Can you cut a cake fairly to solve this middle school math problem?

[brewer]

Change the cutting implement to have multiple blades. One action = one cut with multiple pieces. Think outside the box.

[Anomalocaris]

Change the cutting implement to have multiple blades. One action = one cut with multiple pieces. Think outside the box.

you will have to have a configurable multi-blade implement that can change blade spacing according to cake size. A bit of stretch to say that is an elegant solution.

[Whateverist]

You'll have to excuse Mr Brewer. I think he is one of those rebellious, liberal thinkers.

[brewer]

Change the cutting implement to have multiple blades. One action = one cut with multiple pieces. Think outside the box.

you will have to have a configurable multi-blade implement that can change blade spacing according to cake size.  A bit of stretch to say that is an elegant solution.

You may be underestimating Lostys culinary skills.

[popeyespappy]

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5 cuts. Each running from the center to the outside edge. In the case of a 20 x 20 x 5 cake each cut needs to be 16 linear centimeters along the outside edge. That would give everyone 160 square centimeters of icing and 400 cubic centimeters of cake.

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[account_inactive]

[brewer]

Tell Rob it will make his Crohns flair, make 2 cuts.

[Faith No More]

Goddammit, Whateverist, this is one of those problems that I know has to be fairly simple to solve, but I'm sitting here racking my brain over it and loathing myself for not being able to figure it out.

[Jackalope]

* Jackalope eats cake.

There, all five get an equal piece.

[Whateverist]

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5 cuts. Each running from the center to the outside edge. In the case of a 20 x 20 x 5 cake each cut needs to be 16 linear centimeters along the outside edge. That would give everyone 160 square centimeters of icing and 400 cubic centimeters of cake.

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Exactamundo.  But to elaborate

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Of course every one of the triangles have the same height regardless of where their base may be.  For those which straddle a corner, each can be dissected into two triangles with a common height whose bases sum to the same as the other.  So for a square with side length = 20, the base of each triangle must be 4(20)/5 = 16  and the combined sides of a piece which includes a corner, like your red one, where 12 + 4 = 16.  Since the height of both triangles remains 8, the combined area of the two triangles would be 12(8)/2 + 4(8)/2 which can be rewritten as (12 + 8)(8)/2.

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