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Re: Orthocenter to centroid distance
Posted:
Jan 2, 2005 4:13 AM


On 1 Jan 05 13:42:24 0500 (EST), Terry wrote: > > > >I would like to know the distance between orthocenter and centroid in >a traingle and the area of the triangle when the length of a side is >a given and distances from centroid to the sides are given. > >the length of side a, and distances from the centroid to sides a, b >and c are given. how to find the area and distance between >orthocenter and centroid in a scalene triangle. > > >What is the distance between orthocenter and centroid . If the >coordinates are known then i can write the equations. > > > Let ABC be the trinagle with sides a ,b ,c and centroid is I and >orthocenter is O. I can find the area of BCI because i know the >distances of 3 sides. But how to calculate the area of the triangle ? >The second question is how to calculate the distance between I and O.
1. I will denote the centroid G, the orthocenter H and the circumcenter O. Let H_a, H_b, H_c be the feet of the altitudes to the sides a = BC, b = CA, c = AB, M_a, M_b, M_c be the midpoints of the sides a, b, c and N_a, N_b, N_c be the feet of normals from the centroid G to the sides a, b, c. If the distances a = BC, GN_a, GN_b, GN_c are all known, the problem is overdetermined, but I will assume that the data is consistent.
The medians m_a = AM_a, m_b = BM_b, m_c = CM_c meet at the centroid G and
AM_a = 3*GM_a, BM_b = 3*GM_b, CM_c = 3*GM_c
It follows that the altitudes
h_a = AH_a = 3*GN_a, h_b = BH_b = 3*GN_b, h_c = CH_c = 3*GN_c
are also known. The area of the triangle ABC is then
S(ABC) = 1/2*a*h_a = 3/2*GN_a*a
The remaining sides b and c of the triangle ABC are
b = 2*S(ABC)/h_b = 2/3*S(ABC)/GN_b c = 2*S(ABC)/h_c = 2/3*S(ABC)/GN_c
The radius R of the circumcircle is equal to
R = abc/[4*S(ABC)]
and the distance HO between the orthocenter H and the circumcenter O is known to be
HO = sqrt[9*R^2  (a^2 + b^2 + c^2)]
The points H, G, O are collinear (Euler line) and HG = 2*GO. This is easily seen from the fact that the circumcenter O of the triangle ABC is the orthocenter of the medial triangle M_aM_bM_c, this medial triangle is centrally similar to the triangle ABC with the similarity center G and coefficient 1/2. The similarity coefficient is considered to be negative because the corresponding points of the 2 triangles lie on the opposite sides from the point G.
Consequently,
HG = 2/3*HO = 2/3*sqrt[9*R^2  (a^2 + b^2 + c^2)] = = 2/3*sqrt[9*a^2*b^2*c^2/[16*S(ABC)]  (a^2 + b^2 + c^2)]
where a is known and b, c, S(ABC) have been calculated previously.



