More Pizza!

Claim: For any point D, Omar has a straight cut which gives him at least 5/9 of the pizza.

Proof: Take a triangle plot with vertices A, B, C, and grid lines parallel to AB, BC, CA.
For any point D within the triangle, draw in these grid lines, which cut AB, BC and CA.
Let $A_B$ be the point where the line parallel to the side opposite $B$ cuts the side opposite $A$ , with the other points similarly defined.
Define $x = \frac{ | A_B A_C | } { |BC | }$ to be the fractional part that is cut out along line $BC$ , $y = \frac{ |B_A B_C | } { |CA| }$ and $z = \frac{ |C_B C_A | } { |AB| }$ respectively.

Observe that $x + y + z = \frac{ |A_B A_C | } { |BC| } + \frac{ |B_A B_C | } { |CA| } + \frac{ |C_B C_A | } { |AB| } = \frac{ |A_B A_C | } { |BC| } + \frac{ |CA_B| } { |BC| } + \frac{ |A_C B| } { |BC| } = 1.$

Without loss of generality, $x \geq \frac{1}{3}$ . Then, Omar makes the cut $C_A B_A$ ), and picks the slice $BCB_A C_A$ . The fractional area of $A B_A C_A$ is $(1-x)^2$ , so the fractional area of Omar's slice is given by $1 - ( 1-x)^2 = 2x - x^2 \geq \frac{5}{9}$

Hence, Omar has a cut that yields at least 5/9 of the pizza.

Note: We see that if Omar is restricted to these cuts, then he can only guarantee 5/9 of the pizza if $x = y = z = \frac{1}{3}$ , which happens when $D$ is the centroid of the pizza. This leads to the next claim

Claim: If Ahmed picks the centroid of the pizza, the best that Omar can do is 5/9 of the pizza.

Proof: Say Omar picks a cut, which intersects the perimeter of the triangle at $E$ and $F$ . If $DE \neq EF$ , then by tilting the cut further, Omar can increase the size of the slice that he picks.

As an explicit example, suppose Omar makes the cut EF and picks the slice $BCEF$ . Then, since $DE < DF$ , we can tilt the line about D, to $E'F'$ . Comparing areas, since $\triangle D E E' < \triangle D F F'$ , hence $BCE'F' > BCEF$ . Note: If Omar picks the slice $AEF$ , then he should tilt in the opposite direction.

This tilting no longer results in a larger area when $DE = DF$ . Hence, when Omar maximizes his slice, we must have $DE = EF$ .

Since $D$ is the centroid, this implies that $EF$ is parallel to one of the sides of the triangle. Then, from the above analysis, since we have $x = y = z = \frac{1}{3}$ , thus the largest slice that Omar can pick has fractional area $1 - (1 - \frac{1}{3} ) ^2 = \frac{5}{9}$ .