The Lemon Meringue Pie Problem
 

You may assume that the game takes place in an unobstructed horizontal plane.

Axioms
1) A sentient human being holding one and only one Lemon Meringue Pie is a player.

2) There are more than 1 players.

3) If any two players are n feet apart, they are the only two players who are n feet apart.

4) At the signal, each player will throw their pie at the player who is nearest to them. All thrown pies are direct hits.

Questions:
1) What is the maximum number of pies that any player can be hit by? (Pies only count if they are thrown by players).

2) If A_1 hits A_2, who hits A_3 ... A_n who hits A_1 forms an n-gon, prove (or dis-prove) that no n pie throws can form a closed n-gon.

3) Prove (or dis-prove) that no two pie throws can cross.

I have a couple of further questions but I haven't been able to prove the answers yet, so try these for now and we'll see how it goes.

For (3), define "cross". For instance, with two players, they will obviously throw their pies at each other. Does that mean their throws cross completely? :unsure:

[quote=ken_g6]Define "cross"[/quote]

Good question ken_g6, I thought I had everything covered but obviously not.

If A is the set of points constituting the path of pie a, and B is the set of points constituting the path of pie b, and A contains exactly one element of B, then a and b cross. If not, then not. So two pies with identical paths with opposite velocities do not cross.

[QUOTE=Numbers]If A is the set of points constituting the path of pie a, and B is the set of points constituting the path of pie b, and A contains exactly one element of B, then a and b cross. If not, then not. So two pies with identical paths with opposite velocities do not cross.[/QUOTE]

So if A & B both throw at C, then do they (line segments AC & BC), cross path at the single point C?

Part (2):

[spoiler]If A_2 is hit by A_1, but hits A_3, then A_3 must have been closer to A_2 than A_1 was.

Therefore, if there are N players 1..N, N>=3 such that for n in {2..N} A_n-1 hits A_n, n<=N, for n in {3..N}
the distance between A_n-1 and A_n must be less than the distance between A_n-2 and A_n-1.

Now assume A_N is A_1. But due to the sequence above, A_N-1 is closer to A_1 than A_2 is. So no N pie
throws can form a closed circuit for N>2. But a closed n-gon must have 3 or more sides. So no N pie
throws can form a closed n-gon.[/spoiler]

[quote=axn1]So if A & B both throw at C, then do they (line segments AC & BC), cross path at the single point C?[/quote]
I tried quite hard to make the wording of the problem as accurate as I could, but the first two people to respond have shown (correctly) that I didn't try hard enough.

No, these two pies do not cross. We could consider that poor C will get hit by two pies thrown from different directions, and each will land on a different side of his face. So that although the Meringue from one pie will meet the Meringue from the other, the pies as pies do not cross. But that isn't either very mathematical or completely obvious. So, a new definition of cross.

When a and b are two pies thrown at different players, A is the set of points constituting the path of pie a, and B is the set of points constituting the path of pie b. If A contains exactly one element of B, then a and b cross.

I had actually hoped you might try to prove some special cases first. For example, when there are four players the question asks you to prove that no four pies will form a rectangle. The proof of this special case could then be considered a 4-gon conclusion.

[QUOTE=Numbers]I had actually hoped you might try to prove some special cases first. For example, when there are four players the question asks you to prove that no four pies will form a rectangle. The proof of this special case could then be considered a 4-gon conclusion.[/QUOTE]

[spoiler]Well, that's easy. Four players can't be on a rectangle :wink:
If we consider some "trapezoidal" pattern, than we can show that in the set of distances N, there
must be a smaller one. If that smaller one does exist, than both players in the extremes would
throw the apples at each other, breaking the sequence. [/spoiler]

This makes me think of that famous painting with the rectangles that 'surround' each other in an outgoing spiral because each is bigger then the last. I'll have to look that one up and post a link. Your model doesn't have the rectangles but it does behave in a similar manner.

Part 1: Maximal Number of Pies
[SPOILER]
The most pies any one person can be hit with is 5.

Start with the target "T" and someone "A" who will hit him. Consider the
possible placement of another person "B" to also hit target "T". B must be
closer to T than to A; if not, A is a better target for B than T. Hence B
and T must be on the same side of the perpendicular bisector of AT.

B must also be further from A than T is from A, else B would be a better
target for A that T is. Hence B must be outside the circle centered on A that
goes through T.

