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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 $m,n\in\mathbb{Z}$ 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.

Rand al'Thor
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  • So, the trees are on a rectangular grid? What stops you from looking straight down between a single row of trees? It sounds much too straightforward so it makes me think I'm missing something... – tmpearce Sep 13 '16 at 00:08
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    @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 :-) – Rand al'Thor Sep 13 '16 at 00:09
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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. – gtwebb Sep 13 '16 at 00:10
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    @randal'thor aah I see, when I first read it I thought the trees were spaced on an integer grid but with 0,0 not falling on the grid. In this case, I guess I'll have to suppose that I can't lean at least 20cm while keeping my feet on 0,0 ;) – tmpearce Sep 13 '16 at 00:13
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    @tmpearce i would probably also assume that the width between you eyes are negligable. – gtwebb Sep 13 '16 at 00:14
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    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... – Nij Sep 13 '16 at 00:28
  • @Nij you are in the exact center, so doesn't that mean you are either in no sectors or 4 (or 8) of them? –  Sep 13 '16 at 00:51
  • @EasterlyIrk You are, but the direction you're looking in isn't. – Rand al'Thor Sep 13 '16 at 00:58
  • @EasterlyIrk wot 'e sed. – Nij Sep 13 '16 at 01:00
  • @randal'thor oh, sorry. –  Sep 13 '16 at 01:01
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    lateral-thinking solution: yes, just look up :) (although I guess these perfect cyliners extend to infinity :/) – Jonathan Allan Sep 13 '16 at 01:17
  • @JonathanAllan No [tag:lateral-thinking] tag! – Rand al'Thor Sep 13 '16 at 01:19
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    ...hence why it's a comment not an answer – Jonathan Allan Sep 13 '16 at 01:34
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    Can you edit this question to provide a "at most" radius for the trees, too? I feel like just writing "at least", the answer is trivially "no" by making all trees 100m in diameter. – Ian MacDonald Sep 13 '16 at 16:57
  • with radius at least 20 centimetres doesn't that prevent a solution from being given? if all 8 first trees have 10m diameters (at least 20cm), the anser is obviously no, and a better question is "how do I get out of this" – njzk2 Sep 13 '16 at 17:07
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    @IanMacDonald In theory the answer could still be "we don't know". And even if the wording of the problem does suggest that it's "no", that doesn't make it trivial - you still need to prove that it's always no! – Rand al'Thor Sep 14 '16 at 02:18
  • Scratches head. "Now, wasn't there a tree on this very spot last time I was here?" – Dr Xorile Mar 17 '19 at 06:07
  • And what kind of tree grows as a perfect cylinder with radius 20cm? – Florian F Jul 12 '21 at 14:55
  • @FlorianF A tree of the species Mathematicae puzzlensis. – Rand al'Thor Jul 12 '21 at 14:57

5 Answers5

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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.

Anon
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    @randal'thor After submitting this solution, I had a look at the source you gave and oh wow the solution there seems ridiculously overcomplicated. – Anon Sep 13 '16 at 03:23
  • That's a very beautiful solution. I guess the original one is with square brackets, cos/sin, and the pigeonhole principle? – Puzzle Prime Sep 13 '16 at 03:37
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    @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. – Anon Sep 13 '16 at 03:53
  • Oh wow, can you send me the full proof to take a look? Then I will also add my solution below, even though it uses a bit standard approach. – Puzzle Prime Sep 13 '16 at 03:56
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    @ArturKirkoryan Here you go – Anon Sep 13 '16 at 03:57
  • Here's a better link, without missing pages. Scroll to page 167. – Anon Sep 13 '16 at 04:29
  • The same observation using Pick's theorem can be used to show that all front-row trees (the ones with (m,n) coprime) are actually visible. Here's what the visible grid points roughly look like (for a different value of tree radius): https://4.bp.blogspot.com/-GnkO0B17uYE/V8wzDH_x99I/AAAAAAAA6hI/2KZ9aJQNq38wMxI1SOGnx-3LOcRXyrqcQCLcB/s1600/grid.png – ffao Sep 13 '16 at 04:56
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    +1 for avoiding Minkowski's theorem. And for "the line $UC$" :) – Rosie F Sep 13 '16 at 05:50
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    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. – Rand al'Thor Sep 13 '16 at 12:26
  • I don't find this proof very convincing for two reasons. Firstly, "you can see their midpoint $C$, since the line $UC$ (pun intended) has equal distance to $A$ and $B$." But $UC$ only has equal distance to $A,B$ if $|UA| = |UB|$ (such that the midpoint of $AB$ is also the angle bisector of $AUB$), and no justification of that is given. Secondly, the definition of front-row as "iff there are no other trees directly between you and it" seems to me to imply that if you can't see between two trees, they aren't both front-row, since the nearer of the two is directly between you and the farther. – Peter Taylor Sep 13 '16 at 15:02
  • @PeterTaylor Any line through $C$ has equal distance to $A$ and $B$. And "directly between" means that the front tree's center lies on the line segment connecting you and the rear tree's center. In other words, a tree is front-line iff it has coprime coordinates. – Anon Sep 13 '16 at 15:08
  • I don't a bit confused by "we know that UAB has 0.5 times the area of a grid square, so in this case the area of UAB is 50m2." The grid is 1mx1m so should these values and the one in the next paragraph be reduced by 100? – gtwebb Sep 13 '16 at 16:22
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    Nope. The question says that the grid is $10m \times 10m$. – Anon Sep 13 '16 at 18:47
  • The technique I thought we could use turned out to be flawed. And you are right about the proof in the book. It may be possible to get it fixed if narrow the strip, make it with length 1000m, and check if it is possible to have any trees in the 4 little extra regions outside the circle. Haven't worked on that though, and even if it is correct, wouldn't be very pretty imo. Once again, congrats for your solution. – Puzzle Prime Sep 14 '16 at 02:02
  • @ArturKirkoryan Actually the part with the extra bits is rather easy to prove: The squared distance of any tree to the origin must be a multiple of $100m^2$. The squared distance of any point inside the extra bits lies between $250000m^2$ and $250000.04m^2$, so there can't be any trees in there. Unfortunately this variant does not generalize to every forest radius, only most of them. – Anon Sep 14 '16 at 03:42
  • Yeah, that's correct. Maybe it is worth to sketch/add the second solution to your answer as well. It is a natural and rather direct application of Minkowski, which is a nice theorem to know (or refresh, as in my case:). – Puzzle Prime Sep 14 '16 at 04:51
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    McFry: I agree with @PeterTaylor, and I disagree with the conclusion of your first paragraph below your illustration (“you can see their midpoint $\rm C$”).  While the puzzle says that the trees are regularly planted in a precise rectangular lattice, it doesn’t say that all the trees are all the same size.  Without that assumption, your statement is false. – Peregrine Rook Sep 19 '16 at 03:17
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    @PeregrineRook If you're a stickler for those details, how about you just assume the paragraph "Wlog. assume that the trees are all the same size" at the beginning of my proof. – Anon Sep 19 '16 at 03:34
  • @Anon One way to avoid that problem would be to consider C as any visible point on AB. – Krad Cigol Mar 17 '19 at 11:55
  • @KradCigol but in that case I don't think you could prove that the distance |UC| is greater than 250m, could you ? – Evargalo Jul 08 '21 at 14:08
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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.

