Little explorations with HP calculators (no Prime)

[Han]

Here is a proof (using linear algebra) that reduces the general case to the special case provided by Gerson.

Define the barycentric coordinates \(\mathbf{\lambda} = (\lambda_1, \lambda_2, \dotsm, \lambda_n ) \) of a point \( \mathbf{P}\in \mathbf{R}^m \) relative to a set of points \( \{ \mathbf{Q}_1, \mathbf{Q}_2, \dotsm, \mathbf{Q}_n \} \) as the solution to the equation
\[ \mathbf{P} =\sum_{k=1}^n \lambda_k \cdot \mathbf{Q}_k \]
(This is actually a system of equations in the coordinates of the points.) We can even require that \( \sum \lambda_k = 1 \) by embedding these points in a hyperplane \( x_{m+1} = 1 \) of \( \mathbb{R}^{m+1} \) though this is not necessary for this problem.

Let \(f \) be the linear transformation \( f(\mathbf{x}) = \mathbf{A}\cdot \mathbf{x} \) with \( |\mathbf{A}| = 1 \) (determinant equal to 1; i.e. \(f\) is a unimodular transformation). Observe that \( f \) preserves volume (since \( | \mathbf{A}| = 1 \)), and \( f \) preserves barycentric coordinates because:
\[ f(\mathbf{P}) = \mathbf{A}\cdot \mathbf{P} =
\mathbf{A} \cdot \sum_{k=1}^n \lambda_k \cdot \mathbf{Q}_k
= \sum_{k=1}^n \lambda_k \cdot \mathbf{A}\cdot \mathbf{Q}_k
= \sum_{k=1}^n \lambda_k \cdot f(\mathbf{Q}_k) \]
Thus, it suffices to show that every non-degenerate triangle ABC with
\[
\begin{align*}
D & = \frac{1}{3}\cdot A + \frac{2}{3}\cdot B + 0 \cdot C\\
E & = \frac{1}{2}\cdot A + 0\cdot B + \frac{1}{2}\cdot C\\
F & = 0\cdot A + \frac{1}{4}\cdot B + \frac{3}{4}\cdot C\\
\end{align*}
\]
can be transformed into a 30-60-90 right triangle (with points A'=f(A), B'=f(B), and C'=f(C)) having the same area and D', E', and F' satisfying the three equations analogous to those for D, E, and F above. Let \( \mathbf{u} \) be the vector from A to C, and \( \mathbf{v} \) be the vector from A to B. Similarly let \( \mathbf{u}' \) be the vector from A' to C', and \( \mathbf{v}' \) be the vector from A' to B'. Then the transformation matrix \( \mathbf{A} \) is the solution to
\[\mathbf{A}\cdot \begin{bmatrix} \mathbf{u} & \mathbf{v} \end{bmatrix}
= \begin{bmatrix} \mathbf{u}' & \mathbf{v}' \end{bmatrix} \]
This solution always exists provided that ABC (and likewise A'B'C') is non-degenerate (i.e. vertices are in general position).