"Given three parallel straight lines. Construct a square three of whose vertices belong to these lines." Are all three lines required?

Question

Given three parallel straight lines. Construct a square three of whose vertices belong to these lines.

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

Answer 1

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.

Answer 2

HINT: see diagram below to find that side $l$ of the square is equal to $\sqrt{a^2+b^2}$.

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

Answer 3

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 :

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).

Fig. 2.

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.

Answer 4

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?

Answer 5

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 $D$.

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 $C'$.