geometry
Two friends A, B want to share a (circular) pizza, by playing a game.
A does a (straight line) cut
B also does a cut
A does another cut and
B does the last cut.
Now they alternate turns picking one slice at a time until last, starting from A.
Does B have a strategy (at least) not to eat less pizza?
I think the key is to make equal pieces, i.e. splitting the pizza in pairs of pieces of equal area.
Lemma. Consider $n\geq2$ equally spaced chords through the point $(a,0)$ of the unit disk, having slopes $\phi_k:=\phi_0+{k\pi\over n}$ $\>(1\leq k\leq n)$. Then the sum of the squares of the $2n$ resulting chord pieces is $2n$, independently of $\phi_0$.
Proof. Intersecting the line $s\mapsto (a+s\cos\phi, s\sin\phi)$ with the unit circle we get the equation $s^2+2as\cos\phi+a^2-1=0$. If $s_1$ and $s_2$ are its two solutions we can say that $$s_1^2+s_2^2=(s_1+s_2)^2-2s_1s_2=4a^2 \cos^2\phi-2(a^2-1)=2+2a^2\cos(2\phi).$$ Noting that $\sum_{k=1}^n\cos(2\phi_k)=0$ we conclude that the sum of the $2n$ considered squares is $2n$.
Proof of the Theorem. Assume that we have $4n\geq8$ equally spaced blades, and let $S(\phi)$ denote the shaded pizza area when one blade points in direction $\phi$. When $\phi\mapsto r(\phi)$ is the polar representation of the circular pizza boundary with respect to the center of the cutter then it is easy to see that $$S'(\phi)=\pm{1\over2}\sum_{k=1}^{4n} (-1)^k r^2\left(\phi+{k\pi\over2n}\right)\ .$$ Here the right side is $\equiv0$, since the alternating sum of the $4n$ blade-length-squares vanishes according to the Lemma. It is then easy to see that $S(\phi)\equiv$ half the area of the pizza.
Applying the Theorem to two concentric pizzas of radii $1$ and $1+\epsilon$ one concludes that the "crust area" is halved as well, and in the limit $\epsilon\to0+$ it follows that the shaded areas together cover half the circumference of the pizza.
The approach uses polar coordinates. The pair $(\theta,r)$ define a point in that coordinate system, with $P$ in the origin. So for each angle $\theta$ you can compute the corresponding distance $r(\theta)$ of a point on the circle in the direction denoted by $\theta$.
The area element in polar coordinates is
$$\mathrm dA = \frac12 r^2\mathrm d\theta$$
which explains the area computation formula
$$A=\int_{\theta_1}^{\theta_2}\frac12r(\theta)^2\mathrm d\theta$$
Now in your specific case, you integrate over four areas at the same time. These areas are parametrized by angles $\theta$ which differ by multiples of $\frac\pi2=90°$. The boundaries denote one single slice.
So what is $r(\theta)$? If the pizza has its center at $(C_x,C_y)$ relative to $P$, and has a radius of $R$, then $r$ has to satisfy the equation
\begin{align*} (r\cos\theta - C_x)^2 + (r\sin\theta - C_y)^2 &= R^2 \\ r^2 - 2(C_x\cos\theta+C_y\sin\theta)r + (C_x^2+C_y^2-R^2) &= 0 \\ C_x\cos\theta+C_y\sin\theta+\sqrt{(C_x\cos\theta+C_y\sin\theta)^2-C_x^2-C_y^2+R^2} &= r(\theta) \end{align*}
I'm choosing the greater of the two solutions of the quadratic equation since the other would lead to a negative $r$, the other point where the line for angle $\theta$ intersects the circle. Polar coordinates use $r>0$.
We now want to obtain a relationship between the integrand at angles $\theta$, $\theta+\tfrac12\pi$, $\theta+\pi$ and $\theta+\tfrac32\pi$, i.e. corresponding parts from each of the four slices. In order to compute these, let's use some abbreviations.
\begin{align*} P(\theta)&:=C_x\cos\theta+C_y\sin\theta\\ D(\theta)&:=P(\theta)^2-C_x^2-C_y^2+R^2\\ r(\theta)&=P(\theta)+\sqrt{D(\theta)}\\ r(\theta)^2&=P(\theta)^2+2P(\theta)\sqrt{D(\theta)}+D(\theta)\\ P(\theta+\pi)&=-P(\theta)\\ r(\theta)^2+r(\theta+\pi)^2&=2P(\theta)^2+2D(\theta)\\ r(\theta)^2+r(\theta+\pi)^2&= 4(C_x\cos\theta+C_y\sin\theta)^2+2(R^2-C_x^2-C_y^2)\\ r(\theta+\tfrac12\pi)^2+r(\theta-\tfrac12\pi)^2&= 4(C_x\sin\theta-C_y\cos\theta)^2+2(R^2-C_x^2-C_y^2)\\ P(\theta)^2=(C_x\cos\theta+C_y\sin\theta)^2&= C_x^2\cos^2\theta+2C_xC_y\sin\theta\cos\theta+C_y^2\sin^2\theta\\ P(\theta+\tfrac12\pi)^2=(C_x\sin\theta-C_y\cos\theta)^2&= C_x^2\sin^2\theta-2C_xC_y\sin\theta\cos\theta+C_y^2\cos^2\theta\\ P(\theta)^2+P(\theta+\tfrac12\pi)^2&=(C_x^2+C_y^2)(\sin^2\theta+\cos^2\theta) =C_x^2+C_y^2\\ \sum_{k=0}^3r(\theta+\tfrac k2\pi)^2&=4(C_x^2+C_y^2)+4(R^2-C_x^2-C_y^2)=4R^2 \end{align*}
So the sum of the area elements will always be $2R^2\mathrm d\theta$, which explains why all $n$ people get equal shares of this pizza.
Best Answer
Yes, $B\;$can achieve equality.
Assume the players are not allowed to make no cut (i.e., pass), but are allowed to make a cut which coincides with a previous cut.
After $A$'s first cut, let $B\;$make a cut, $l\;$say, along the perpendicular bisector of $A$'s cut.
Thereafter, for every cut that $A\;$makes, let $B\;$make a cut symmetrical about $l\;$to $A's$ cut (i.e., the same cut as $A$ except reflected about the line $l$). Note: It might coincide with $A$'s cut.
It follows that every final piece appears with even multiplicity, up to congruence, so in the worst case, as $A\;$chooses pieces, $B$'s next choice can at least match.