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?

 A: *

*If(assumption) the line passes through any of the vertices, then it must be the median (passing through the centroid) which is the only possible line (that passes through the vertices and) that would divide the triangle in two triangle of equal area and since all sides are unequal the line would never divide the perimeter into half. This proves that the line doesn't pass through any of the vertices.

*Another useful theorem is the Haider's theorem:

Haider's Theorem: For any triangle ABC and any line $\ell$, $\ell$ divides$\\$ the area and the perimeter of $\Delta$ ABC in the same$\\$ ratio if and only if it passes through the triangle's incenter.


*Let us call a line that simultaneously bisects the area and perimeter a bisecting line. Every triangle has exactly one, two, or three bisecting lines; no other values are possible. It is not too hard to prove this claim

*For a detailed solution see this and for the resource I used see this.

A: 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, $x = \sqrt{\frac32}.$
A: Consider triangle $\mathrm {OAB}$. Let line $l$ be the line that is supposed to divide the area and permieter in half. 
Intial perimeter=$12$, initial area=$6$. Let $\angle OBA=\theta$
We get $3$ scenarios. I am discussing one here.

$l$ intersects $AB$ in point $Q$ and $OB$ in point $P$. Let $PB$=a, $BQ$=b
$\therefore a+b=6\; (\mathrm {half\;of\;12})$ 
From AM-GM inequality, $a+b \ge 2\sqrt{ab} \implies ab \le9$
Area($\triangle PBQ$)= $\frac12absin\theta=3\; (\mathrm {half\; of\; 6})$
$\therefore \frac12(ab)\frac45=3 \implies ab=7.5 \lt 9$
So, this line is possible.
Check with the other two scenarios where the line intersects the other two sides. Then see whether you get any contradicting conditions on $a$ and $b$.
