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Given three parallel straight lines. Construct a square three of whose vertices belong to these lines. enter image description here

What does "belongs" mean in the context of this question? Do the three lines have to be used to construct the square?

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    $\begingroup$ As stated, the condition is ambiguous, so perhaps one can argue that there's a "trivial" solution with two vertices on one line and two vertices on another line. However, it is almost-certianly the intention of the problem that each line contains (at least) one vertex. $\endgroup$ – Blue Feb 19 at 22:22
  • $\begingroup$ Oh, and there's another trivial solution: two opposite vertices on one line, and one vertex on another line. :) $\endgroup$ – Blue Feb 19 at 22:34
  • $\begingroup$ Probably the lines have to have the same distance between them. $\endgroup$ – NoChance Feb 19 at 22:43
  • $\begingroup$ @NoChance: Same distances aren't necessary. $\endgroup$ – Blue Feb 19 at 22:45
  • $\begingroup$ @Blue, it would be nice if you could provide a quick sketch...Thanks. $\endgroup$ – NoChance Feb 19 at 22:46
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We are going to show two solutions, one geometric and the other algebraic :

Geometric solution :

Yes, the 3 lines are required. Consider the following figure :

enter image description here

Fig. 1.

Take any point $O$ on the line which is in-between the two others, called extreme lines. Define $A$ and $A'$ as the perpendicular projections of $O$ on these extreme lines. Let $d'=OA$ and $d=OA'$. Let $B$ and $B'$ be the points on the two extreme lines such that $AB=d$, and $A'B'=d'$, taken in the same direction (see remarks below). let us establish that :

$B,O,B'$ are solution points for this problem.

Proof : Triangles $OAB$ and $OA'B'$ are isometrical right triangles ; thus they possess complementary acute angles $\alpha$ and $\beta$ (see remarks below) with

$$\alpha+\beta=90° \tag{1}.$$

Consider flat angle $A'OA$. We can write, with the notations given in the figure :

$$\text{angle} \ A'OA \ = \ \alpha+\beta+\gamma \ = \ 180°$$

Using (1), we deduce that $\gamma=90°$. As the hypotenuses are equal, i.e., $OB=OB'$, the rectangle that can be built on $OB$ and $OB'$ is a square.

Remarks : 2 more rigorous definitions :

1) "taken in the same direction" means that $\vec{AB}=k \vec{A'B'}$ with $k>0$.

2) $\alpha$ could be formally defined as the acute angle such that $\tan(\alpha)=d/d'$.

Other remarks :

1) The above figure is reminiscent of the (now classical) proof of Pythagoras' theorem : see for example http://qed.sbytes.com/pythagoras.

2) There are two other solutions (@greedoid has found one of them, and I am grateful to her/him to have pointed in this direction). Let us recall that figure 1 was based on a right triangle with legs $(d,d')$. But using an auxiliary (dotted) line whose definition we leave to the reader, we get triangles with legs $(d+d',d)$ (Fig. 2) and $(d+d',d')$ (Fig. 3).

enter image description here

Fig. 2.

enter image description here

Fig. 3.

Algebraic solution :

Consider that coordinate axes have been chosen in such a way that the current points on the different lines are resp.

$$A_0(x_0,a), \ \ A_1(x_1,b), \ \ A_2(x_2,c).$$

with $a,b,c$ fixed. As the issue is translation invariant, one may assume WLOG that $x_0:=0$.

The constraint of the problem are twofold :

$$\begin{cases}\|\vec{A_0A_1}\|^2=\|\vec{A_0A_2}\|^2\\ \vec{A_0A_1} \perp \vec{A_0A_2} \end{cases}\tag{*} $$

As $\vec{A_0A_1}=\binom{x_1}{u}$ and $\vec{A_0A_2}=\binom{x_2}{v}$ with

$u=:b-a$ and $v:=c-a$, we can transform (*) into the following system of 2 equations with 2 unknowns :

$$\begin{cases}x_1^2+u^2=x_2^2+v^2\\ x_1x_2+uv=0\end{cases}\tag{**} $$

This system has clearly two solutions

$$\begin{cases}x_1=u\\ x_2=-v \end{cases} \ \ \text{and} \ \ \begin{cases}x_1=-u\\ x_2=v \end{cases} \tag{***}$$

and there are no other solutions because system (**) is equivalent to a quadratic equation which has but 2 solutions.

