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Is it possible to determine a triangle given its three perpendicular bisectors (meeting at a point which will be the circumcenter) and, say, a point of an edge, or any condition that can make the solution unique, using compass and straightedge? Of course I could put a system of equations, but I'm looking for a graphical procedure.

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I'd like to add two things: the first is that this may sound like a homework question, but in fact it isn't, I was thinking about it and don't know the answer. And secondly, I had problems looking for a solution on the Internet as it seems that in English the word "bisector" is both used for the lines which meet at the circumcenter and the lines which meet at the incenter. – Juanlu001 Feb 14 '12 at 20:17
This is because in-center is the intersection point of the angular bisectors. – Aryabhata Feb 14 '12 at 20:39
And circumcenter is the intersection of the perpendicular bisectors, yes. And I was looking for these. – Juanlu001 Feb 14 '12 at 21:04

1 Answer

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Yes, it seems like the question you have in mind is enough.

Suppose $P$ is the given point and $l,m,n$ are the perpendicular bisectors. Also assume that it is known that $P$ lies on the side which is perpendicular to $l$.

Now draw a line $q$ which is perpendicular to $l$, passing through $P$. One side of the triangle lies along this line.

Now reflect line $q$ on $m$ to give a new line $q_m$.

Similar reflect line $q$ on $n$ to give a new line $q_n$.

The intersection point of $q_m$ and $q_n$ is a vertex of the triangle and we can construct the whole triangle, given that point.

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Sorry, maybe I have been unable to explain myself, but this procedure gives me the triangle whose angular bisectors are $l$, $m$ and $n$, and I'm looking for the triangle whose perpendicular bisectors are those three lines. – Juanlu001 Feb 14 '12 at 21:17
@Juanlu001: You are right. I am having a braindead day today. – Aryabhata Feb 14 '12 at 23:17
@Juanlu001: Updated with a possible answer. – Aryabhata Feb 15 '12 at 2:20
Excellent! I think I see how it works. Thanks! – Juanlu001 Feb 15 '12 at 9:13

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