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Cut a pentagonal cake into 7 equal pieces


bonanova
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It's your birthday, and a cake has been brought to your table, frosted on the top and sides.

The cake is 3" high and has a base in the shape of a regular pentagon, 3" on a side.

It's your task to cut the cake into pieces for yourself and your six guests.

You want each piece to have the same volume and the same amount of frosting.

The cutting is conventional: as seen from above, the cuts are like spokes radiating from the pentagon's center.

The cutting planes are perpendicular to the cake's base.

You realize that if you had invited two fewer guests (5 people in all) your job would be much easier.

But after some thought you devise a plan that gets the job done.

How do you cut your cake to provide seven pieces with equal volume and frosting?

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tongue in cheek :thanks:

cook an additional cake that completely encapsulates the pentagon with the exact same type of icing as the other cake. place the cake around the cake to make a square. cut the cake into 7 equal strips.

Op solved

tongue removed from cheeck

Edited by BMAD
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Bonus question for those who are at ease with formulae: Using this proceeding, in how many parts can you cut the cake at most?


- a pentagon has a center: cut a concentric cylinder representing 2 parts
- the remaing object can be easily cut in 5 parts
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A polygon of n sides can be divided into equal N areas using ribbon..because a triangle can be divided into halves when cut straight from a vertex to the opposite side's midpoint.

A C



B



D



E

If ACD is cut from C thru mid pt of AD by CB , ACB=BCD
so BCD=DCE if BD=DE (see above where ABDE is linear) .
Here we see that triangle ACE is divided into 3 equal parts.

When the pentagon's fifths are divided into seven making 35 triangles,around its perimeter are also 35 equal lengths sides.
5 length marks is equivalent to 1/7.

For circle we can think of it as having many equal sides of many similar triangles with common vertex at center of circles.


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