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You are serving a cake to $10$ children. The the cake is shaped like a box, whose top face is square. The top and sides are covered with a thin layer of frosting.

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Every child demands a fair share, so you must cut the cake into $10$ pieces where every piece has the same volume of cake and surface area of frosting. How do you do this?

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  • $\begingroup$ The whole cake must be passed out, so you can't throw away scraps? $\endgroup$ – mmking Jul 22 '15 at 0:00
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    $\begingroup$ @mmking Correct. You divide it into 10 pieces, then each child gets a piece. $\endgroup$ – Mike Earnest Jul 22 '15 at 0:02
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    $\begingroup$ Do we have a blender? $\endgroup$ – Milo Brandt Jul 22 '15 at 0:15
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    $\begingroup$ I don't respond well to children making demands. Send them each to bed with an equal share of zero and scoff the cake yourself! $\endgroup$ – James Webster Jul 22 '15 at 14:46
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    $\begingroup$ cut off icing, as it too sugary and not healthy for kids, and rest of cake divide in same size rectangles $\endgroup$ – user902383 Jul 22 '15 at 14:55
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We will make all of our cuts vertical, so we can treat this as a square which we need to divide into $10$ pieces with equal slices of the area and the perimeter. This is reasonably easy: Choose $10$ points dividing the perimeter into $10$ equal lengths, and then make cuts inwards from each point to the center. The area of a triangular slice is half the altitude of the triangle (distance of the outer line to the center) times its base (the length on the perimeter). Since altitude relative to the center is constant along all four segments, the area of each slice is proportional to the amount of the square's perimeter it contains (even if it contains multiple edges) - so given that the perimeter was divided evenly, so must be the area.

$\hskip 1in$enter image description here

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    $\begingroup$ Worthy of note: The same construction works with any polygon that can have a circle inscribed (tangent to all edges, of course) $\endgroup$ – Milo Brandt Jul 22 '15 at 0:30
  • $\begingroup$ What I love about this is that as long as the points are evenly spaced, you can rotate to your heart's content! $\endgroup$ – corsiKa Jul 23 '15 at 3:09
  • $\begingroup$ Hm.... is there some way to cut using only non-vertical cuts.... hmmm... $\endgroup$ – Bojidar Marinov Jul 23 '15 at 7:49
  • $\begingroup$ Should you discuss the particular cases where a corner is included in your cut? Not that it’s more difficult (split the shape into two triangles, one with base x, the other with base l - x where l is the 1/10 of the perimeter), but your answer is incomplete. $\endgroup$ – Édouard Jul 23 '15 at 11:11
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One of the possible solutions is:

Facing the square, mark the perimeter of the square into 10 equal pieces and cut through center of the square. So each piece is 0.4L long along the square. Please note that the corner pieces will have two triangular pieces but the calculations below hold.
Mock up approx picture shown (i didnt have tooling to cut the perimeter into bits). enter image description here

Calculations

Lets say square length is L and Height H
Frosting on top = 0.5 * 0.4L * 0.5L
Frosting on bottom = same as top
Frosting on side = 0.4L * H
Total frosting area = 0.2L2 + 0.4LH
Cake volume = top area * height = 0.5 * 0.4L * 0.5L * H = 0.1L2H

Validation

Total vol = 1L2H is 10 times above
Total surface area of frosting = 2L2+4LH which is 10 times what we found above

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  • $\begingroup$ By the time I typed it, looks like Meelo answered $\endgroup$ – beginner 101 Jul 22 '15 at 0:33
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    $\begingroup$ I only accepted the other answer since it was first, this is still a perfect answer! Welcome to the site, by the way! $\endgroup$ – Mike Earnest Jul 22 '15 at 1:44

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