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.

  • $\begingroup$ Welcome to Math.SE! Typically, the community here likes to see your thoughts about the problem, as this helps answerers target their responses to your experience level, while avoiding wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) As you are interested more in thought process than solution, perhaps you can say something about your motivation: are you conducting research on problem-solving techniques? trying to be a better problem-solver? (doing homework that asks about process?) $\endgroup$
    – Blue
    Nov 11, 2018 at 12:00
  • 1
    $\begingroup$ I've edited the question.I'm interested in general problem solving techniques because I typically solve them by doing everything that can be done which certainly isn't an efficient way of solving problems. $\endgroup$
    – JoeDough
    Nov 11, 2018 at 12:12
  • $\begingroup$ Consider you found the square and draw the two horizontal lines from the vertexes not laying on the vertical lines you already have, the intersections of the four lines will give another figure, that can give many insights. $\endgroup$
    – N74
    Nov 11, 2018 at 13:07

2 Answers 2


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!


Two parallel lines are: $x=1$ and $x=6$
distance between them is $5$ units.
So, you can construct a square having $(1,5)$ as one co-ordinate, and another lying on line $x=6$, opposite to $(1,5)$ ,i.e., $(6,5)$

Take other $2$ points on lines $x=1$ and $x=6$ at a distance of $5$ units from $2$ points already taken.

So, there are $2$ possible squares: $$(1,5),(6,5),(1,0),(6,0)$$ and $$(1,5),(6,5),(1,10),(6,10)$$

  • $\begingroup$ I've made a mistake, first line isn't $x=2$ but $x=1$ $\endgroup$
    – JoeDough
    Nov 11, 2018 at 11:10
  • $\begingroup$ no problem, just shift the points, and now length of side will be 5 $\endgroup$ Nov 11, 2018 at 11:11
  • $\begingroup$ @JoeDough is your point not (1,5) now? $\endgroup$ Nov 11, 2018 at 11:16
  • $\begingroup$ But where's the point (2,5)?It's the only given point of the square, other 2 are somewhere on the given lines and 4th one has to be found.I've edited the question to be more clear. $\endgroup$
    – JoeDough
    Nov 11, 2018 at 11:17
  • $\begingroup$ ohk, then you will have to plot all points and proceed accordingly $\endgroup$ Nov 11, 2018 at 11:18

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