2
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As a continuing series of "divided by".

Today is my Birthday! I wil be cutting my own birthday cake and share equally to 11 of my friends.Assuming that I do straight cuts and there is no topping (I don't have money for topping :/) on a non-flat round cake, what is the minumum cuts needed for me to cut equally the cake for myself and friends?

Note:No, its not "0" this time! It's a legitminate answer!

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    $\begingroup$ Easy: 0 cuts. Your 11 friends all refuse to eat your toppingless cake, leaving you with the whole cake to yourself. Happy birthday! $\endgroup$ – Ian MacDonald Apr 29 '15 at 15:52
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    $\begingroup$ @IanMacDonald God dammit lol.Did I not put that the answers is not "0".And of course assume they eat.Otherwise they come to my birthday party for what? Nice joke though $\endgroup$ – ministic2001 Apr 29 '15 at 15:55
  • $\begingroup$ I cant believe you would waste the extra 6 slices you could have had! $\endgroup$ – Ewan Apr 30 '15 at 13:25
  • $\begingroup$ Also, using a straight knife is criminally wasteful! think of the starving children! $\endgroup$ – Ewan Apr 30 '15 at 13:27
6
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The minimum is

$4$ cuts

First you draw $2$ parallel lines to divide the cake in 3 parts, then you draw a perpendicular line, getting 6 equal pieces. At the end, you make a longitudinal cut to split all the pieces, getting 12 parts (one for you and 11 for friends). See this picture:
enter image description here

Another equivalent solution, perhaps simpler (and symmetrical), is this:
enter image description here

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  • $\begingroup$ How do you ensure is equal? I mean isnt there an EASIER way to make it equal? Because it seems estimating it 3 equal pieces in the first place is already difficult.The solution in my head is simple.Plus all same size coincidencely. $\endgroup$ – ministic2001 Apr 29 '15 at 15:58
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    $\begingroup$ That second pic is the same as my answer. Just saying :'you could cut the cake in half vertically and then make 3 cuts to divide the halves into 6 pieces each. But you'd have to be a walking computer to do that' $\endgroup$ – Spacemonkey Apr 29 '15 at 16:03
  • $\begingroup$ Yeah.You alternative answer is correct.My solution was 4 cuts too.Cut the cake into 4 equal pieces.One cut is horizontal, one cut is vertical.Then cut the thickness of the cake 2 times.That makes 3 layers of 4 pieces.Therefore thats 12.Your answer is easier though. $\endgroup$ – ministic2001 Apr 29 '15 at 16:04
  • $\begingroup$ @Spacemonkey But you didn't mention how to do those 3 cuts! $\endgroup$ – leoll2 Apr 29 '15 at 16:10
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    $\begingroup$ @leoll2, true but you could be fancy and make them slanted if you were so inclined (pun). The whole thing just seemed really obvious so I didn't write it up super well. I was just commenting because ministic2001 said I was wrong and I was a bit confused when your answer got accepted. $\endgroup$ – Spacemonkey Apr 29 '15 at 16:13
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I'm definitely missing something

You need 12 pieces. You said you wanted to share equally to 11 of your friends, sharing implies yourself. Note that nothing implies needing to eat the whole cake, however the fewest amounts of cuts will have you eating the whole cake. The answer would be 6 cuts, or if you are really awesome at gauging pieces (since its a non-flat cake), you could cut the cake in half vertically and then make 3 cuts to divide the halves into 6 pieces each. But you'd have to be a walking computer to do that

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  • $\begingroup$ Nope not right. But good attempt $\endgroup$ – ministic2001 Apr 29 '15 at 16:01
1
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You need 4 cuts are needed because you need 12 pieces (11 friends and you)
1- cut the cake in half vertically (you have now 2 pieces)
2- cut each half in third vertically (you have now 6 pieces)
3- cut the cake in half horizontally (you have now 12 pieces)

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  • $\begingroup$ If your answer is exactly the same as leoll2, let me imply the same thing. (Imagine u r in real life) $\endgroup$ – ministic2001 Apr 29 '15 at 15:59
  • $\begingroup$ How do you ensure is equal? I mean isnt there an EASIER way to make it equal? Because it seems estimating it 3 equal pieces in the first place is already difficult.The solution in my head is simple.Plus all same size coincidencely. $\endgroup$ – ministic2001 Apr 29 '15 at 16:00
1
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This one would be easier, you don't really need to think their size equal or not :D enter image description here

Divide them with one single cut vertically, and another cut horizontally. Then add two cuts to its thickness as picture above.

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  • $\begingroup$ Just 4 cuts, for minimum! $\endgroup$ – Nai Jun 8 '15 at 9:51

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