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Math Puzzle: Integer Points

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Re: Math Puzzle: Integer Points

#32

Post by nimnam »

choose two integers x and y. These represent the first point on a plain. Only adjacent points can be added i.e. any point can only be 1 integer away from this origin point otherwise there will exist an integer point between the origin point the chosen point. so now we have (x, y), (x+1, y), (x, y+1), and (x+1, y+1). Adding a fifth point anywhere on the graph will make it such that an integer point will exists between two of the selected points.

Re: Math Puzzle: Integer Points

#33

Post by fahadkhan »

nemo1618 wrote: post #28

While we're all solving puzzles, can anyone help me with this one? No doubt you've doodled this shape in your graph paper before: http://i.imgur.com/oY29sBc.png What function does this slope approximate? It almost a circle, but not quite.

sqrt(x)+sqrt(y)=C

Re: Math Puzzle: Integer Points

#36

Post by chnuschper »

nemo1618 wrote: post #28

While we're all solving puzzles, can anyone help me with this one? No doubt you've doodled this shape in your graph paper before: http://i.imgur.com/oY29sBc.png What function does this slope approximate? It almost a circle, but not quite.

It's y=(sqrt(x) - 1)^2

Consider the function l(t) which gives the appropriate line function f(x) for the line going through (t,0). The function we are looking for can then be defined as c(x) := max(l(t)(x), t in (0;1]). Just solve for t and pop it into l(t), et voila.

Re: Math Puzzle: Integer Points

#37

Post by thejteam »

If I am remembering correctly, this was on the ARML(American Regions Math League) competition "power question" roughly, say 16-17 years ago. The answers given here from nilkn and the children comments are correct.

Re: Math Puzzle: Integer Points

#38

Post by jules »

chnuschper wrote: post #36
nemo1618 wrote: post #28

While we're all solving puzzles, can anyone help me with this one? No doubt you've doodled this shape in your graph paper before: http://i.imgur.com/oY29sBc.png What function does this slope approximate? It almost a circle, but not quite.

It's y=(sqrt(x) - 1)^2 Consider the function l(t) which gives the appropriate line function f(x) for the line going through (t,0). The function we are looking for can then be defined as c(x) := max(l(t)(x), t in (0;1]). Just solve for t and pop it into l(t), et voila.

Sorry, I accidentally down-voted you when I tried to up-vote you. This answer is absolutely correct. Here is my very similar derivation:

All lines go through (t,0) and (0,1-t). Now take two such lines, one with constant t and another with constant s and find their intersection. We have:

    y = 1-t - (1-t)/t * x
    y = 1-s - (1-s)/s * x
So:

    1-t - (1-t)/t * x = 1-s - (1-s)/s
    s-t = (s-t)/(st) * x
So:

    x = st
The points on the curve will be generated by the intersection of two almost adjacent lines (this is easy to see geometrically), so we take s =~ t. Then we have:

    x = t^2
    y = (1-t)^2
So we have:

    y = (1 - sqrt(x))^2
Or, my favorite form:

    sqrt(x) + sqrt(y) = 1

Re: Math Puzzle: Integer Points

#40

Post by fahadkhan »

jonsen wrote: post #35
fahadkhan wrote: post #34

It doesn't hold. Pick (0,1), (0,2), (0,3), (0,4), (0,5) on the plane z = 0

Well, you have five integer points on the same line.

"Suppose we arbitrarily choose 5 integer points in a plane.

Show that we can always find 2 among these 5 integer points such that the line segment joining the 2 points contains at least 1 more integer point."

I am not sure what you mean. I have 'arbitrarily' picked 5 integer points on a plane. Yes, they happen to lie on line. All their line segments contain only these integer points.

Perhaps you mean that is line not on "a plane"? In that case (0,0), (0,1), (0,2), (0,3), (0,1) is another counter example.

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