[EventHorizon]

Mike was at a store before the big game.

He gets a call from his coach who says that the lucky game ball (a sphere) has 5 small holes in it (each a single point). Without the lucky game ball the team is sure to lose.

Mike goes to the ball repair aisle and finds the following:

5 single hole (fixes 1 point) repair kits for $2 each

2 half ball (fixes all points on a closed hemisphere) repair kits for $5 each

What is the least amount of money Mike can spend and still repair the lucky game ball?

[Guest]

I'm sure I'll get this wrong, but here goes.

He could spend a minimum of $5 dollars, assuming all points are on a closed hemisphere. Or, he could buy a new game ball for under $10 and convince the coach he's fixed the old one, and the power of suggestion does the rest....

[Guest]

$9

[bonanova]

Dividing the ball into hemispheres along a dividing line that does not go through one of the holes divides the holes into groups of 0-5, 1-4, 2-3, 3-2, 4-1 or 5-0 holes. Worst case is 2-3 or 3-2. In either of these cases a half-ball patch can cover 3 holes. Two single hole patches cover the others. With luck, 0-5 or 5-0, the half-ball patch works by itself. But we want to ensure success.

[Guest]

9 dollars

5 for a hemisphere

2 $2 single hole patches.

This will cover them all.

it's possible all holes are on one hemisphere and thus would be able to return the 2 pathces.

[EventHorizon]

Mike was at a store before the big game.

He gets a call from his coach who says that the lucky game ball (a sphere) has 5 small holes in it (each a single point). Without the lucky game ball the team is sure to lose.

Mike goes to the ball repair aisle and finds the following:

5 single hole (fixes 1 point) repair kits for $2 each

2 half ball (fixes all points on a closed hemisphere) repair kits for $5 each

What is the least amount of money Mike can spend and still repair the lucky game ball?

I wrote this up quick. Change the last sentence to...

Without knowing where the holes are, what is the least amount of money Mike can spend and still be guaranteed to repair the lucky game ball?

With that said.....here's a quick spoiler for the answer....but can _you_ prove it?

$7

[EventHorizon]

A little fyi, a closed hemisphere is one that includes the circumference. So it's like if you chose the northern hemisphere of the earth, it would include the points on the equator.

[Guest]

Mike was at a store before the big game.

He gets a call from his coach who says that the lucky game ball (a sphere) has 5 small holes in it (each a single point). Without the lucky game ball the team is sure to lose.

Mike goes to the ball repair aisle and finds the following:

5 single hole (fixes 1 point) repair kits for $2 each

2 half ball (fixes all points on a closed hemisphere) repair kits for $5 each

What is the least amount of money Mike can spend and still repair the lucky game ball?

7$ in worst case.We can have all the holes from one side at the ball 5-0 or 0-5(5$),or 4-1 or 1-4(6$),or 2-3 or 33-2(7$)-worst case.

[Guest]

A little fyi, a closed hemisphere is one that includes the circumference. So it's like if you chose the northern hemisphere of the earth, it would include the points on the equator.

At first I had the same answer as bonanova, for the same reason. After reading your answer, I actually grabbed a ball, and tried to space out 5 dots, so that there was the most space between them. Which turned out to be, looking face on, one dot on top, one on the bottom, one on the rightmost edge, one on the leftmost edge, and the fifth being either the closest point toward you, or farthest point from you. And using that as a worst case scenario for spacing.. a half a ball still covers 4 dots.

Getting 4 dots on a sphere, the best case spacing is 1/4 the circum., which stays the same up to 6 dots. So if you center the half ball repair kit on one of the dots, it reaches 1/4 the circum. in all directions, covering all but the one dot on the oppisite side of the sphere. so for his 7$.. he could fix anywhere from 4 to 6 holes.

