## Riddle (sort of)

Anything goes, but keep it seemly...
On a 24m x 24m field there is 65 unicorns. The evil Dalton brothers are approaching with a circular net, diameter 4,25m. Is it possible for the unicorns to spread out on the field in such a way that the Dalton brothers couldn't catch at least 2 horns in the net? The tips of the unicorn horns are always pointing up on the same height above ground.

I suppose in "4,25m" you mean 4.25m not 425m (otherwise the net is as large as a baseball stadium and can catch all the unicorns )... So the problem is to pack a 24x24 square with 65 circles of radius 2.125, right?
udosuk

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You only need half circles near the edges, but still I cannot find room for more than 42.

So I must say: No, it can't be done thinking inside the 24x24 box.

Ruud.
Ruud

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You shun seducement and a snare
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You will not play ‘cos you surmise
I’m luring you with tricky lies
But it’s a riddle - r.ae –
On the line, upfront, straight as can be

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It might even prove some fun to be
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emm

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udosuk wrote:I suppose in "4,25m" you mean 4.25m not 425m (otherwise the net is as large as a baseball stadium and can catch all the unicorns )... So the problem is to pack a 24x24 square with 65 circles of radius 2.125, right?

Yes, I mean 4.25m.
Ruud wrote:You only need half circles near the edges, but still I cannot find room for more than 42.

In your picture I see a lot of free space between the circles for the unicorns to hide in, so I'm not quite convinced by the proof. There is an easier way to solve the problem than packing the square with circles. Also, packing it with 65 circles wouldn't prove that one of them must catch 2 unicorns.

RW
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Here's a picture to help you visualize the situation (proportions are slightly wrong):

Code: Select all
`     24m                       O*-------------*               -|-           O   |. .        . |                ^    net    -|-|      .     .|                    _____    ^ \| .       .   |                   /     \      The Evil Dalton brothers|  .  .       |                  |       | O  /      (sneaking on their knees)|       .     | 24m              |       |-|-| .     .     |               O   \_____/  ^|   .       . |              -|-|     .    .  |               ^   -------|  .     .    |                    4.25m*---------\---*                                                     65 unicorns`

[Edit: corrected the size of the field]

RW
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The only thing that can get anyway near an unicorn is a virgin - only female virgins need apply - so that excludes the evil Dalton brothers.

But their association with virgins does appear to have been more recent. In fact at some point they became inextricably linked. Supposedly the Romans discovered that unicorns were lovers of purity and innocence and as a sort of test, they placed a virgin at the feet of a unicorn. The unicorn was unavoidably attracted and fell asleep on her lap.

Later the act of standing a woman in front of a unicorn would prove to be a test of her virginity (this may have originated in England, I don't know). Frankly as tests go, this one's a bit rugged. Cos if the woman wasn't a virgin, the unicorn's horn would pierce her body and kill her. If she was a virgin, theoretically the horn would pass through her body unharmed.

Furthermore, it didn't always pay to be a virgin who associated with a unicorn, because they were very jealous and possessive. If the virgin took up with a suitor and married him, the unicorn would kill the husband and gut the virgin.

But if RW is assuming the unicorns don't flee, and I'll leave the math to others, if you just consider the horns as points, can you spread out these points over the 24m x 24m field such that if you throw a 4.25m net it would land on at lease two of them or is the spread such that the net would only land on one.

MCC
MCC

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Come on, this isn't that hard... Hint number 2: the mathematical proof is so simple that even the average 13-year old can understand it (at least in the Finnish schools). You don't need a ruler, graphical calculator or even the constant pi.

MCC wrote:if you just consider the horns as points, can you spread out these points over the 24m x 24m field such that if you throw a 4.25m net it would land on at lease two of them or is the spread such that the net would only land on one.

The question was actually "can you spread out these points over the 24m x 24m field such that if you throw a 4.25m net it would only land at one of them, or is the spread such that it is always possible to hit two points in some part of the field."

