Very easy but tricky

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]

18 Answers

1
Rohan Ghosh ·

still i am not convinced ..

will the bullets from the upper vertices only hit the sides below the diagonal ?

my figure doesnt give the result ...

1
Rohan Ghosh ·

not 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

9
Celestine preetham ·

yes but still cant make sense of #3

x -1 things repeated x times is x(x-1) right ??

1
satan92 ·

i considered the vectors as the ones joining the fox and the vertices of the polygon

9
Celestine preetham ·

rohan can u explain #3 more clearly ? dint get u pls

1
Rohan Ghosh ·

oh i took it carelessly earlier

now i got the fact :) .. good approach

1
ith_power ·

yeah. 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.

1
Rohan Ghosh ·

if the fox is little above then the line from n will intersect in the upper half ...

1
ith_power ·

let me give a figure:

Check that the polygon is convex.

1
Rohan Ghosh ·

hmm a good problem ..

wait posting soln

1
ith_power ·

Sorry my mistake in typing . consider bullets from upper vertices.

1
Rohan Ghosh ·

but ithpower didnt you consider the bullets fired by the ones at the diagonal ?

they will also hit one of the n sides wont they?

1
ith_power ·

Dont 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.

1
Rohan Ghosh ·

a more elegant one?

hmm let me think ..

1
ith_power ·

Come on b555. I would like to see your solution.
Anyway, there is an elegant non-counting mathematical solution

39
Dr.House ·

oh! sorry !

1
Rohan Ghosh ·

this was not done ..

1
Rohan Ghosh ·

this 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

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