AMC12 2021 A

Let $\ell$ be the directrix of $\mathcal P$; recall that $\mathcal P$ is the set of points $T$ such that the distance from $T$ to $\ell$ is equal to $TF$. Let $P$ and $Q$ be the orthogonal projections of $F$ and $A$ onto $\ell$, and further let $X$ and $Y$ be the orthogonal projections of $F$ and $V$ onto $AQ$. Because $AF < AV$, there are two possible configurations which may arise, and they are shown below. Set $d = FV$, which by the definition of a parabola also equals $VP$. Then as $AQ = AF = 20$, we have $AY = 20 - d$ and $AX = |20 - 2d|$. Since $FXYV$ is a rectangle, $FX = VY$, so by Pythagorean Theorem on triangles $AFX$ and $AVY$, \begin{align*} 21^2 - (20 - d)^2 &= AV^2 - AY^2 = VY^2\\ &= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2 \end{align*} This equation simplifies to $3d^2 - 40d + 41 = 0$, which has solutions $d = \tfrac{20\pm\sqrt{277}}3$. Both values of $d$ work - the smaller solution with the bottom configuration and the larger solution with the top configuration - and so the requested answer is $\boxed{\textbf{(B)}\ \tfrac{40}{3}}$.