Bob and his mom love pizza. Hence, they buy a pizza which happens to be shaped as an equal sided triangle. Bob, being a spoiled brat, demands to not only slice the pizza (with a straight cut), but to then choose first which slice he wants. The mother grudgingly agrees, but she requires that she is at least allowed to pick a point on the pizza through which the straight cut must pass.
How large a fraction of the pizza's area can the mother be assured she will get, if she chooses the best point?
Slicing pizza

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Possible spoiler
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If she marks the centroid of the triangle, then the brat can give her as much as half of the pizza (by making the cut through a corner and the centroid), or as little as 4/9 (by making the cut through the centroid, parallel to one of the sides).
Symmetry considerations make me suspect that the centroid is the best place for her to mark, but I've not proven it...
The centroid is 2/3 of the way along a line from a corner to the centre of the opposite side. The triangle she would get is similar in shape to the original, but 2/3 of the linear dimensions. The 4/9 figure is simply (2/3)^2
I've also not proven that, for the centroid, 1/2 is the maximum area, and 4/9 the minimum, but that seems a good bet.
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If she marks the centroid of the triangle, then the brat can give her as much as half of the pizza (by making the cut through a corner and the centroid), or as little as 4/9 (by making the cut through the centroid, parallel to one of the sides).
Symmetry considerations make me suspect that the centroid is the best place for her to mark, but I've not proven it...
The centroid is 2/3 of the way along a line from a corner to the centre of the opposite side. The triangle she would get is similar in shape to the original, but 2/3 of the linear dimensions. The 4/9 figure is simply (2/3)^2
I've also not proven that, for the centroid, 1/2 is the maximum area, and 4/9 the minimum, but that seems a good bet.
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 Posts: 1462
 Joined: Wed Jun 02, 2004 11:04 pm
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I'm no longer sure the centroid is the best point.
If you draw a line from one corner of the pizza to the centre of the opposite side, and mark off a point h / sqrt(2) along the line, where h is its length, then I think that might be a better spot than the centroid.
The centroid is 2/3 of the way down the line (0.6667) the new spot is 1/ sqrt(2) (0.7071). The best point might be somewhere between. More math needed... :?
If you draw a line from one corner of the pizza to the centre of the opposite side, and mark off a point h / sqrt(2) along the line, where h is its length, then I think that might be a better spot than the centroid.
The centroid is 2/3 of the way down the line (0.6667) the new spot is 1/ sqrt(2) (0.7071). The best point might be somewhere between. More math needed... :?

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 Posts: 1462
 Joined: Wed Jun 02, 2004 11:04 pm
 Location: UK

 Posts: 2608
 Joined: Mon Jun 07, 2004 4:58 pm
 Location: Copenhagen

 Posts: 2608
 Joined: Mon Jun 07, 2004 4:58 pm
 Location: Copenhagen
Here come the geometric proof. (See attached figure)
The proof is divided into 3 cases:
Case 1: The mother points at the centroid G and Bob makes a cut parallel with a side of the pizza
Bob's cut is shown on the figure as MN. The cut divides the triangle into a trapezoid and a smaller equilateral triangle. The relationship of the sides of the smaller triangle to the original is 2/3 and hence the area is 4/9 of the original. The mother is thus assured 4/9.
Case 2: The mother points at the centroid G and Bob doesn't make a cut parallel to one side
Let Bob's cut be the line DE. Let fA be a function which returns the area of a geometric figure. We now want to show that fA(AMNB) > fA(ADEB) >= 0.5*fA(ABC). Let the line through M and parallel to BC cut the line DE in the point F and cut the angle bisector AI in the point H. As triangle GMF = triangle GNE, we get that fA(AMNB) = fA(ADEB) + FA(MFD). Similarly, fA(AMNB) = 0.5*fA(ABC) + fA(MHA). Since fA(MFD) <= fA(MHA) we get the required result: fA(AMNB) > fA(ADEB) >= 0.5*fA(ABC). Hence, the mother will get more than 4/9 of the pizza, and hence Bob wont make that cut.
Case 3: The mother picks a point different from G
Let this point P be inside CMN. A cut through P parallel to MN will result in less than 4/9 for the mother. A cut through P not parallel with MN will by symmetry likewise be less than 4/9.
QED.
The proof is divided into 3 cases:
Case 1: The mother points at the centroid G and Bob makes a cut parallel with a side of the pizza
Bob's cut is shown on the figure as MN. The cut divides the triangle into a trapezoid and a smaller equilateral triangle. The relationship of the sides of the smaller triangle to the original is 2/3 and hence the area is 4/9 of the original. The mother is thus assured 4/9.
Case 2: The mother points at the centroid G and Bob doesn't make a cut parallel to one side
Let Bob's cut be the line DE. Let fA be a function which returns the area of a geometric figure. We now want to show that fA(AMNB) > fA(ADEB) >= 0.5*fA(ABC). Let the line through M and parallel to BC cut the line DE in the point F and cut the angle bisector AI in the point H. As triangle GMF = triangle GNE, we get that fA(AMNB) = fA(ADEB) + FA(MFD). Similarly, fA(AMNB) = 0.5*fA(ABC) + fA(MHA). Since fA(MFD) <= fA(MHA) we get the required result: fA(AMNB) > fA(ADEB) >= 0.5*fA(ABC). Hence, the mother will get more than 4/9 of the pizza, and hence Bob wont make that cut.
Case 3: The mother picks a point different from G
Let this point P be inside CMN. A cut through P parallel to MN will result in less than 4/9 for the mother. A cut through P not parallel with MN will by symmetry likewise be less than 4/9.
QED.

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