Result (***) is in complete agreement with any of the 3 solutions found in the geometrical solution.

What we have done for one of the straight lines can be done for any of the two others (we haven't made any assumption on the order of numbers $a,b,c$) : this explains why we have 3 essentially different solutions.

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  • $\begingroup$ This is a bit complex for me. I was wondering what implies that the angle BOB' is a 90-degrees? $\endgroup$ – NoChance Feb 20 at 10:51
  • $\begingroup$ Angle AOA' is a flat angle (180°). If you subtract from it angles AOB ($\alpha$ degrees) and A'OB' ($90° - \alpha$), what remains is angle BOB' = $180°- (\alpha+90°-\alpha)=90°$. Remark : conversion degrees - radians is done using $\frac{\pi}{2}=90°$. $\endgroup$ – Jean Marie Feb 20 at 11:08
  • $\begingroup$ Thanks for your clarification. What I meant to ask is, the steps used in the construction of the square don't imply that BOB' is 90 degrees or that $A'OB'=\frac{\pi}{2}-\alpha$ $\endgroup$ – NoChance Feb 20 at 11:35
  • $\begingroup$ No, all what I do is to construct right triangles OAB and OA'B' which are identical triangles, thus with say the most acute angle $\alpha$, and the less acute angle $(90°-\alpha)$ for both of them. Nothing more. Of course, I have, afterwards, completed my drawing by displaying the "finished" square. $\endgroup$ – Jean Marie Feb 20 at 11:42
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    $\begingroup$ I have added an algebraic solution. $\endgroup$ – Jean Marie Feb 24 at 14:46
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HINT: see diagram below to find that side $l$ of the square is equal to $\sqrt{a^2+b^2}$.

enter image description here

EDIT. As pointed out by greedoid, there are two more solutions:

enter image description here

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First case, $A$ is on inner line (left picture). Since the rotation of $B$ for $90^{\circ}$ around $A$ takes $B$ to $D$ so it takes line which has $B$ to new line which cuts the other outher line in $D$.

enter image description here

Second case, $A'$ is on outer line (right picture). Since the rotation of $A'$ for $90^{\circ}$ around $D'$ takes $A'$ to $C'$ so it takes line which has $A'$ to new line which cuts inner line in $C'$.

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  • $\begingroup$ [+1] I hadn't thought to this second solution. $\endgroup$ – Jean Marie Feb 21 at 0:08
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    $\begingroup$ There is a third solution where the right angle is situated on the third line. $\endgroup$ – Jean Marie Feb 22 at 10:43
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A vertex is a point. A line is a set of points. The problem asks of you to construct a square such that the given lines "pass through" three vertices of the square.

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  • $\begingroup$ Good. A drawing would help. $\endgroup$ – NoChance Feb 19 at 22:40
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Let $l$, $k$ and $m$ be given parallel lines such that $k$ placed between $l$ and $m$.

For example, see the Aretino's picture.

Let m be the upper line, k be the middle line and l be the lower line.

Let $A\in k$ and $R^{90^{\circ}}_A$ be a rotation around $A$ by $90^{\circ}.$

Now, let $R^{90^{\circ}}_A(l)\cap m=\{B\}$ and $R^{-90^{\circ}}_A(m)\cap l=\{D\}$.

Thus, $AB\perp AD$ and $AB=AD$.

Can you end it now?

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  • $\begingroup$ Can down-voter explain me why he made it? $\endgroup$ – Michael Rozenberg Feb 20 at 8:31
  • $\begingroup$ @greedoid It was your down-voting? $\endgroup$ – Michael Rozenberg Feb 20 at 8:39
  • $\begingroup$ Your solution is incomplete and very hard to read. Also, how you came up with those rotations? Fix that and I will upvote. $\endgroup$ – Aqua Feb 20 at 10:21
  • $\begingroup$ @greedoid I fixed, but you can say me before and I am ready to explain. $\endgroup$ – Michael Rozenberg Feb 20 at 10:42
  • $\begingroup$ Basicly you change nothing, also this problem has 2 nocongruent solutions. $\endgroup$ – Aqua Feb 20 at 10:47

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