[EventHorizon]

7$ in worst case.We can have all the holes from one side at the ball 5-0 or 0-5(5$),or 4-1 or 1-4(6$),or 2-3 or 33-2(7$)-worst case.

don't you mean

... 4-1 or 1-4(7$),or 2-3 or 33-2(9$)-worst case.

?

For $6, the only way to spend that much is 3 single point patches....and you can only fix 3 holes that way.

[EventHorizon]

At first I had the same answer as bonanova, for the same reason. After reading your answer, I actually grabbed a ball, and tried to space out 5 dots, so that there was the most space between them. Which turned out to be, looking face on, one dot on top, one on the bottom, one on the rightmost edge, one on the leftmost edge, and the fifth being either the closest point toward you, or farthest point from you. And using that as a worst case scenario for spacing.. a half a ball still covers 4 dots.

Getting 4 dots on a sphere, the best case spacing is 1/4 the circum., which stays the same up to 6 dots. So if you center the half ball repair kit on one of the dots, it reaches 1/4 the circum. in all directions, covering all but the one dot on the oppisite side of the sphere. so for his 7$.. he could fix anywhere from 4 to 6 holes.

That's actually the method I initially approached the version of this problem that I saw. It doesn't really prove it though.

A technique similar to Bonanova's but with being a little more picky about hemisphere placement ends up being a pretty good proof.

[bonanova]

I wrote this up quick. Change the last sentence to...

Without knowing where the holes are, what is the least amount of money Mike can spend and still be guaranteed to repair the lucky game ball?

With that said.....here's a quick spoiler for the answer....but can _you_ prove it?

$7

I think $7 is correct. It also seems the ball could have had six holes.

[Edit. This is only slightly beyond the proof given by Flight in post #9.]

Place points at the north and south poles and place four more points on the equator 90 degrees apart.

Those six points - or equivalent placement that maintains their relative position - is the greatest degree of spacing possible.

Yet a closed hemisphere, placed with the center of its surface over any of those points, also covers four of the others at its rim.

Consider the hemisphere centered on the north pole and rim-covering the equator.

If any of the equatorial points are moved south of the equator, the hemisphere is moved to the south pole to recapture them.

If two equatorial points are moved [say] an inch south and the other two an inch north, then slide the hemisphere down to the equator.

The hemisphere now covers the points at the poles. It can then be slid around the equator to a longitude where it also covers three of those points.

Thus it seems the ball could have had six holes and still be repaired with two patches.

[bonanova]

Part A:

Place four points on a sphere so that they form a regular tetrahedron.

This is the greatest spacing for four points on a sphere.

Any set of three of these points forms a triangle coverable by a hemisphere.

Add a 5th point anywhere on the sphere.

Cover the triangle that encloses the 5th point.

Now let's try to make it more difficult.

Part B:

If the tetrahedron is deformed, at least one of the triangles will become smaller.

Call the triangle points [a b c] the fourth tetrahedron point [d] and the 5th point [e].

Place the hemisphere so that [e] is on its rim.

We need to show that by rotating the hemisphere about point [e] at some angle it will also cover 3 of the other 4 points.

If [e] is placed opposite a point outside of the triangle, then [a b c] and possibly [e] can all be covered at some rotation angle.

So let's assume [e] is placed opposite a point that lies inside the triangle [abc].

Then the three other points are [d] and two of [a b c]. Two things are known:

[1] As the hemisphere rotates, for 180 degrees it covers [d] and the other 180 degrees it doesn't.

[2] As the hemisphere rotates, it covers, in turn, (a b), (b), (b c), [c], (c a), (a), and again (a b).

The ranges of rotation angle where two of the triangle points are covered do not lie within a 180 degree interval.

Therefore, within the 180 degrees of rotation angle where [d] is covered there is a position that also covers two of the points of the triangle.

Thus, out of any 5 points on a sphere, 4 of them can be covered by a closed hemisphere.