Stewbob
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Facts:

  1. I am standing
  2. Trees are not less than 20cm radius

Assumptions:

  1. I am an average human, at least 140cm tall (lo-ball estimate)
  2. My eyes are positioned near the top of my body
  3. 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!

jymbob
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    The trees are cylindrical, not spherical. And this isn't a lateral-thinking puzzle. – Rand al'Thor Sep 13 '16 at 15:32
  • 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 – jymbob Sep 13 '16 at 15:34
  • But you still need to prove that you can't ... – Rand al'Thor Sep 13 '16 at 15:36
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    I'm happy to defer to the proof offered by @McFry above – jymbob Sep 13 '16 at 15:46
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    I don't see how your height is relevant to the radius of the trees – njzk2 Sep 13 '16 at 17:08
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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. – jymbob Sep 14 '16 at 09:18
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Meta answer:

The fact that you say "at least 20cm" suggests that if there is a solution, it must be "no" (that is, the answer is either "it depends on the size" or "no", so if there is a definitive answer, it must be "no").

Lateral thinking answer:

Yes. Just look directly upwards.

risky mysteries
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Acccumulation
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    Both "answers" in your answer seem more like comments - neither is taking the question seriously and solving as is clearly intended: as a mathematical problem. – bobble Jul 06 '21 at 21:23
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I note the description says "each 10 metres from the next"; it does not say "10 metres between centres". And each tree has a radius of "at least 20 centimetres". Which taken together would suggest an interpretation where fatter trees have greater distances between their centres.

It does not say what happens to trees that would intersect the circular boundary. I suggest that any tree that is wholy or partly inside the boundary should conform to the planting lattice, even if centred outside the boundary.

I suggest interpreting "see out of the forest" as simply having a clear line of sight to any point outside the boundary, even if it's the surface of a tree that's partly inside the boundary.

I would then posit a configuration of 4 trees, each 490 metres in diameter, planted on the compass points of the border. Even though the like-sized trees planted on the lattice outside the forest protrude into the forest, there's still outside visibility of around 1°20’ of arc at each of the compass bearings 30°, 60°, 120°, 150°, 210°, 240°, 300° & 330°.

The requirement for a regular lattice suggests there could be two sizes of tree, but the fatter trees would still have to be small enough be 10 metres apart on the diagonals (since otherwise that would be the nearest tree).

Enlarging the compass-point trees to a diameter of 690 metres means that the the "small" trees on the alternate lattice points are only 290 m in diameter, and lie entirely outside the forest, visible at the compass bearings of 45°, 135°, 225°, & 315°.

Conversely, reducing the compass point trees to 290m means that the 10 m gaps to the nearest trees are now nearer to square-on to the observer at the centre, leaving wider views beyond the forest.

Martin Kealey
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  • Someone voted this down and didn't even leave a comment explaining why. All of the answers have had to specify additional assumptions, so I don't see why mine is any worse. I intentionally chose different assumptions to make it interesting; there's no point just re-doing the same answer to the same assumptions. – Martin Kealey Jul 15 '21 at 08:29