The lines with equations $x+y=3$ and $2 x-y=0$ meet at point $A$. The lines with equations $x+y=3$ and $3 x-t y=4$ meet at point $B$. The lines with equations $2 x-y=0$ and $3 x-t y=4$ meet at point $C$. Determine all values of $t$ for which $A B=A C$.
We proceed by (very carefully!) calculating the coordinates of points $A, B$ and $C$, and then equating $A B$ and $A C$ to solve for $t$.

Point $A$ is the point of intersection of the lines with equations $x+y=3$ and $2 x-y=0$.

Since $x+y=3$ and $y=2 x$, then $x+2 x=3$ and so $x=1$ which gives $y=2 x=2$.

Therefore, $A$ has coordinates $(1,2)$.

Point $B$ is the point of intersection of the lines with equations $x+y=3$ and $3 x-t y=4$.

Adding $t$ times the first equation to the second, we obtain $t x+3 x=3 t+4$ and so $x=\frac{3 t+4}{t+3}$, with the restriction that $t \neq-3$.

Since $x+y=3$, then $y=3-\frac{3 t+4}{t+3}=\frac{3 t+9}{t+3}-\frac{3 t+4}{t+3}=\frac{5}{t+3}$.

Therefore, $B$ has coordinates $\left(\frac{3 t+4}{t+3}, \frac{5}{t+3}\right)$.

Point $C$ is the point of intersection of the lines with equations $2 x-y=0$ and $3 x-t y=4$.

Subtracting $t$ times the first equation from the second, we obtain $3 x-2 t x=4-0$ and so $x=\frac{4}{3-2 t}$ with the restriction that $t \neq \frac{3}{2}$.

Since $y=2 x$, then $y=\frac{8}{3-2 t}$.

Therefore, $C$ has coordinates $\left(\frac{4}{3-2 t}, \frac{8}{3-2 t}\right)$.

We assume that $t \neq-3, \frac{3}{2}$.

Since $A B$ and $A C$ are non-negative lengths, then the following equations are equivalent:

$$
\begin{aligned}
A B & =A C \\
A B^{2} & =A C^{2} \\
\left(\frac{3 t+4}{t+3}-1\right)^{2}+\left(\frac{5}{t+3}-2\right)^{2} & =\left(\frac{4}{3-2 t}-1\right)^{2}+\left(\frac{8}{3-2 t}-2\right)^{2} \\
\left(\frac{(3 t+4)-(t+3)}{t+3}\right)^{2}+\left(\frac{5-(2 t+6)}{t+3}\right)^{2} & =\left(\frac{4-(3-2 t)}{3-2 t}\right)^{2}+\left(\frac{8-(6-4 t)}{3-2 t}\right)^{2} \\
\left(\frac{2 t+1}{t+3}\right)^{2}+\left(\frac{-2 t-1}{t+3}\right)^{2} & =\left(\frac{2 t+1}{3-2 t}\right)^{2}+\left(\frac{4 t+2}{3-2 t}\right)^{2} \\
(2 t+1)^{2}\left[\left(\frac{1}{t+3}\right)^{2}+\left(\frac{-1}{t+3}\right)^{2}\right] & =(2 t+1)^{2}\left[\left(\frac{1}{3-2 t}\right)^{2}+\left(\frac{2}{3-2 t}\right)^{2}\right]
\end{aligned}
$$

$$
\begin{aligned}
(2 t+1)^{2}\left[\frac{2}{(t+3)^{2}}-\frac{5}{(3-2 t)^{2}}\right] & =0 \\
(2 t+1)^{2}\left[\frac{2(3-2 t)^{2}-5(t+3)^{2}}{(t+3)^{2}(3-2 t)^{2}}\right] & =0 \\
(2 t+1)^{2}\left[\frac{2\left(4 t^{2}-12 t+9\right)-5\left(t^{2}+6 t+9\right)}{(t+3)^{2}(3-2 t)^{2}}\right] & =0 \\
(2 t+1)^{2}\left[\frac{3 t^{2}-54 t-27}{(t+3)^{2}(3-2 t)^{2}}\right] & =0 \\
3(2 t+1)^{2}\left[\frac{t^{2}-18 t-9}{(t+3)^{2}(3-2 t)^{2}}\right] & =0
\end{aligned}
$$

Therefore, the values of $t$ for which $A B=A C$ are $t=-\frac{1}{2}$ and the two values of $t$ for which $t^{2}-18 t-9=0$, which are $t=\frac{18 \pm \sqrt{18^{2}-4(1)(-9)}}{2}=\frac{18 \pm \sqrt{360}}{2}=9 \pm 3 \sqrt{10}$.

$$
\text { ANSWER: }-\frac{1}{2}, 9 \pm 3 \sqrt{10}
$$