A rectangle has dimensions $a$ units by $b$ units with $a > b$. A diagonal divides the rectangle into two triangles. A square, with sides parallel to those of the rectangle, is inscribed in each triangle. Find the distance between the vertices (of the squares) that lie in the interior of the rectangle.
In the solution, we find that one square has an 'interior' vertex $(\frac{ab}{a+b},\frac{ab}{a+b})$, because the vertex must lie on both $y=x$ and $y = -\frac{b}{a}x + b$. Then the author asserts that symmetry permits us to conclude that the coordinates of the other vertex of the other inscribed square are $(\frac{a^2}{a+b}.\frac{b^2}{a+b})$. I have drawn several pictures trying to see the symmetry, but I can't seem to discern it, so I left no other option but a (relatively) brute-force method. Perhaps someone with a more geometric eye could kindly point out the symmetry being appealed to.
Best Answer
If we let rectangle have vertices $(0,0), (a,0), (0,b), (a,b)$, then we have central symmetry with respect to $(\frac a2,\frac b2)$ (this is the intersection of diagonals). This central symmetry is given by formula $(x,y)\mapsto (a-x, b -y)$, and thus $(\frac{ab}{a+b},\frac{ab}{a+b})\mapsto (\frac{a^2}{a+b},\frac{b^2}{a+b}).$
To find formula for central symmetry with respect to $(x_0,y_0)$, remember that central symmetry is affine transformation, and thus of the form $f(v) = Av + b$, where $A$ is linear map. To find central symmetry, first observe that $f(0,0) = 2(x_0,y_0)$, while $Av = -v$ (because that is central symmetry with respect to origin). Thus, central symmetry is given by $f(x,y) = (2x_0-x,2y_0-y)$.