RW
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It's very shaky ground when I start answering maths riddles ... but since all is quiet on the Northern Front -

24 x 24 = 576 sq m ÷ 65 = 8.86 each x 2 unicorns = 17.72sqm = 4.21 x 4.21m

So the 4.25m net will catch 2 at once ... unless they can bunch up a bit.
emm

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emm wrote:24 x 24 = 576 sq m ÷ 65 = 8.86 each x 2 unicorns = 17.72sqm = 4.21 x 4.21m

So the 4.25m net will catch 2 at once ... unless they can bunch up a bit.

But the net was circular, not square... Seems this isn't so easy after all. To be honest with you, I couldn't solve this either (on a 30min time limit), but I also felt really, really stupid when I heard the obvious solution.

hinter #3 wrote:---------- (approximately 580 BC–500 BC, Greek: ΠŸαγόqς) was an Ionian (Greek) mathematician and philosopher, founder of the mystic, religious and scientific society called ------------, and is known best for the ----------- theorem which bears his name.

RW
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Since my answer with circles has been rejected, I will now try an answer with triangles.

2 unicorns need at least a distance of 4.25 m from horn to horn, otherwise the net will catch them both.

For 3 unicorns, they need to stand in a triangle with each side at least 4.25 m, otherwise the net will catch 2 of them.

This triangle has a surface of 4.25 * (4.25 / 2) = 8.60625 m2

Each subsequent unicorn adds a similar triangle.
The total surface for N unicorns equals (N-2) * 8.60625 m2
For 65 unicorns, we need: 542,19375 m2
The total surface equals: 24 * 24 = 576 m2

So, if we can pack the triangles with no more than 33.81 m2 waste, we can save 64 unicorns.

It is possible to place 5 triangles side by side, with 5 triangles upside down. The width is 23.375 m, this leaves a strip of 0.625 m wide.
The waste of this strip is 0.625 * 24 = 15 m2.

Our margin is down to 18.81 m2...

Unfortunately, there is more waste for our 10 unicorns. Both sides produce half a triangle waste. That is 8.60625 m2 in total.

For 65 unicorns, we need 6 such bands of 10 unicorns. 6 * 8.60625 = 51.6375 m2.

It is not possible for the unicorns to evade the net. I advice them to aim their horns at the Daltons and settle their differences that way.

Ruud.
Ruud

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Interesting solution Ruud, that could even work as a proof, must think about it for a while. But now when you tried both circles and triangles, maybe you still could find the 3 row solution that uses squares.

RW
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Take the 4.2m squares and divide them into right triangles.

The hypotenuse = 5.94m = the diameter of the circumcircles which is bigger than the net so it won't catch them both.

I don't really know what I'm doing here. I'm just having a go!
emm

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...............o0O0o...............
the omen has been seen
a warning was on my screen
in the post count of the she-lob
...............o0O0o...............
Ruud

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Give up?
emm

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Creepy... 666 <=> Evil Madame Medusa...

Anyway for RW's riddle, I got a non-water tight approach...

Suppose we arrange the unicorns on the perimeter of the field, which has a total length of 24x4=96m, where we can put at most 22 unicorns (in practice only 21, but let's say 22)... Now consider an inner square within the field, of a distance of at least 4.25xsqrt(3)/2~=3.68m from the edges... All points outside this inner square is bound to be closer than 4.25m to the 22 unicorns along the edges of the field, because they're included in one of the equilateral triangles formed by 2 adjacent outside unicorns... So for this inner square, the side is 24-3.68x2=16.64m, with a perimeter of 16.64x4=66.56m, where we can put no more than 15 unicorns. Similarly, forming another "layer" within, the next level is a 9.28m square with perimete=37.12m, allowing 8 more unicorns. At last, we have a 1.92m small square right in the middle, which allow us to put 1 unicorn only. So in total we could at most put 22+15+8+1=46 unicorns...

But I cannot be sure if this approach cover the maximum number of unicorns we could put within the field...
udosuk

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