6
$\begingroup$

How can 11 trees be planted in a garden, such that there are 6 rows that each contain exactly 4 trees?

A row consists of some number of trees in a straight line. The same tree can be part of multiple rows, but rows can’t contain more than 4 trees. Rows can be horizontal, vertical or slanted at any angle.

Spoiler/attribution:

https://erich-friedman.github.io/packing/trees

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
12
$\begingroup$

A slightly different solution from AxiomaticSystem's:

Solution

Edit: skewed the lines a bit so it's clear there is no line of 3.

Improve this answer
$\endgroup$
11
$\begingroup$

You could arrange the trees like this:

enter image description here

Improve this answer
$\endgroup$
10
  • 1
    $\begingroup$ @WeatherVane Perhaps I haven’t worded the puzzle clearly. But AxiomaticSystem seems to have understood my intention. I don’t wish to respond further. $\endgroup$
    – Will.Octagon.Gibson
    Commented Jul 5 at 0:14
  • 2
    $\begingroup$ Obviously, no arrangement of trees can have only rows of four - technically, every tree exists in infinitely many rows of one, and rows of two are also inevitable. The literal interpretation is flawed, and I'm editing the question to fix it. $\endgroup$
    – AxiomaticSystem
    Commented Jul 5 at 3:37
  • 2
    $\begingroup$ @WeatherVane One could restrict the puzzle as follows: For any two trees the row defined by passing through them needs to be either one of the exactly 6 rows with 4 trees or contain exactly 2 trees. With this definition this solution would have an extra row of 3 along the x=y line. I'm not sure whether one could adjust this solution slightly by removing the symmetry to avoid that extra row. $\endgroup$
    – quarague
    Commented Jul 5 at 6:37
  • 1
    $\begingroup$ @quarague You can, indeed, move up the higher of the two rows passing through the rightmost tree. This forces the lower row through the leftmost tree to move right, and makes the two points on y=x move in opposite directions, breaking that row of three. $\endgroup$
    – AxiomaticSystem
    Commented Jul 5 at 14:06
  • 2
    $\begingroup$ Yea, Weather Vane's concern is too nit-picky. Though the improved question wording makes it better too. $\endgroup$
    – justhalf
    Commented Jul 7 at 2:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.