analytic geometrygeometrytriangles
There is a triangle with $A$, $B$ and $C$ points. And there is another small triangle inside with $D$, $E$ and $F$ points. All of three sides of each triangle are parallel and all of the distances between these sides are equal: $|GH| = |IJ| = |KL| = k$
I have $A$, $B$ and $C$ points $(x_a,y_a)$,
$(x_b,y_b)$, $(x_c,y_c)$ and distance between sides $k$. How can I find the smaller triangle's points $(D, E, F)$.
$\def\A{{\bf A}} \def\B{{\bf B}} \def\C{{\bf C}} \def\R{{\bf R}} \def\D{{\bf D}} \def\f{\phi}$As others have mentioned, if the task is to find the coordinates of $B$ and $C$, this problem is underdetermined.
Let's first consider the triangle with point $A$ located at the origin and point $C$ lying along the positive $x$-axis. Then the coordinates of the points can be found with simple trigonometry, $$\begin{eqnarray*} \A_0 &=& (0,0) \\ \B_0 &=& (c \cos A, c\sin A) \\ \C_0 &=& (b,0). \end{eqnarray*}$$ The angle $A$, and the sides $b$ and $c$ have been given. This is an SAS triangle. The triangle can be solved by finding $a$ with the law of cosines and one of the other angles with the law of sines.
(Added: For completeness, $A = 72^\circ$, $b = 2.61r$, and $c=r$. The law of cosines gives $a = \sqrt{b^2+c^2-2bc\cos A} = 2.49r$. Then $\frac{\sin A}{a} = \frac{\sin B}{b}$ implies $B = 86^\circ$. Lastly, $A+B+C = 180^\circ$ implies $C = 22^\circ$.)
The collection of triangles you are interested in have vertices of the form $$\begin{equation*} \D = \A + \R(\f)\D_0 \tag{1} \end{equation*}$$ where $\A = (A_x,A_y)$ is the given location of $A$, and where $\R(\f)$ is a rotation matrix. The transformation (1) is a counterclockwise rotation by the angle $\f$, followed by a shift so the point $A$ has the given coordinates. In components, $$\begin{eqnarray*} A_x &=& A_x \\ A_y &=& A_y \\ B_x &=& A_x + c\cos A\cos\f - c\sin A\sin\f \\ &=& A_x + c\cos(A+\f) \\ B_y &=& A_y + c\cos A\sin\f + c\sin A\cos\f \\ &=& A_y + c\sin(A+\f) \\ C_x &=& A_x + b\cos \f \\ C_y &=& A_y + b\sin \f. \end{eqnarray*}$$
Below we plot the triangle before rotation and translation in black.
(We set $r = 1$ in the figure.)
The dotted triangle has been rotated counterclockwise by $\f = 30^\circ$, and then translated so the new location of point $A$ is $(2,1)$.
For reference,
$$\begin{eqnarray*}
A_x &=& 2 \\
A_y &=& 1 \\
c &=& r = 1 \\
b &=& 2.61 \\
A &=& 72^\circ \\
\f &=& 30^\circ.
\end{eqnarray*}$$
Hint:
using elemental geometrical properties of equilateral triangles you can see that:
$$ PC=PA=PB=\frac{10\sqrt{3}}{3} $$
and the distance of $P$ from the sides is half this value.
Can you do from this adding or subtracting such values from the coordinates of $P$ to find the vertices?
Best Answer
If you can find the coordinates of $D$ relative to $A$ you can find the other coordinates by the same method.
Write $\overrightarrow{AB}=\underline{a}$ and $\overrightarrow {AC} = \underline{b}$ and let $\overrightarrow{AD}=\underline{x}$
Then since D lies on the angle bisector of angle $BAC$, we have $$\underline{x}=\lambda(\underline{\hat{a}}+\underline{\hat{b}})$$
Here the hats denote the unit vectors.
Let angle $BAD=\theta$ and $\underline{\hat{n}}$ be the unit vector normal to the plane. Then, using cross-product,$$\underline{x}\times\underline{a}=|\underline{x}||\underline{a}|\sin\theta\underline{\hat{n}}=k|\underline{a}|\underline{\hat{n}}$$ $$\implies\lambda(\underline{\hat{a}}+\underline{\hat{b}})\times\underline{a}=k|\underline{a}|\underline{\hat{n}}$$ $$\implies\lambda=\frac{k}{|\underline{\hat{b}}\times\underline{\hat{a}}|}=\frac{k|\underline{a}||\underline{b}|}{2\Delta}$$
Here $\Delta$ denotes the area of the triangle $ABC$
Thus the position of $D$ relative to $A$ is given by $$\underline{x}=\frac{k}{2\Delta}\left(\underline{a}|\underline{b}|+\underline{b}|\underline{a}|\right)$$
This can be written easily in terms of the given coordinates.