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Gardener's Woe (Posted on 2006-06-24) Difficulty: 4 of 5
An unlucky gardener planted a 10x10 square array of 100 old seeds out in the garden. Only 5 of these seeds have germinated including one at the southwest corner (0,0) where a slug is currently reducing it to ground level.

When it finishes it will head directly to the next closest doomed plant. After it eats that one it will again leave a slime trail to the closest remaining plant and so on until the garden is no more.

Where are the 4 remaining seedlings if the path crawled by the slug is the longest possible and it never has to choose between two equidistant snacks?

Note: Although the slug will never have to choose between two equidistant seedlings, this doesn't imply that no two are equidistant.

Next find the locations if 6 seedlings had germinated instead of 5.

See The Solution Submitted by Jer    
Rating: 4.0000 (1 votes)

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Some Thoughts 'Arcs' for part 1 | Comment 8 of 26 |
So that the slug must only go to the next closet plant, that plant must lie on the circumference of a circle whose centre is the plant just consumed; no other may lie within that circumferece (otherwise it would be the next closest)

[I note the 'this not imply ....' condition.  Two plants could lie on the same arc but their distance apart would need to be greater than the radius of their 'circle'.  I am envisaging an example where one of these is the second consumed and the other is the fifth which means suitably spacing the 3rd and 4th; the path then could begin either 'clockwise' or 'anticlockwise' but the slug does have to make a choice]

My choice is (0,0), (4,6), (0,9), (9,9) and (8,0).

If my second placement was at (4,7) this would be slightly greater than 8 units invalidating my choice of (8,0).

A choice of (5,8) is greater than 9 units and would invalidate either (0,9) or (9,0) as choices further down the chain.

The distance from (9,9) to (8,0) is greater than to (9,0); not I cannot choose (7,0) as it falls within the radius of (the (4,6) plant.

Distance:≈30.2665 units

Just noted that I concur with Dej Mar. Now for 6 plants.
 Thought: What would the max be if slug did a "knight's tour", ie all intervals equidistant - the hypotenuse of congruent triangles, which is what I initially thought this problem to be?

Edited on June 24, 2006, 8:53 pm
  Posted by brianjn on 2006-06-24 20:46:15

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