I'm going to solve this geometry problem without using rotational , any other way like coordinates :
Problem :
Given an equilateral triangle $\Delta ABC $ such that :
$P$ point inside triangle with distance from the vertices of triangle as following in picture :
How I use coordinates here ?
Can you assist!
Thanks!
On
It's rather a comment to the Carl Schildkraut's answer, it's not in a way an answer itself. I already mentioned a link in the comments to the question for which this could be a duplicate, except this question conversely wants coordinates. Vectors are more neat here, IMHO.
Following the steps in the answer we have
0. Expand the things
$$\begin{cases}
4 t^2 - 4 t x + x^2 + y^2 = 100\\
4 t^2 + 2 t x - 2 \sqrt{3} t y + x^2 + y^2 = 64\\
4 t^2 + 2 t x + 2 \sqrt{3} t y + x^2 + y^2 = 36
\end{cases}$$
1.
$$\begin{cases}
4 t^2 - 4 t x + x^2 + y^2 = 100\\
4 t^2 + 2 t x - 2 \sqrt{3} t y + x^2 + y^2 = 64\\
4 \sqrt{3} t y = 36-64
\end{cases}$$
2.
$$\begin{cases}
4 t^2 - 4 t x + x^2 + y^2 = 100\\
6 t x - 2 \sqrt{3} t y = 64-100\\
2 \sqrt{3} t y = -14
\end{cases}$$
$$\begin{cases}
4 t^2 - 4 t x + x^2 + y^2 = 100\\
6 t x = 64-100-14=-50\\
2 \sqrt{3} t y = -14
\end{cases}$$
3.
$$\begin{cases}
4 t^4 - 4 t^2 ( t x ) + t^2 x^2 + t^2 y^2 = 100t^2\\
4 t x =-\frac{100}{3}\\
2 \sqrt{3} t y = -14
\end{cases}$$
$$\begin{cases}
12 t^4 - 3 t^2\cdot (4 t x ) + 3 t^2 x^2 + 3 t^2 y^2 = 300t^2\\
4 t x =-\frac{100}{3}\\
\sqrt{3} t y = -7
\end{cases}$$
$$12t^4-200t^2+\frac{625}{3}+49=0$$
$$36t^4-600t^2+772=0$$
$$9t^4-150t^2+193=0$$
$$\frac{D}{4}=75^2-9\cdot 193=36^2\cdot 3$$
$$t^2=\frac{75\pm 36\sqrt{3}}{9}$$
$$t=\pm_2\sqrt{\frac{25}{3}\pm_1 4\sqrt{3}}$$
Any questions welcome.
Orient the triangle to be centered at $(0,0)$ and let $t$ be $\frac{s}{2\sqrt{3}}$ so that $$A=(2t,0), B=(-t,t\sqrt{3}), C=(-t,-t\sqrt{3}).$$ We have some point $P=(x,y)$ so that $$(2t-x)^2+y^2=100,\ (t+x)^2+(y-t\sqrt{3})^2=64,\ (t+x)^2+(y+t\sqrt{3})^2=36.$$ I won't do out the algebra here because it's rather messy, but here's the general strategy: