<s>[INST] How many lines bisect both perimeter and area of a 3-4-5 triangle? 
How many lines exist that divide both the perimeter and the area of a triangle with sides $3$, $4$ and $5$ into half?

 [/INST] If the line passes through the sides of length $3$ and $4$,
and its intersection with side $3$ is $x$ units from the acute angle on that side,
then the line cuts off a right triangle of base $3-x$ and height $3+x$.
The area of this triangle is $\frac 12 (9-x^2) \leq \frac 92.$
Setting this equal to $3$, we would have $x = \pm\sqrt3,$
but the construction of this line requires $0\leq x\leq 1,$
so there is no such line that cuts the triangle's area in half.
If the line passes through the sides of length $4$ and $5$,
and its intersection with side $4$ is $x$ units from the acute angle on that side,
then it cuts off a triangle with base $x$ and height $\frac 35 (6-x)$.
The area of this triangle is $\frac 12 \cdot x \cdot \frac 35 (6-x) = \frac{3}{10} x(6-x)$,
which takes a maximum value $\frac{27}{10}<3$ at $x=3,$
so no such line can cut the triangle's area in half.
The remaining case is a line through sides $3$ and $5$.
Let the line intersect side $3$ at a point $x$ units from the right angle.
Then it cuts off a triangle of base $3-x$ and height $\frac 45 (3+x),$
which has area $\frac 25 (9-x^2).$
Setting this equal to $3$, we find that $x = \pm\sqrt{\frac32},$
but $0\leq x\leq 2$ by the construction of the line,
so we have one solution, $x = \sqrt{\frac32}.$
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