Let $a$ and $b$ be $2$ real and strictly positive numbers. The
points $B(a, 0)$ and $D(0, b)$ are $2$ opposite points of the square
$ABCD$. Find the other $2$ points.
This is what the information given would look like on a graph.
Where $B(a,0)$ can slide along the $Ox$ axis (but not at or to the left of $O(0,0)$ since $a$ is strictly positive) and $D(0, b)$ can slide along the $Oy$ axis (but not at or lower than $O(0,0)$ since $b$ is strictly positive).
So now we have to find the other $2$ points of the square $ABCD$ (I assume in terms of $a$ and $b$). This is what I tried:
Because $ABCD$ is a square I thought we should use the many special properties of the diagonals of a square. So the first thing I did was to find the equation of the line $d$ that passes through the diagonal $[DB]$. We know that this is a diagonal since the statement of the problem says that $B$ and $D$ are opposite points of the square. This is that line:
$$d: y = -\dfrac{b}{a}x + b$$
Then , we know that the diagonals intersect at their mid-points, so using the points $B(a, 0)$ and $D(0, b)$ of the diagonal, we can find the center of the square, $M$:
$$x_M = \dfrac{x_B + x_D}{2} = \dfrac{a + 0}{2} = \dfrac{a}{2}$$
$$y_M = \dfrac{y_B + y_D}{2} = \dfrac{0 + b}{2} = \dfrac{b}{2}$$
So we have that the intersection of the diagonals (center of the square) is:
$$M \bigg (\dfrac{a}{2}, \dfrac{b}{2} \bigg )$$
Then , knowing that the diagonals of a square are perpendicular, knowing the equation of the line of one of the diagonals and knowing that the point $M$ is on this other diagonal (since $M$ is on both of the diagonals), we can find the equation of the line on which the second diagonal is located. After some calculations I found this equation to be:
$$g : y = \dfrac{a}{b}x + \dfrac{b^2 – a^2}{2b}$$
Now our picture looks like this:
And now all that is left is that we have to find the $2$ points $A$ and $C$ that are located on the blue line, $g$, such that $ABCD$ is a square. Because we have that $A \in g$ and $C \in g$, the $2$ points are:
$$A \bigg( x_A, \dfrac{a}{b}x_A + \dfrac{b^2-a^2}{2b} \bigg )$$
$$C \bigg( x_C, \dfrac{a}{b}x_C + \dfrac{b^2-a^2}{2b} \bigg )$$
And I am lost here. I can't find $x_A$ and $x_C$ in terms of $a$ and $b$. I tried equating segment lengths, finding those segment lengths using the general formula:
$$NM = \sqrt{(x_N – x_M)^2 + (y_N – y_M)^2}$$
for $2$ points $N(x_N, y_N)$ and $M(x_M, y_M)$. But that resulted in some monstrous calculations with a quadratic of $x_A$ and a bunch of $a's$ and $b's$.
So how can I find $A$ and $C$?
Best Answer
I made an interactive graph on Desmos that allows you to drag points $B$ and $D$ and watch how points $A$ and $C$ change.
While it might not be the quickest way to get there, you can certainly get the solution in the way that you were attempting - with your function $g$ and the distance function, as you were on track to do!
To solve the problem this way, just set up the two equations, and solve. You want the two points $X \in \{A, C\} = \{(x_A, y_A), (x_C, y_C)\}$ that satisfy two conditions. First, they lie along the line $g$, i.e. \begin{align} y &= \frac{a}{b}x + \frac{b^2 - a^2}{2b} \end{align}
Second, the distance from $B$ to $D$ is twice the distance from each $X$ to $M$, i.e. \begin{align} \sqrt{a^2 + b^2} &= 2 \sqrt{\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b}{2}\right)^2} \\ \Rightarrow a^2 + b^2 &= 4 \left(x - \frac{a}{2}\right)^2 + 4 \left(y - \frac{b}{2}\right)^2 \\ \Rightarrow a^2 + b^2 &= \left(2x - a\right)^2 + \left(2y - b\right)^2 \end{align}
Substitute the value for $y$ from the first equation into the second, and solve for $x$. You need two points, but the equation is quadratic, so that's a hint that you're probably gonna be OK and end up with 2 possible solutions which are your two points. \begin{align} a^2 + b^2 &= \left(2x - a\right)^2 + \left(2\left( \frac{a}{b}x + \frac{b^2 - a^2}{2b} \right) - b\right)^2 \\ a^2 + b^2 &= \left(2x - a\right)^2 + \frac{a^2}{b^2} \left(2x - a\right)^2 \\ a^2 + b^2 &= \frac{a^2 + b^2}{b^2} \left(2x - a\right)^2 \\ b^2 &= \left(2x - a\right)^2 \Rightarrow \pm b = 2x - a \end{align} $$ \Rightarrow x = \frac{a \pm b}{2} $$
The two solutions ($\pm$) give you the two values for $x$: \begin{align} x_A &= \frac{a + b}{2}\;,\; x_C = \frac{a - b}{2} \end{align}
Substitute each solution for $x$ into the equation for $y$, to find the two associated values of $y$. \begin{align} y_A &= \frac{a}{b}\left(\frac{a+b}{2}\right) + \frac{b^2-a^2}{2b} = \frac{a + b}{2} \\ y_C &= \frac{a}{b}\left(\frac{a-b}{2}\right) + \frac{b^2-a^2}{2b} = \frac{-a + b}{2} \end{align} Therefore points $A$ and $C$ are given by \begin{align} A &= (x_A, y_A) = \left(\frac{a + b}{2} , \frac{a + b}{2}\right) \\ C &= (x_C, y_C) = \left(\frac{a - b}{2} , \frac{-a + b}{2}\right) \end{align}
Looking at the interactive graph, the way we found $A$ and $C$ was to solve for the intersections between the circle centered at $M$ - which is the set of points at the same distance from $M$ as $B$ and $D$ - and the red line $g$ which is perpendicular to $\overline{BD}$ and which passes through $M$.