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.

  • $\begingroup$ @lioness99a Yes. I mentioned that in the body of my question. $\endgroup$ – user193319 Jul 24 '17 at 13:08

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 $f(x,y) = (2x_0-x,2y_0-y)$.


You've set $(0,0)$ as the bottom left vertex and the diagonally opposite vertex as $(a,b)$. In addition, the diagonal drawn has negative gradient.

I believe 'by symmetry' just simply means that if $(p,q)$ is the interior vertex of one square, since the diagram has a rotational symmetry of $180°$ about the centre of the rectangle, then the interior vertex of the other square will be $(a-p, b-q)$.

Here, $(p,q)=(\frac{ab}{a+b},\frac{ab}{a+b})$, so

$$(a-p, b-q)=\left(a-\frac{ab}{a+b},b-\frac{ab}{a+b}\right)=\left(\frac{a^2+ab-ab}{a+b},\frac{b^2+ab-ab}{a+b}\right)=\left(\frac{a^2}{a+b},\frac{b^2}{a+b}\right)$$


The figure should be drawn in the following way:-

enter image description here

The result of $P = (\dfrac {ab}{a + b}, \dfrac {ab}{a + b})$ can be found by solving the two mentioned equations.

We use the vertical line through Q as the line of symmetry (for P and R). Its equation is $x = \dfrac {a}{2}$. The symmetrical distance between P (and also R) from that line is $\dfrac {a}{2} - \dfrac {ab}{a + b}$.

Then the x- coordinate of the other vertex (T) = that of R is given by:- $\dfrac {ab}{a + b} + 2 \times [\dfrac {a}{2} - \dfrac {ab}{a + b}] = … =\dfrac {a^2}{a + b}$.

The y-coordinate of T can be found in a similar fashion.


We can draw a picture of this situation as follows:


We need to find $c,d,e,f$ to know the interior vertices of the squares.

Can you see that this is rotationally symmetric, if we spin it around by $180^\circ$, it will be exactly the same problem?

This means that the distance between $(0,b)$ and $(c,f)$ is equal to the distance between $(e,d)$ and $(a,0)$. Equally, the distance from $(c,f)$ to $(a,0)$ is equal to the distance from $(e,d)$ to $(0,b)$.

We can see that $$(c,f)=\left(\frac{ab}{a+b}, \frac{ab}{a+b}\right)$$ by the definition of a square, and then we can find the distance from here to $(a,0)$:

\begin{align}D_1&=\sqrt{\left(a-\frac{ab}{a+b}\right)^2+\left(0-\frac{ab}{a+b}\right)^2}\\ &=\sqrt{\left(\frac{a^2}{a+b}\right)^2+\left(-\frac{ab}{a+b}\right)^2}\\ &=\sqrt{\frac{a^4}{(a+b)^2}+\frac{a^2b^2}{(a+b)^2}}\\ &=\frac{a\sqrt{a^2+b^2}}{a+b}\end{align}

Now we know that the distance from $(e,d)$ to $(0,b)$ is equal to this:


We can repeat this with the other distances to get \begin{align}D_2&=\sqrt{\left(\frac {ab}{a+b}-0\right)^2+\left(\frac{ab}{a+b}-b\right)^2}\\ &=\frac{b\sqrt{a^2+b^2}}{a+b}\end{align}

and therefore \begin{align}\sqrt{(e-a)^2+d^2}&=\frac{b\sqrt{a^2+b^2}}{a+b}\end{align}

Now we have two simultaneous equations in $e,d$ to solve:

\begin{align}\sqrt{e^2+(d-b)^2}&=\frac{a\sqrt{a^2+b^2}}{a+b}\\ \sqrt{(e-a)^2+d^2}&=\frac{b\sqrt{a^2+b^2}}{a+b}\end{align}

If we solve these while remembering that $0<e<a$ and $0<d<b$ then we find that \begin{align}e&=\frac{a^2}{a+b}\\ d&=\frac{b^2}{a+b}\end{align}

  • 2
    $\begingroup$ Your figure does not correspond to the disposition considered by the OP. $\endgroup$ – Christian Blatter Jul 24 '17 at 13:53
  • $\begingroup$ @ChristianBlatter Why not? What is wrong with it? $\endgroup$ – lioness99a Jul 24 '17 at 14:09
  • $\begingroup$ Well, neither of the vertices in your figure have coordinates $(\frac{ab}{a+b},\frac{ab}{a+b})$. You must consider the other diagonal. $\endgroup$ – Ennar Jul 24 '17 at 14:15
  • $\begingroup$ Corrected my answer $\endgroup$ – lioness99a Jul 24 '17 at 15:15
  • $\begingroup$ You cannot say without loss of generality that $(e,d) = (ab/(a+b),ab/(a+b))$ since it is false. You see, $(ab/(a+b),ab/(a+b))$ lies on the line $y = x$, and upper right vertex of the left square by definition of square lies there, while the other vertex simply cannot in that case, since $a>b$. $\endgroup$ – Ennar Jul 24 '17 at 17:51

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