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Construct (with ruler and compass) a square given one point from each side.

I see a very interesting question.Answer, but there is no resolution, I do not know the reason why such a mapping?Is not there other mapping methods?

enter image description here

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I'm not sure to understand the question... Do you want a proof that this solution works? –  T. Verron Sep 29 '12 at 8:42

1 Answer 1

Well... Assuming it's a proof you want, here you go:

Draw a square, and let $A, B, C$ be 3 points, on 3 different sides of the square, in that order (so that they are in the same order as on your picture above). Draw the perpendicular to $(AC)$ passing through $B$, and let $D$ denote the intersection of this perpendicular and the 4th side of the square.

Now what we want to prove is that $BD = AC$. I'm sure there is a way to do this formally with Pythagore's theorem, but the easiest way is to deduce it from geometrical transformations : translate the $(AC)$ line until $A$ is in one corner of the square, and the $BD$ line until $B$ is in the same corner. Now $C$ is the image of $D$ by rotation of center $A\equiv B$ and angle $90°$. This shows that $BD=AC$.

To be 100% rigorous, we need to check that you indeed get a square with the construction, but the argument is identical (translate the $[AC]$ and $[BD']$ segments until they are two sides of the square).

This proves that there are actually 6 squares going through $A$, $B$, $C$ and $D$ in the wanted way, and that you can build all of them with the method given : for the first line drawn, you can choose either $(AB)$ or $(AD)$ as well as $(AC)$. And when you draw the circle of center $B$ and radius $AC$, there are 2 intersections, leading to 2 different $D'$ and 2 different squares.

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