Rotate a triangle to face a fixed point

Question

my problem consists of rotating a triangle around the T axis so that the point A faces the point P. The only known points are A, B, C and the T (the center of the triangle), look at the left figure below.

My question is how to make this transformation and have the new coordinates (x, y) of the 3 points (A,B,C) that form the triangle as in the right figure below :

EDIT:

VertexArray rotatedTri(Vector2f T,Vector2f P,Vector2f A,Vector2f B,Vector2f C)
{

VertexArray mtri(Triangles);

double s = atan2(P.y-T.y,P.x-T.x) - atan2(A.y-T.y,A.x-T.x);

double aX = P.x + cos(s) * (A.x-P.x) + sin(s) * (A.y - P.y);
double aY = P.y + sin(s) * (A.y-P.y) + cos(s) * (A.x - P.x);

double bX = P.x + cos(s) * (B.x-P.x) + sin(s) * (B.y - P.y);
double bY = P.y + sin(s) * (B.y-P.y) + cos(s) * (B.x - P.x);

double cX = P.x + cos(s) * (C.x-P.x) + sin(s) * (C.y - P.y);
double cY = P.y + sin(s) * (C.y-P.y) + cos(s) * (C.x - P.x);


mtri.append(Vector2f(aX,aY));
mtri.append(Vector2f(bX,bY));
mtri.append(Vector2f(cX,cY));

return mtri;
}

Answer 1

Write each point like $A$ as $(a_x, a_y)$.

Let $$ s = atan2(p_y- t_y, p_x - t_x) - atan2(a_y - t_y, a_x - t_x); $$ That's the angle ($\bmod 2\pi$) between the rays $TA$ and $TP$. Then $$ a'_x = t_x + \cos(s) (a_x - t_x) - \sin(s) ( a_y - t_y) \\ a'_y = t_y + \sin(s) (a_x - t_x) + \cos(s) ( a_y - t_y) $$ and similar formulas hold for $B'$ and $C'$, where primed points indicate "post-transform". So $$ b'_x = t_x + \cos(s) (b_x - t_x) - \sin(s) ( b_y - t_y) \\ b'_y = t_y + \sin(s) (b_x - t_x) + \cos(s) ( b_y - t_y) $$ etc.

(Note: atan2 is a standard function in most computer math libraries; roughly, $atan2(y, x)$ is a slight generalization of $\arctan(\frac{y}{x})$ to handle angles of magnitude greater than $\pi/2$.)

Matlab code:

function mse2()
% pick three points A, B, C, and let T be their average, to give a point to
% rotate around. Then pick $P$ that's farther from $T$ than any of the
% three points:
clear;
close all;

A = rand(2, 1); B = rand(2, 1); C = rand(2, 1); 
T = (A + B + C)/3; 
m = max([dot(T-A, T-A), dot(T-B, T-B), dot(T-C, T-C)]); 
d = 2 * sqrt(m); % twice the distance to the closest point. 
v = rand(1); 
P = T + [d * cos(v), d* sin(v)]; 

% show this 
hold on; 
tplot(A, 'A');
tplot(B, 'B');
tplot(C, 'C');
tplot(T, 'T');
tplot(P, 'P');
plot([A(1), B(1), C(1), A(1)], [A(2), B(2), C(2), A(2)], '-r');

%
% OK, now compute the primed points:
s = atan2( P(2) - T(2), P(1) - T(1)) - atan2(A(2) - T(2), A(1) - T(1)); 
cc = cos(s);
ss = sin(s); 
M = [cc, -ss; ss, cc];

Ap = T + M*(A-T);
Bp = T + M *(B-T);
Cp = T + M * (C-T)
tplot(Ap, 'Ap');
tplot(Bp, 'Bp');
tplot(Cp, 'Cp');
plot([Ap(1), Bp(1), Cp(1), Ap(1)], [Ap(2), Bp(2), Cp(2), Ap(2)], '-g');
hold off;
figure(gcf)

function tplot(X, name)
plot(X(1), X(2), 'o');
text(X(1) + 0.01, X(2), name);

Here's a typical (graphics) output of the program, showing before (Red) and after (green):

Answer 2

[This is a variation of John Hughes’ answer that doesn’t require explicitly computing any angles or trigonometric functions of them.].

Let the coordinates of the points be $A(a_x,a_y)$, $B(b_x,b_y)$, etc. A rotation about the point $T$ is accomplished by first translating $T$ to the origin, rotating about the origin, and then translating back. The first and last steps are done by subtracting/adding, respectively, the coordinates of $T$ to those of the other points, so the only tricky part is working out the rotation in between.

We can break that rotation down into two steps: first, rotating so that the point $A$ is on the positive $x$-axis, and then rotating from there to align it with $P$. Using the fact that the columns of a transformation matrix are the images of the basis, we can write down a matrix for the second rotation without explicitly computing the rotation angle. The $x$-axis gets sent to a unit vector in the direction of $P-T$, so the first column is $P-T$, normalized. The second column is the image of the $y$-axis, which is just $P-T$ rotated 90° counterclockwise and normalized. So, the second rotation is represented by the matrix $$R_2={1\over\sqrt{(p_x-t_x)^2+(p_y-t_y)^2}}\begin{bmatrix}p_x-t_x & -(p_y-t_y) \\ p_y-t_y & p_x-t_x \end{bmatrix}.$$ We can construct the first rotation matrix $R_1$ in a similar way. However, for that we want to rotate $A-T$ to to $x$-axis, so we want the inverse matrix, which for a rotation is its transpose. That is, instead of $A-T$ being the first column, it’s the first row. That gives $$R_1={1\over\sqrt{(a_x-t_x)^2+(a_y-t_y)^2}}\begin{bmatrix} a_x-t_x & a_y-t_y \\ -(a_y-t_y) & a_x-t_x \end{bmatrix}.$$ The matrix for the total rotation is the product of these two matrices: $$R=R_2R_1={1\over\sqrt{({a'_x}^2+{a'_y}^2)({p'_x}^2+{p'_y}^2)}}\begin{bmatrix} a'_xp'_x+a'_yp'_y & -(a'_xp'_y-a'_yp'_x) \\ a'_xp'_y-a'_yp'_x & a'_xp'_x+a'_yp'_y\end{bmatrix}$$ where $A'=(a'_x,a'_y)=A-T=(a_x-t_x,a_y-t_y)$ and so on. Comparing this to John Hughes’ answer, we can see that $$\begin{align} \cos(s) &= {a'_xp'_x+a'_yp'_y \over \sqrt{({a'_x}^2+{a'_y}^2)({p'_x}^2+{p'_y}^2)}} \\ \sin(s) &= {a'_xp'_y-a'_yp'_x \over \sqrt{({a'_x}^2+{a'_y}^2)({p'_x}^2+{p'_y}^2)}} \end{align}$$ so you can use these formulas to get the sine and cosine of the total rotation angle directly without needing to compute any of the angles explicitly. From there, you can proceed as in the other answer. (These two expressions can also be derived from the formulas for the sine and cosine of the difference of two angles.)

It’s also useful to perform a couple of sanity checks to make sure that you haven’t made a sign error or some other mistake along the way. Since you want $A$ to end up aligned with $P$, you can compute the resulting point directly: it’s just $T+{|A-T|\over|P-T|}(P-T)$. Compute that for a couple of sample configurations and compare it to the result you’re getting from the rotation. Another sanity check is to compute the determinant $$\begin{vmatrix}a_x&a_y&1\\b_x&b_y&1\\c_x&c_y&1\end{vmatrix}$$ before and after rotating the points. The values should be equal.