Prove or disprove {wtw^R | |w| = |t|} is context free
Suppose $S_c$ is context free. Then let $p$ be the pumping length and choose $s = 0^{2p}0^p1^p0^{2p}$ in $S_c$ that satisfies the pumping lemma. Thus, we can break $s$ into $uvxyz$, with $|vxy| \leq p$ and $|vy| \geq 1$. And for all $i$, $uv^ixy^iz$ is a string in $S_c$.
We have three cases.

$vy$ contains all 0s and these 0s can be chosen from either the beginning $0^{2p}$ of $s$, middle $0^p1^p$ of $s$ or the last $0^{2p}$ of $s$. Now, by pumping lemma, since $uv^ixy^iz \in S_c$, for all $i$. We choose $i = 2$, so we have $s' = uv^2xy^2z = uvvxyyz$. Clearly, since $vy$ contains all 0s, and regardless which part of $s$ are chosen from, this string is the form of $wtw^R$ such that $|w|$ = $|t|$ = $w'$ with either $w$ is all 0's and $w'$ is not all 0s or vice verse. This implies that $w' \neq w^R$ or $w' \neq w$.

$vy$ doesn't contain any 0s in last $0^{2p}$ of $s$ (We choose in the last part of $s$, but it's essentially the same by choosing first or middle part of $s$). Similarly, either the length of $uv^2xy^2z$ is not multiple of 3 (since pumping will increase the number of 1's), or this string is the form of $wtw'$ such that $|w|$ $=$ $|t|$ $=$ $|w'|$ with $w$ consists all 0s but $w'$ contains some 1s. Thus $w' \neq w^R$.

$vy$ contains some 0s and some 1s in last $0^{2p}$ of $s$. From pumping lemma, since $vxy \leq q$, $vxy$ must be sub-string of $0^p1^p$. Then, by pumping the string $s' = uv^ixy^iz$ for all $i$, the length is string is not multiple of 3 since pumping increases number of 0s or 1s, or the string is in the form of $wtw'$ such that $|w|$ $=$ $|t|$ $=$ $|w'|$ with $w$ is all 0s but $w'$ is not all 0s.


Thus, we can conclude that all three cases above reach contradiction and hence $S_c$ is not context free.