really meets the spirit of the question) but I don't have the rep.
NNB: I don't perceive this as lateral thinking, just making reasonable assumptions from the stated question, but happy to be argued down!
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
$endgroup$
1
$begingroup$
The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 15:32
$begingroup$
Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
$endgroup$
– jymbob
Sep 13 '16 at 15:34
2
$begingroup$
I'm happy to defer to the proof offered by @McFry above
$endgroup$
– jymbob
Sep 13 '16 at 15:46
1
$begingroup$
I don't see how your height is relevant to the radius of the trees
$endgroup$
– njzk2
Sep 13 '16 at 17:08
1
$begingroup$
Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
|
show 1 more comment
$begingroup$
Facts:
- I am standing
- Trees are not less than 20cm radius
Assumptions:
- I am an average human, at least 140cm tall (lo-ball estimate)
- My eyes are positioned near the top of my body
- The majority of trees have a radius below 70cm
Therefore I can easily see out of the forest over the tops of the trees.
NB: This would have been a comment (I don't feel it really meets the spirit of the question) but I don't have the rep.
NNB: I don't perceive this as lateral thinking, just making reasonable assumptions from the stated question, but happy to be argued down!
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
$endgroup$
Facts:
- I am standing
- Trees are not less than 20cm radius
Assumptions:
- I am an average human, at least 140cm tall (lo-ball estimate)
- My eyes are positioned near the top of my body
- The majority of trees have a radius below 70cm
Therefore I can easily see out of the forest over the tops of the trees.
NB: This would have been a comment (I don't feel it really meets the spirit of the question) but I don't have the rep.
NNB: I don't perceive this as lateral thinking, just making reasonable assumptions from the stated question, but happy to be argued down!
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
edited Sep 13 '16 at 15:33
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
answered Sep 13 '16 at 15:30
jymbobjymbob
1312
1312
1
$begingroup$
The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 15:32
$begingroup$
Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
$endgroup$
– jymbob
Sep 13 '16 at 15:34
2
$begingroup$
I'm happy to defer to the proof offered by @McFry above
$endgroup$
– jymbob
Sep 13 '16 at 15:46
1
$begingroup$
I don't see how your height is relevant to the radius of the trees
$endgroup$
– njzk2
Sep 13 '16 at 17:08
1
$begingroup$
Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
|
show 1 more comment
1
$begingroup$
The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 15:32
$begingroup$
Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
$endgroup$
– jymbob
Sep 13 '16 at 15:34
2
$begingroup$
I'm happy to defer to the proof offered by @McFry above
$endgroup$
– jymbob
Sep 13 '16 at 15:46
1
$begingroup$
I don't see how your height is relevant to the radius of the trees
$endgroup$
– njzk2
Sep 13 '16 at 17:08
1
$begingroup$
Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
1
1
$begingroup$
The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 15:32
$begingroup$
The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle.
$endgroup$
– Rand al'Thor
Sep 13 '16 at 15:32
$begingroup$
Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
$endgroup$
– jymbob
Sep 13 '16 at 15:34
$begingroup$
Ah. My bad. So, assuming a cylindrical tree with a reasonable height (let's say, taller thank an average human), no, you can't. That'll teach me to scan questions
$endgroup$
– jymbob
Sep 13 '16 at 15:34
2
2
$begingroup$
I'm happy to defer to the proof offered by @McFry above
$endgroup$
– jymbob
Sep 13 '16 at 15:46
$begingroup$
I'm happy to defer to the proof offered by @McFry above
$endgroup$
– jymbob
Sep 13 '16 at 15:46
1
1
$begingroup$
I don't see how your height is relevant to the radius of the trees
$endgroup$
– njzk2
Sep 13 '16 at 17:08
$begingroup$
I don't see how your height is relevant to the radius of the trees
$endgroup$
– njzk2
Sep 13 '16 at 17:08
1
1
$begingroup$
Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
$begingroup$
Having misread the question (and assumed that all trees were spherical, not cylindrical), I was merely pointing out that one can see over the top of short things while standing.
$endgroup$
– jymbob
Sep 14 '16 at 9:18
|
show 1 more comment
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1
$begingroup$
@tmpearce Because you're also on the same rectangular grid, so there's no two rows you can look directly between. And it's not straightforward :-)
$endgroup$
– Rand al'Thor
Sep 13 '16 at 0:09
2
$begingroup$
IMO the trees should be given a radius of exactly 20cm. The phrase at least implies it doesnt matter if they are bigger in which case i would say you cant see out. And if you are making them perfectly cylindrical on a perfect grid they might as well be an exact size as well.
$endgroup$
– gtwebb
Sep 13 '16 at 0:10
1
$begingroup$
To start with, suppose the grid is aligned NS-EW. You only have to consider the angle from North through to North-East (or any of the equivalent sectors between adjacent primary and secondary cardinal directions) because of the eight-fold symmetry. Either it can be done in one and thence all of the sectors, or it can't be done in any of them...
$endgroup$
– Nij
Sep 13 '16 at 0:28
12
$begingroup$
lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/)
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:17
2
$begingroup$
...hence why it's a comment not an answer
$endgroup$
– Jonathan Allan
Sep 13 '16 at 1:34