At vertices of a convex polygon with 2n sides, sit hunters. while in its interior and not on any diagonal,sits a fox. The hunters shoot at the fox all simultaneously, but the fox ducks. the bullets hit the sides of the polygon.
Prove that at least one side is not hit. ;)
Source:[CRP]
1still i am not convinced ..
will the bullets from the upper vertices only hit the sides below the diagonal ?
my figure doesnt give the result ...
1not x times celestine
see we have to cover all the vertices of the polygon wont we ?
we see that when we choose x we automatically have to choose another x-1
so let us complete choosing all vertices when we choose let a ..
then automatically we get another a-1
so
a+a-1=2n
thus a contradiction
9yes but still cant make sense of #3
x -1 things repeated x times is x(x-1) right ??
1i considered the vectors as the ones joining the fox and the vertices of the polygon
1oh i took it carelessly earlier
now i got the fact :) .. good approach
1yeah. let it intersect.
then number of lines going below the diagonal shown decreases, isn't it . so we have at most n-1 lines going below.
1if the fox is little above then the line from n will intersect in the upper half ...
1but ithpower didnt you consider the bullets fired by the ones at the diagonal ?
they will also hit one of the n sides wont they?
1Dont see the solution now:
Solution:
The fox , not on any diagonal is in one half of the polygon(i.e. above or below the diagonal joining 1st and (n+1)th vertices).let it be in upper half.
Then there are at most (n-1) bullets from above hitting n sides below. so atleast one side unharmed.
1Come on b555. I would like to see your solution.
Anyway, there is an elegant non-counting mathematical solution
1this problem essentialy requires that we cant have an even number of radial vectors where each vector when extended will pass through a region unique to it ..
this can be proved by contradiction ..
suppose we have ..
then observe that for any two vectors we will have one vector only in the region formed by extending them
so for any three we will have 2 ..
for any x we will have x-1
we will cover all the points in this way ..
suppose we do it x times and get all the points ..
we get
x+x-1=2n
2x-1=2n
contradiction