Now start at T and consider the minimal possible angle for BTA. This would
place B at the intersection of the circle and the line, which is 60 degrees. B
cannot be AT the intersection because that would violate the equal distance
rule, so the angle BTA must be greater than 60 degrees.

You can repeat this analysis, placing a new thrower that hits T at 60 +
epsilon degrees, placing 5 throwers. You cannot place a sixth thrower
because that thrower would be less than 60 degrees from A.
[/SPOILER]

I don't know where Fetofs got his apples from, but his answer is otherwise correct, as it is more or less a restatement of ken-g6's generalisation which is also correct.

wblipp is to be commended for his perfect solution to question 1.
[spoiler]In my solution I only considered the throwers to be on the perimeter of a circle,[/spoiler]
[spoiler]and the distance between them has to be greater than the radius of the circle,[/spoiler]
[spoiler]leading quickly to wblipp's solution without the need to consider perpendicular[/spoiler]
[spoiler]bisectors, but it is essentially the same solution.
[/spoiler]

Now, anyone fancy a go at question 3?

part 3 - LaTeX does not like spoilers

[spoiler]Suppose that B throws at A who is R away, and that C is to throw at D such that CD crosses BA.[/spoiler]

[spoiler]Let us view this geometricly, creating a circle of radius R about B[/spoiler]
[spoiler](which nobody except B can be within) centered at (0,0) with A at (R,0).[/spoiler]

[spoiler]Suppose that C is at a position (X1,Y1) with Y1>0 outside the circle,[/spoiler]
[spoiler]and D is at a position (X2,Y2) with Y2<0 outside the circle.[/spoiler]

[spoiler]We then break this into three cases for C.[/spoiler]

[spoiler]1) If X1<0 then C is closer to B than D (smaller dx, dy)[/spoiler]

[spoiler]2) If 0<X1<R then C is closer to A than D,[/spoiler]
[spoiler]which can be seen by noting D is on the opposite side of the line through A[/spoiler]
[spoiler]perpendicular to AC (or the arc through A with vertex C and radius AC).[/spoiler]

[spoiler]3) If R<X1 then C is closer to A than D (smaller dx, dy)[/spoiler]

[spoiler]Edit: Note that the 1) and 3) reasoning hold because CD is said to intersect BA[/spoiler]

[spoiler]I have a picture to better illustrate the middle case,[/spoiler]
[spoiler]which can be posted at a later time if there is an interest.[/spoiler]

I got a little confused by your second case.
[spoiler]Since all you say about C and D is that they are both outside[/spoiler]
[spoiler]the circle centred on B, then they could both be equidistant[/spoiler]
[spoiler]from A and B so that a perpendicular to AC could in fact be AD.[/spoiler]
[spoiler]Which means that D is not on the opposite side of this line as[/spoiler]
[spoiler]you claim. But this is irrelevant to your answer because in that[/spoiler]
[spoiler]case CD would be a perpendicular bisector of AB at the midpoint,[/spoiler]
[spoiler]which by definition is the points equidistant from A and B. So if[/spoiler]
[spoiler]either (or both) C and D are on this line then they are closer to[/spoiler]
[spoiler]either A or B than they are to each other. I'm sure you considered[/spoiler]
[spoiler]this and simply failed to include it in your post.[/spoiler]

[spoiler]For my solution I considered AB to be a cord of a circle centred on C[/spoiler]
[spoiler]from where it was pretty straightforward to show that D (who has to be[/spoiler]
[spoiler]inside the circle) must be closer to one end of the cord than he is to[/spoiler]
[spoiler]any point on the opposite side of it, unless he is sufficiently far away[/spoiler]
[spoiler]that C must suffer the same fate.[/spoiler]

[QUOTE=Numbers]I got a little confused by your second case.
[spoiler]....in that case CD would be a perpendicular bisector of AB at the midpoint,[/spoiler]
[spoiler]which by definition is the points equidistant from A and B.[/spoiler]
[/QUOTE]

[spoiler]From here we know this cannot be the case since we are told that[/spoiler]
[spoiler]"If any two players are n feet apart, they are the only two players who are n feet apart." (Rule 3)[/spoiler]
[spoiler]so this need not be considered further - but was an oversight originally on my part when posting[/spoiler]