Let $A,B,C,D$ be four points in the affine plane. Suppose there is an affine transformation $F$ that maps those points cyclically ($A$ to $B$, $B$ to $C$ and so on). Prove that $ABCD$ is a parallelogram. What if $F$ swapped $A$ with $C$ and $B$ with $D$, instead?

My attempt. My goal is to prove that $\overrightarrow{AB}=-\overrightarrow{CD}$ and $\overrightarrow{BC}=-\overrightarrow{DA}$. I know that $F$ has order $4$, but I'm not able to prove that $dF^2$ is $-Id$. I also tried to write a matrix $$\begin{pmatrix} 0&a \\ 1&b \end{pmatrix}$$ with respect to the basis $(\overrightarrow{AB},\overrightarrow{BC})$, raised to the 4th power, but I don't get a unique solution (a=-1,b=0).


Wlog, we can place the origin at $A$ and use $\vec{AB}$ and $\vec{AD}$ as our basis. In that basis, the homogeneous coordinates of the four points are $$\begin{align}A&=(0,0,1)^T\\B&=(1,0,1)^T\\C&=(x_C,y_C,1)^T\\D&=(0,1,1)^T\end{align}$$ and the problem reduces to showing that $x_C=y_C=1$.

Let $M$ be the matrix of the affine transformation $F$ relative to this basis. We can see immediately that its last column must be equal to $B$. The first column of the matrix is also fairly easy to deduce: Since $\vec{AB}\mapsto\vec{BC}$ and $\vec{AB}$ lies along the first coordinate axis, this column of $M$ must be the direction vector that corresponds to $\vec{BC}$, i.e., $(x_c-1,y_c,0)^T$. Similarly, since $\vec{AD}\mapsto\vec{BA}$, the second column of $M$ must be $(-1,0,0)^T$ (note the sign change). Thus, we have $$M=\pmatrix{x_c-1&-1&1\\y_c&0&0\\0&0&1}.$$ Multiplying $C$ by this matrix and setting the result equal to $D$ yields the following system of equations: $$x_C^2-x_C-y_C+1=0\\x_Cy_C=1$$ for which the solution is $x_C=y_C=1$, as desired. Alternatively, you could set $M^2B=D$, which results in the same equations.

For the second part of the problem, similar considerations yield the transformation matrix $$\pmatrix{-1&0&x_C\\0&-1&y_C\\0&0&1}$$ and applying this matrix to either $B$ or $D$ eventually leads to $x_C=y_C=1$ again.

  • $\begingroup$ +1. The use of homogeneous coordinates makes the problem much easier. $\endgroup$ – Erel Segal-Halevi Feb 25 '17 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.