# Construction of a Triangle from Circumcenter, Orthocenter and Incenter

### Jack D'Aurizio

30 September 2008

Looking at the The many ways to construct a triangle page I was asking myself how to find the vertices of ABC, with straightedge and compass, knowing the positions of

(Incidentally, I underline that (A, a, R) does not fix a triangle, since the sine law holds.)

There are some remarkable facts. By calling N the midpoint of OH we have that

- IN = R/2 - r due to Feuerbach's theorem.
- OI² = 2R (R/2 - r) due to Euler's formula.

so we know (R, r) and can easily draw the circumcircle, the incircle, the 9-points circle, the polar circle, the ortic axis and so on.

Moreover, due to Poncelet's lemma, if we take a point U on the circumcircle, draw the tangents to the incircle and intersect them with the circumcircle, we determine a chord VW that is also tangent to the incircle. (Incidentally, I've noted that one of the proofs of the Butterfly theorem on this site proves Poncelet's lemma too.)

For all these UVW triangles, I, O, R and r are fixed, so, for example, when P travels on the circumcircle, the nine-point-center N(UVW)
travels on a circumference with center I and radius

Otherwise, we can build G (centroid), Na (Nagel point) and F (tangency point of the incircle and the 9-point-circle) and try to intersect the Feuerbach hyperbola (we know its center F and three points on it:

This considerations raise the question of the possibility, or, more probably, the impossibility of the construction.

Using trilinears, one may say that it is hard to determine unsymmetric things like the coordinates of the vertices

So things become even harder, and I ask you:

- Does the (I, O, H) problem admit a simpler reverse construction (lower degree) than mine?
- WHEN, given three proper centers of ABC, is it possible to find
(A, B, C) with straightedge and ruler?

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