The
Mark of Zorro
With
just 3 straight lines through the zed above (fig.
1) form the LARGEST possible number of triangles (see example).
You must also prove that your solution is the best. Hasta
luego!
Just for an anecdote, few people know
that Guy
Williams, the best Zorro ever, was fond of mathematics,
chess and astronomy...
Difficulty
level: ,
basic geometry knowledge. Category:
dividing-the-plane puzzle. Keywords:
triangles, line segments. Related
puzzles:
- Red monad,
- Stairs to square.
- Sotto
il segno di Zorro
Disegna 3 linee rette sulla zeta qui sopra in modo
da formare il maggior numero di triangoli possibile
(vedi esempio).
Devi anche provare che la tua soluzione è la
migliore possibile... Hasta luego!
- Zorro
est arrivéééé...
Trace trois droites sur le Z de façon à former
le plus grand nombre de triangles possible (selon
l’example).
Tu dois également prouver que ta solution
est bien la meilleure... Hasta luego!
As
you can observe in the fig. 3 above, for obtaining
the greatest number of regions when dividing a surface
with straight lines each line segment MUST intersect
ALL the other ones, and by any intersection
point should pass ONLY two lines. The example
in fig. 3C meets all the criteria we have outlined,
thus it contains the greatest number possible of
regions: 7 regions instead of 6.
Applying
to our problem the empirical criteria above, we obtain
the following diagram:
Therefore,
the maximum number of triangles obtainable by intersecting
the large Z with 3 segment lines is SEVEN.
Notice that each of the 3 segment lines touches the
two others + the diagonal and both parallel lines forming
the capital letter Z.
The
Winners of the Puzzle of the Month are: Serhat Duran, Turkey - J.
R. Odbert, USA - Vishal
Dixit, India - Congratulations!
Math
fact behind the puzzle
Calling
the number of line segments n,
it can be demonstrated that the maximum number of
portions a convex polygon can be divided with them
is: (n2 + n + 2)/2 [see
fig. 3 above]
but in the case of a concave polygon such as a crescent-like
figure (see fig. 5) it will be: (n2 +
3n + 2)/2
The numbers generated by the first formula (e.g. 2, 4,
7, 11, 16, 22, ...) are called “Pizza
numbers”.