Guide:
It is know that for a $\triangle ABC$, suppose its length is $a,b,c$, with the vertices being $(x_i, y_i)$ where $i\in \{A,B,C\}$.
Then the formula is given by
$$\left( \frac{ax_A+bx_B+cx_C}{a+b+c},\frac{ay_A+by_B+cy_C}{a+b+c}\right)$$
A proof of the formula can be found here.
You have found the coordinates, hence it should be possible for you to find the lenght of the sides easily.
Note that the edges are given by $\overline{AB}=\{(x+35,2x):x\in[0,35]\}$, $\overline{BC}=\{(x,70):x\in[0,70]\}$ and $\overline{AC}=\{(35-x,2x):x\in[0,35]\}$. The linepieces on centimetre from the edges are given by
$$\hat{AB}=\{(35+x-\frac{2}{5}\sqrt{5},2x+\frac{1}{5}\sqrt{5}):x\in[0,35]\},$$
$$\hat{BC}=\{(x,69):x\in[0,70]\}$$
and
$$\hat{AC}=\{(35-x+\frac{2}{5}\sqrt{5},2x+\frac{1}{5}\sqrt{5}):x\in[0,35]\}$$
which have intersections $(70-\frac{1}{2}(\sqrt{5}+1),69)$, $(35,\sqrt{5})$ and $(\frac{1}{2}(\sqrt{5}+1),69)$, which are the coordinates of the red triangle.
This process can be generalised for different borders. Note that
$$\hat{AB}=\overline{AB}-(\frac{2}{5}\sqrt{5},\frac{1}{5}\sqrt{5})$$
here $(\frac{2}{5}\sqrt{5},\frac{1}{5}\sqrt{5})$ is a unit vector which is orthogonal to $\overline{AB}$. So if we want a border of length $l$, then we can just take
$$\hat{AB}_{l}=\overline{AB}-l\cdot(\frac{2}{5}\sqrt{5},\frac{1}{5}\sqrt{5})$$
This gives us
$$\hat{AB}_{l}=\{(35+x-\frac{2l}{5}\sqrt{5},2x+\frac{l}{5}\sqrt{5}):x\in[0,35]\},$$
$$\hat{BC}_{l}=\{(x,70-l):x\in[0,70]\}$$
and
$$\hat{AC}_{l}=\{(35-x+\frac{2l}{5}\sqrt{5},2x+\frac{l}{5}\sqrt{5}):x\in[0,35]\}.$$
To calculate the coordinates of the new triangle we have to find the intersections of the triangles, these are given by:
$$\hat{AB}_{l}\cap\hat{AC}_{l}=\{(35,l\sqrt{5})\}$$
as due to symmetry $35=35-x+\frac{2l}{5}\sqrt{5}\Leftrightarrow x=\frac{2l}{5}\sqrt{5}$. From which it then follows that $2x+\frac{l}{5}\sqrt{5}=l\sqrt{5}$.
For $\hat{AB}_{l}\cap\hat{BC}_{l}$ we can solve
$$70-l=2x+\frac{l}{5}\sqrt{5}\Leftrightarrow x=35-\frac{l}{2}(1+\frac{1}{5}\sqrt{5})$$
which then gives $35+x-\frac{2l}{5}\sqrt{5}=70-\frac{l}{2}(1+\sqrt{5})$. Hence
$$\hat{AB}_{l}\cap\hat{BC}_{l}=\{(70-\frac{l}{2}(1+\sqrt{5}),70-l)\}.$$
Finally by symmetry we find that
$$\hat{AC}_{l}\cap\hat{BC}_{l}=\{(\frac{l}{2}(1+\sqrt{5}),70-l)\}.$$
Best Answer
I am afraid the co-ordinates cannot be easily determined(though, theoretically they can be if you are willing to calculate!).Here's a geometric insight into the problem.
Firstly,we shall use the angle bisector theorem and the property of the Apollonius circle.
Here in the problem,find $K'$ and $K''$ on $BC$ and $BC$ produced such that $\frac{PB}{PC}=\frac{BK'}{K'C}=\frac{BK''}{-K''C}$(we are using directed segments here).Now draw the circle $\gamma_1$ with diameter $K'K''$Now repeat the same trick with side $CA$ and draw another such circle $\gamma_{2}$. So, $\gamma_{1}\cap\gamma_{2}=P$.This suggest that you may theoretically get the co-ordinate of $P$ but I am not sure if that is a worthwhile exercise.I have no idea as to whether this will help you.