Professor Erasmus has returned from his saturday walk in the park. He has counted the number of trees in the park and also the number of lines formed by these trees. Professor Erasmus claims that

  • there are exactly 26 trees in the park,
  • there are exactly 306 (distinct) lines formed by these trees,
  • and that none of these lines does contain four or more trees.

(When the professor speaks of trees on a line, then he means that the centerpoints of the trees lie on a common line.)

We wonder: Has the professor once again made one of his well-known mathematical blunders, or does such a collection of 26 trees indeed exist?

  • 1
    $\begingroup$ Swords at sundown? $\endgroup$ Mar 22, 2015 at 16:43

2 Answers 2


The maximum possible lines is $\sum 25=\frac{25*26}2=325$

The only way to reduce the line count is to put a point into a pre-existing line, which reduces exactly $2$ lines. Instead of $3$ lines $AB$, $BC$ and $CA$, we have a single line $ABC$. We are not allowed to add another point to the line.

Therefore, we can not have an even number of lines, as subtracting $2$s will only result in another odd number. Professor Erasmus has made a blunder.


Expanding on the previous answer - the 325 is right but the second paragraph is wrong - using the 325 lines, you could merge two lines into one - reducing the total amount by 1. Such merger need to be carried out 19 times. By selecting 2 points out of the 26 it seems that you could select 19 points of the remaining 24 to create 19 mergers.

  • 3
    $\begingroup$ The question specifies that no line contains four or more trees. A merged line must contain at least four. $\endgroup$
    – Callidus
    Feb 14, 2015 at 22:11
  • $\begingroup$ Merging two lines that have a common point will have three trees! $\endgroup$
    – Moti
    Feb 16, 2015 at 7:24
  • $\begingroup$ Meaning the common point to be a tree! $\endgroup$
    – Moti
    Feb 16, 2015 at 7:25
  • 2
    $\begingroup$ Three non-collinear points define a triangle, which has three lines. In making them collinear, you merge all three lines into one, reducing the number of lines by two. That's the scenario that @ghosts_in_the_code outlined. $\endgroup$
    – Callidus
    Feb 16, 2015 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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