Does there exist a terminal $3$-fold $X$ with a curve $C\subset X$ such that $K\_X\cdot C < 0$ admitting a Mori flip $X\dashrightarrow Y$, flipping $C$ to a curve $C'\subset Y$, where the singular locus of $Y$ consists of a single point of type $\frac{1}{2}(1,1,1)$, $C'$ passes through the singular point of $Y$, and $-K\_Y\cdot C' < -\frac{1}{2}$?


Is there any classification result of $3$-fold terminal flips that could answer to this question?


Thank you.



Existence of terminal $3$-fold flips
Yes - there are very many such examples, and you can cook up examples by a procedure called 'Mori's algorithm'.


A k2A flipping neighbourhood is a 3-fold flipping contraction $f\colon(C\subset X)\to (P\in Y)$, where the exceptional curve $C$ is irreducible and $X$ has two $cA\_{d\_i-1}/\tfrac1{r\_i}(1,-1,a\_i,0)$ hyperquotient singularities along $C$. You can construct such neighbourhoods as a 1-parameter $\mathbb Q$-Gorenstein smoothing of the contraction of a rational curve in a surface $f\colon (C\subset H\_X)\to (P\in H\_Y)$, where $H\_X$ has two $T$-singularities of type $\tfrac{1}{r\_i^2d\_i}(1,r\_id\_ia\_i-1)$ for $i=1,2$ along $C$.


Mori's algorithm says that if $\delta=r\_1a\_2+r\_2a\_1-r\_1r\_2>0$ (which is the integer such that $K\_XC=-\frac{\delta}{r\_1r\_2}$), then set $d\_1=d\_3=d\_5=\ldots$ and $d\_2=d\_4=d\_6=\ldots$ and write down a table
$$ \begin{array}{ccccccc}
d\_1\delta & d\_2\delta & d\_1\delta & d\_2\delta & \ldots & d\_n\delta & d\_{n+1}\delta \\ \hline
r\_1 & r\_2 & r\_3 & r\_4 & \cdots & r\_n & r\_{n+1} \\
r\_1-a\_1 & a\_2 & a\_3 & a\_4 & \cdots & a\_n & a\_{n+1} 
\end{array} $$
where $r\_{i+1} = \delta d\_ir\_i - r\_{i-1}$ and $a\_{i+1} = \delta d\_ia\_i - a\_{i-1}$. If $r\_1>r\_2$ then eventually the sign of $r\_i$ will become negative. Let $n$ be the first $n$ such that $r\_n>0$ and $r\_{n+1}<0$. Then the flip $(C^+\subset X^+)$ will have singularities of type $cA\_{d\_i-1}/\tfrac1{r\_i}(1,-1,a\_i,0)$ along $C^+$ for $i=n,n+1$. Moreover $K\_{X^+}C^+=\frac{\delta}{r\_n(-r\_{n+1})}>0$.


Note that if $d=1$ then a $cA\_{d-1}/\tfrac1{r}(1,-1,a,0)$ singularity is simply a $\tfrac1{r}(1,-1,a)$ quotient singularity, and if $r=1$ then this quotient singularity is actually just a smooth point. Therefore, to give a fairly arbitrary example which address your question, we can take $(r\_1,a\_1,d\_1)=(34,25,1)$ and $(r\_2,a\_2,d\_2)=(7,2,1)$ with $\delta=5$ and write down the table
$$ \begin{array}{cccc}
5 & 5 & 5 & 5 \\ \hline
34 & 7 & 1 & -2 \\
9 & 2 & 1 & 3 
\end{array} $$
which tells me that $(C\subset X)$ has $\tfrac1{34}(1,9,25)$ and $\tfrac17(1,2,5)$ singularities with $K\_XC=-\frac{5}{238}$ and the flip $(C^+\subset X^+)$ has one $\tfrac12(1,1,1)$ singularity along $C^+$ and $K\_{X^+}C^+=\frac{5}{2}$. If you look at the resolution of the surface singularity $(P\in H\_Y)$ downstairs, you will find that it is a $\tfrac1{15}(1,4)$ singularity whose resolution is a chain of two rational curves of self-intersection $-4$. The flip $X^+$ is obtained by resolving one of the $-4$ curves and taking a $\mathbb Q$-Gorenstein smoothing of the $\tfrac14(1,1)$ point on the resulting surface.


You make arbitrarily complicated examples by cooking up the two rightmost columns of the table correctly so that $\delta>1$, and then extending the table to the left as far as you want.


