Hard math problem. help me out please?

So A month ago, my teacher assigned us a challenge problem (I'm only in 8th grade so it shoudn't be too hard for you) I've spent about 15minutes a day on it and I still haven't finished it. So a friend of mine talked about posting it on on Yahoo! answers, and I thought, hey! I wonder if someone from southperry can figure it out? The problem is below. help me out pl0x

Tommy's mom is making him a cake for his birthday. The cake will be square (but not necessarily a cube- the height is irrelevant) and will be frosted on the top and on all four sides. Tommy and four of his friends want to share the cake among the five of them so that each person will get the same amount of cake and also the same amount of frosting. How should he cut the cake? (all cuts should be made perpendicular to the base)

8th Grade challenge prolem, I know you guys can do it

like that?

 Spoiler

how many cuts is he allowed to make?

like that?

 Spoiler

how many cuts is he allowed to make?

I guess you can do it with 5. This one's quite hard... let me think about it a bit. But in the meantime:

-The perimeter MUST be divided by 5.
-The Area, too.

EDIT: GOT IT! doodling... please stand by...
reEDIT: Oh wait... nevermind... Need to rethink it a bit...

FINAL EDIT: AHA! YEAH! LET'S PLAY SOME TETRIS, MOTHER PINEAPPLER

like that?

 Spoiler

how many cuts is he allowed to make?

That sould work for a circle, but the cake is square. you are allowed to make as many cuts as you want

I guess you can do it with 5. This one's quite hard... let me think about it a bit. But in the meantime:

-The perimeter MUST be divided by 5.
-The Area, too.

EDIT: GOT IT! doodling... please stand by...
reEDIT: Oh wait... nevermind... Need to rethink it a bit...

I did it by dividing a square into 25 squares. each piece must have 5 squares, and be attached to 4 frosted edges. But corners are 2 edges, and that's what threw me off

I did it by dividing a square into 25 squares. each piece must have 5 squares, and be attached to 4 frosted edges. But corners are 2 edges, and that's what threw me off

Look my final edit ;D

I guess you can do it with 5. This one's quite hard... let me think about it a bit. But in the meantime:

-The perimeter MUST be divided by 5.
-The Area, too.

EDIT: GOT IT! doodling... please stand by...
reEDIT: Oh wait... nevermind... Need to rethink it a bit...

FINAL EDIT: AHA! YEAH! LET'S PLAY SOME TETRIS, MOTHER PINEAPPLER

you are my hero

oops didnt see square.

anyway, here you go

http://mathforum.org/library/drmath/view/55161.html

oops didnt see square.

anyway, here you go

http://mathforum.org/library/drmath/view/55161.html

Uhm... Does it have to be cut from the center point too...?

Anyways, my solution also cuts from the middle, technically

I had to think a bit.

I did it myself but this contains my answer: http://mathforum.org/library/drmath/view/55161.html

and explains it much better then I could. I split it up into triangles and then realised I could do this.

The areal and volume must then be equal for all persons. As we don't know the cake's height, we have to give all people the same amount of frosting from the sides. Apart from that, the volume they gain should also be equal.

This means that, as long you cut vertically, all people should have 1/5 of the areal of frosting from the sides, and 1/5 of the areal of the frosting from the top. They will thus get the same amount of cake.

However, how do we find the areal of the frosting from the sides? Well, as we may write the areal of the sides as 4*s*h, where both s and h is unknown, we can, as long we cut vertically, skip the h because this will be fixed by the vertical slices. The amount of frostings on each side each person will get is then 4l/5, or 4/5*s where s is the width/length of the cake.


So Alloy's right, this will then be at least one right possibility. 3 other ways is made by rotating this pattern 90 degrees.

Edit: Well, I just saw perpendicular, but I couldn't exactly figure out where "base" is. If you mean by cutting it vertically, then the upper is correct. If though, they mean that all cuts should be perpendicular, then the upper answer is wrong, though they get the same amount of cake and frosting.
Cut the cake into 25 equal squares (4 cuts on each side). Now, let's say that the cake's sides will equal 1 frosting unit on the sides, and the corners will equal 2 frosting units on the sides. We have 4 corners, and 3 * 4 sides, therefore we got 20 frosting units which we will have to divide to everyone.

There are numerous ways to get the right answer, this is only one of the ways:

Here's one way that all cuts go through the centre

Here's one way that all cuts go through the centre

Fantastic. The 5th kid will starve to death, though.

EDIT: OH I see what you did. Is the same Area on the 4 red parts and the others? There's Icing on the top, too.

If we assume side length of 5 (since it's a square)

The 4 triangles have base 4, height 2.5 = area 5
They also have 4 units of perimeter.

If you subtract 4*5 = 20 from the total area of 25, you find that the red areas total 5. Plus they have 1 unit of perimeter at each corner, for 4 total.

So yes, it is equally split 5 ways.


This technique also leads to a less rotationally symmetric division, but single pieces for each user:
"cut" it into 4 triangles, base 5 height 2.5 (centre point being the middle of each).
Put them in order so the "side" of the cake is along a line.
Mark the slices so that each has a base of 4, and cuts to the centre point. Fold it back up, make the cuts.



(I believe this is the same split used on the link given earlier)

Any split like this, done only with lines from the base to the point above them, with the length of the base for each person totalling 1/5 of the total, will be split equally. For example, cut each triangle into 5 pieces with base 1. Then give each person 4 of those pieces (of 20 total) for 1/5 of the cake each.

Odd, I thought the area was different on those ones. I did think of that one, but discarded it, as you can check in my post, before...

Don't you think the pink area is quite bigger than the green one?

And yes, the 5th getting 4 parts seems the best solution, unless the teacher is a Tetris fan.