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Topic: Orthocenter to centroid distance
Replies: 2   Last Post: Jul 23, 2012 8:33 PM

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Vladimir Zajic

Posts: 319
Registered: 12/3/04
Re: Orthocenter to centroid distance
Posted: Jan 2, 2005 4:13 AM
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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
>co-ordinates 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

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.


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.

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