[Guest]

Hi all, my first post here so go easy

5 points on a sphere. If you draw a (equatorial) circle around the sphere that goes through any two of them, dividing the sphere into two hemispheres, then either the 3 remaining points are in one of those hemispheres ($5) , or there are 2 in one and 1 in the other. place the point patch over the 1 and the hemisphere patch over all the rest, it will have 2 points on the edge of the hemisphere patch and (probably) two fully under it ($7)

Cheers,

Sam

edit: oh crikey, in the time it took me to register we've moved onto six holes? eeek!

Edited March 12, 2008 by Insanimal

[EventHorizon]

Hi all, my first post here so go easy

5 points on a sphere. If you draw a (equatorial) circle around the sphere that goes through any two of them, dividing the sphere into two hemispheres, then either the 3 remaining points are in one of those hemispheres ($5) , or there are 2 in one and 1 in the other. place the point patch over the 1 and the hemisphere patch over all the rest, it will have 2 points on the edge of the hemisphere patch and (probably) two fully under it ($7)

Cheers,

Sam

edit: oh crikey, in the time it took me to register we've moved onto six holes? eeek!

Exactly. Good Job.

[EventHorizon]

I think $7 is correct. It also seems the ball could have had six holes.

...

I believe I have come up with a counter-example for being able to cover any 5 out of 6 points on a sphere. (i.e., 5 is the most holes that can work and still have n-1 covered in a well-placed hemisphere)

Let four of the points form a square (just so that the points are not all on a circumference and so that the square has some area). If you take each of the those points and look at the point opposite them (through the center of the sphere), let this define the area that I will put the other two points. First let the new points be at the average of two consecutive corners of the new square (formed by the opposing points, not by the original points). Now move these two points closer together by just a little.

You can obviously get the four points that form the initial square. To get another point you would need to move the hemisphere enough to uncover at least one of the points that form the square.

So if there were 6 holes it would cost at least $9 to guarantee being able to fix the ball without knowing where the holes are in advance.

[bonanova]

Insanimal,

Kudos - a much simpler proof than mine.

Welcome to the forum!

EventHorizon,

I suspect you're right.

If you put the rim on one of the two opposite points, there would be a 180 degree interval that would cover the other opposite point, but nothing constrains the square to present three of its points inside that interval, or all four of its points outside that interval. In fact, by construction it never presents all four of its points.

Nice job.

[Guest]

I havn't read any solutions but here is a stab...

Since the ball has 5 holes in it. It is impossible for each hole to be equidistant from each other. This means you can make at least a triangular pyramid by connecting 4 points on the sphere. In the case of a perfect pyramid which requires 4 points to be equidistant from each and at the same angles then any 3 points must be in the same hemisphere and the 4th point perpendicular to the plane that is created by your original 3 points. Now the 5th point must be in the area that is "cut" of of the sphere by on of the planes the pyramid makes. You simply choose the plane that slices out the spare point and cover that side of the ball with half and fill in the last point with your single point. Now when you start shifting the points around this is still true. by connecting any 4 points (unless they are in the same plane which would make the issue easier since 4 points on a sphere in a plane HAVE to be in a hemisphere because the worst case the points existing on the circumference which would be covered by a hemisphere.) You still have a pyramid that divides the sphere into 4 sections. One of which has to contain at least 4 of the points and will be able to fit under a hemisphere because if it doesn't you turn the hemisphere over. I am fairly certain this covers all possibilities. It is an extrapolation of the proof of 1/2BH = area of a triangle.

[Guest]

I havn't read any solutions but here is a stab...

