problem snowballs: "With odd number of children in field, distances between are all different."

the distances between the children are all different, so an equilateral triangle is not possible
 
stephanson12 said:
Hello, is there anyone who can help me with this problem?
Thanks!

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First, as has been pointed out, the word "so" is inappropriate. I will suppose that it's supposed to say "and the distances are all different". We have to be told that, not to infer it.

Now, I'd think about the meaning of the statement. As I read it, everyone who is the nearest to somebody will be hit; so you need to prove that there will always be someone who is nobody's nearest target. Do you agree?
 
indeed, you have to prove that there will always be someone who is no one's nearest target.
 
So, what ideas do you have? What have you learned that might be useful? (We have no idea what subject you are studying, or whether this is related to a class.)

I might start by considering a small number, like 1, 3, or 5, and picturing what might happen. I might also consider an even number, and see if there's a reason they require an odd number. This is all part of understanding the problem itself, but might give insight into how to prove it.

I haven't solved it yet; I'm just thinking with you. But it reminds me of the pigeonhole principle, and if that's something you've learned, then you might consider whether it could be used.
 
You can also prove that at least 1 of the children will always have 2 or more snowballs thrown at him. Because because of this there will always be someone who doesn't get hit
 
If you start with a random person, and this person throws the snowball to the person closest to him, and it continues in a chain until the last person (for example if they are in the shape of a circle), where each person is hit only once and then throws at the next person in the chain, the last person will not be able to throw at the first person who has not yet been hit, otherwise the first person would have to have thrown at the last person and that is not the case. The last person then has to throw at the penultimate person, which means he is hit twice and the first person is never hit. So the problem is solved for this part.
But if a person of choice from the chain does not throw to the next person in the chain, but back to the previous one, the previous person will have been hit 2 times and therefore there will always be a person who will never be hit.
But here I have only explained it if they are in a chain, in the shape of a circle. I don't know if this also applies if they are randomly mixed together.
 
stephanson12 said:
If you start with a random person, and this person throws the snowball to the person closest to him, and it continues in a chain until the last person (for example if they are in the shape of a circle), where each person is hit only once and then throws at the next person in the chain, the last person will not be able to throw at the first person who has not yet been hit, otherwise the first person would have to have thrown at the last person and that is not the case. The last person then has to throw at the penultimate person, which means he is hit twice and the first person is never hit. So the problem is solved for this part.
But if a person of choice from the chain does not throw to the next person in the chain, but back to the previous one, the previous person will have been hit 2 times and therefore there will always be a person who will never be hit.
But here I have only explained it if they are in a chain, in the shape of a circle. I don't know if this also applies if they are randomly mixed together.
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I don't really understand the logic here, but if you have a chain [imath]p_1, p_2, ..., p_n[/imath] in which person [imath]p_k[/imath] throws ball at person [imath]p_{k+1}[/imath] for [imath]1 \leq k < n[/imath], and [imath]p_n[/imath] throws the ball at [imath]p_1[/imath], then what can you say about the distances between [imath]p_k[/imath] and [imath]p_{k+1}[/imath]?
 
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