I am not a Math guy. So, if this question is silly, then please forgive me.
I have a triangle which has the following coordinates: (all the measurements are in cm.)
A - (35, 0)
B - (70, 70)
C - (0, 70)
I have filled this triangle with black color.
Now, I want to draw another triangle inside it. I don't know what the co-ordinates of that triangle will be. But after drawing new triangle i will fill it with red color and it should look like there is a black border of 1 cm around it.
Can you please let me know the co-ordinates for new triangle?
Update:
@Floris gave me this value of 3 pairs of vertices:
$(70-\frac{1}{2}(\sqrt{5}+1),69)$, $(35,\sqrt{5})$ and $(\frac{1}{2}(\sqrt{5}+1),69)$
In this case:
A – (35, 0)
B – (70, 70)
C – (0, 70)
P – $(35,\sqrt{5})$
Q – $(70-\frac{1}{2}(\sqrt{5}+1),69)$
R – $(\frac{1}{2}(\sqrt{5}+1),69)$
Which gives me output like:
But now suppose I want to have border of 10 cm:
So, I changed the value to: (please consider x = 10)
A – (35, 0)
B – (70, 70)
C – (0, 70)
P – 35,x√5
Q – 70-((x/2)(√5+x)),70-x
R – (1/2)(√5+x),70-x
But I am getting
I want the formula for inner triangle's vertices. In short I need P, Q, and R having some x.
Best Answer
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 $$\hat{AC}_{l}\cap\hat{BC}_{l}=\{(\frac{l}{2}(1+\sqrt{5}),70-l)\}.$$