Since the ball has 5 holes in it. It is impossible for each hole to be equidistant from each other. This means you can make at least a triangular pyramid by connecting 4 points on the sphere. In the case of a perfect pyramid which requires 4 points to be equidistant from each and at the same angles then any 3 points must be in the same hemisphere and the 4th point perpendicular to the plane that is created by your original 3 points. Now the 5th point must be in the area that is "cut" of of the sphere by on of the planes the pyramid makes. You simply choose the plane that slices out the spare point and cover that side of the ball with half and fill in the last point with your single point. Now when you start shifting the points around this is still true. by connecting any 4 points (unless they are in the same plane which would make the issue easier since 4 points on a sphere in a plane HAVE to be in a hemisphere because the worst case the points existing on the circumference which would be covered by a hemisphere.) You still have a pyramid that divides the sphere into 4 sections. One of which has to contain at least 4 of the points and will be able to fit under a hemisphere because if it doesn't you turn the hemisphere over. I am fairly certain this covers all possibilities. It is an extrapolation of the proof of 1/2BH = area of a triangle.

Nice proof insanimal I think I definately over thought this. My proof seems to match bon's. And yours ...

Has to work since you can rotate the devideing plane along those 2 points until it creates 2 perfect hemisphers.

I agree kudo's.

[Guest]

Not entirely sure I understood the problem, but ...

5$ There's nothing in the problem statement that says the holes can't all be on one side of the ball. In fact, if the point of the holes is to introduce curving behavior, then having all five holes in one side would be logical. However, I suspect there's some other detail here you're not telling us.

[Guest]

Not entirely sure I understood the problem, but ...

I didn't, but as soon as I read a couple replies and re-read the problem, it was obvious what my mistake was. Incidentally, Insaminal's proof was super easy to grasp. Good work!

[Guest]

Well, here's my simple proof]5 points on a sphere. If you draw a (equatorial) circle around the sphere that goes through any two of them, dividing the sphere into two hemispheres, then either the 3 remaining points are in one of those hemispheres ($5) , or there are 2 in one and 1 in the other. place the point patch over the 1 and the hemisphere patch over all the rest, it will have 2 points on the edge of the hemisphere patch and (probably) two fully under it ($7)

I couldn't believe this solution, an d astonished how many of you agreed with it.

If you have a valuable ball having 2 holes on it, settled on north and south poles, would you agree to make a hemisphere shaped patch on it? Can a hemisphere cover both poles? Can anybody rely on that patch?.

Insanimal's solution depends on it. If you agree with me, than all other proofs of 7$ are false. That is my proof:

Simply draw a pentagon on the equator of ball. Make holes on the points. None of the hemispheres you would draw, will cover more then 3 holes. As you admitted, a north (or south) hemisphere will not cover them totally.

Thus, Bonanova's solution is false too. Anyhow he draws a tetrahedron. I think he should draw a pentahedron instead.

[bonanova]

I couldn't believe this solution, an d astonished how many of you agreed with it.

If you have a valuable ball having 2 holes on it, settled on north and south poles, would you agree to make a hemisphere shaped patch on it? Can a hemisphere cover both poles? Can anybody rely on that patch?.

Insanimal's solution depends on it. If you agree with me, than all other proofs of 7$ are false. That is my proof:

Simply draw a pentagon on the equator of ball. Make holes on the points. None of the hemispheres you would draw, will cover more then 3 holes. As you admitted, a north (or south) hemisphere will not cover them totally.

Thus, Bonanova's solution is false too. Anyhow he draws a tetrahedron. I think he should draw a pentahedron instead.

The OP says that a $5 patch fixes all the points on a closed hemisphere.

Your case five points on the equator is a $5 fix.

[Guest]

The OP says that a $5 patch fixes all the points on a closed hemisphere.

Your case five points on the equator is a $5 fix.

I would accept it if we were talking about points. Because a point has no area. But a hole always has an area. Anyway, if OP says so, then OK. Even I don't know who OP is, maybe the "big brother"

[bonanova]

I would accept it if we were talking about points. Because a point has no area. But a hole always has an area. Anyway, if OP says so, then OK. Even I don't know who OP is, maybe the "big brother"

OP = Original Post or Poster.

A riddle is defined by the words used to pose it.

A solution is an answer that fits the conditions of the riddle.

Agree, if a hole had zero area, you wouldn't have to patch it.