**Lets Be Fair**

There are six children want to slice a strawberry pan cake
with chocolate topping, the cake’s shape is square. The rule is: each child
gets the same part of cake that is the same topping and strawberry paste.

Here are my steps to solve this problem:

- I imagine the cake and draw a square representing the cake.

- Divide the square into two same parts by drawing
a diagonal line and each part has the same area (chocolate topping) and
strawberry paste.

- My strategy now is how to divide each part into
three parts which have the same area and strawberry paste. Putting dots will
help me to deal with strawberry paste. Trial and error plays their roles here.
If I put one dot at each side of the triangle, there will be a
part that does not have the same part of strawberry paste.

- Since by having one dot at each side of triangle cannot solve the problem or satisfy the solution, I try to add more dots ( two dots at each side of triangle) which divide a side of square into three same part.

Now, look at the trapezium BCEF, calculating this area will lead us to the area of OBCD, ODE and OEF. Say that the length of the square’s side is S. The square is the area of the trapezium is½ (BC+FE)(CE)½ (1/3*S + 2/3*S)(S)½(S)(S)½ S^{2}Since the triangle ODE and OEF is congruent and it implies that their area are the same,The area of OEF is½(FE*OX)½ (2/3S * ½S)½ (1/3 S^{2})1/6 S^{2}The area of OEF is also 1/6 S^{2}, adding the two area of ODE and OFE results in 2/6 S^{2}.The area of OBCD is1/2 S^{2}-2/6 S^{2}1/6 S^{2}So, the area of OBCD, ODE, and OEF is 1/6 S^{2}.It’s now clear that the area of three chocolate topping is the same and it applies also to trapezium ABFG.We can conclude that the way we cut the cake satisfies fair sharing and each child gets the same parts of chocolate topping and strawberry paste.LETS ENJOY THE CAKE!

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