Construct a square given a point and two lines.

analytic geometrygeometry

The problem:

One point of the square – $(2,5)$, and lines $x=1$, $x=6$ are given (1 point of the square is on each given line), all points of the square are in the first quadrant.
A square needs to be constructed with this information.

How would you solve this problem and thus determine if a square can even be constructed?
I'm interested more in your thought process than the solution.
How do you approach a problem like this?
Which steps do you take and why?

I've considered equating the distances between points (2,5) and (1, a) with the distance between (2,5) and (6,b).I don't think that anything can be done with this, maybe I have to find something else and make a system of equations?I've tried to find the distance between points (1, a) and (6,b) – that would be equal to sqrt(2)*n (n is the length of square's side), connect it with some distance (between the given and unknown point) but I can't extract one variable so I can insert it into the first equation.

Best Answer

Let $A$ be the given point, and let $B$ and $C$ be points on the respective given lines. The key observation is

Because $\overline{AB}$ and $\overline{AC}$ are congruent and perpendicular, the horizontal distance between $A$ and $B$ must equal the vertical distance between $A$ and $C$, and vice-versa.

enter image description here

The image shows that $\triangle ABP \cong \triangle ACQ$, proving the observation immediately. The image also shows that there are clearly two matching choices for $B$ and $C$, so that there are two solution squares.

The idea also works when $A$ is not between the lines (note that how the $B$ and $C$ points match is reversed from above):

enter image description here

(Of course, $A$ on a line works, too, as an obvious special case where $C$ and $C^\prime$ coincide.) So, we see that it's always possible to construct the squares. $\square$

We leave as an exercise to the reader the task of finding the coordinates of the points for the specific problem in the original question.

As for thought process ...

  • My first instinct is usually to abandon specific numbers and to generalize; this prevents important algebraic and geometric patterns from being lost in the arithmetic of numbers. So, instead of $A=(2,5)$, $x=1$, $x=6$, I wanted to consider $A=(p,q)$, $x=r$, $x=s$.

  • These days, I have a bit of a knee-jerk impulse to jump into Mathematica to try a brute-force solution. I quickly assigned some coordinates ---$A=(p,q)$, $B=(r,b)$, $C=(s,c)$--- and entered conditions for making the squares: $\overline{AB}\perp\overline{AC}$ becomes $(A-B)\cdot(A-C)= 0$, while $\overline{AB}\cong\overline{AC}$ becomes $(A-B)\cdot(A-B)=(A-C)\cdot(A-C)$.

  • I let Mathematica do some instantaneous symbol-crunching, which in this case yielded linear constraints on the $y$-coordinates of $b$ and $c$. That told me the problem was actually easy, so I took a step back. The "key observation" above came to me pretty quickly ... shaming me because I didn't think of it sooner. :) (But, hey ... I only lost about two minutes!)

  • Then I went to the GeoGebra app to draw-up some diagrams that make the "key observation" obvious, and double-checking the point-outside-the-lines case. (I ofttimes go right into GeoGebra to experiment, but a first pass through Mathematica seemed quicker in this case.)

That's about it!

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