Stuff involving the nine-point circle, Euler line, reflections, enlargements and stuff

O and H are the circumcentre and orthocentre respectively of triangle ABC. M is the midpoint of BC. Then AH = 2OM.
AH = 2OM 1.png
This might be well-known but I present 4 proofs.
Proof 1.
Let BO meet circle ABC again at T.
AH = 2OM 7.png
Since BT is a diameter of the circle ABC, ∠BAT = 90 and ∠BCT = 90 by Thales.
CH is perpendicular to AB since H is the orthocentre; and as TA is also perpendicular to AB, TA and CH are parallel.
Similarly, AH and TC are both perpendicular to BC and hence they are parallel.
Thus TCHA is a parallelogram.
So AH = TC.
Also, as BC = 2BM and BT = 2BO, triangle BCT is an enlargement of triangle BMO, centre B and scale factor 2. Hence TC = 2OM so AH = 2OM.

Proof 2.
The next three proofs make use of results concerning the nine-point circle and the Euler line.
Recall that the nine-point circle of a triangle passes through: the midpoint of each side (shown orange below); the feet of the three altitudes (green); and the midpoints of the line segments joining each vertex to the orthocentre (red). The orthocentre itself is shown as grey.
AH = 2OM 2.png
Consider enlarging nine-point circle, with the orthocentre as the centre of enlargement, and scale factor two. The red points move away from the orthocentre to land at the vertices of the triangle. Hence this enlargement enlarges the nine-point circle to the triangle’s circumcircle. Therefore, the orthocentre reflects across any point on the nine-point circle to land on the circumcircle.
AH = 2OM 3.png
(Above: The image of H, H’, upon reflection about any point P on the nine-point circle, lies on the triangle’s circumcircle.)
Now we reflect H across two specific points to solve our problem.
Returning to our original diagram, let D be the foot of the altitude from A. Denote the reflections of H across D and M as X and Y respectively. As established, X and Y lie on triangle ABC’s circumcircle, because D and M lie on the nine-point circle.
AH = 2OM 4.png
Since X and Y are reflections of H across D and M respectively, triangle HXY is an enlargement of triangle HDM, scale factor 2 and centre H. As AD is an altitude, ∠HDM = 90, so ∠HXY = 90.
Hence, by the converse of Thales, AY is a diameter of triangle ABC’s circumcircle, so O is the midpoint of AY.
Therefore, since HY = 2MY and AY = 2OY, triangle OMY enlarges to triangle AHY, centre Y and scale factor 2. Therefore 2OM = AH.

Proof 3.
The enlargement carrying the nine-point circle to the circumcircle, centre H and scale factor 2, must also carry the centre of the nine-point circle to the triangle’s circumcentre. Hence, if N is the centre of the nine-point circle, it must be the midpoint of OH.
Let K be the midpoint of AH and let D be the foot of the altitude from A as before. The nine-point circle passes through K, D and M.
AH = 2OM 5.png
Since AD is an altitude, ∠KDM = 90 so KM is a diameter of the nine-point circle by the converse of Thales. Thus N is the midpoint of KM.
Since N bisects OH and KM, OMHK is a parallelogram. Therefore OM = KH so 2OM = AH.

Proof 4.
This final proof is the shortest and makes use of the Euler line. That is, the centroid, G, lies on OH such that OG : GH = 1 : 2.
Since G lies on AM and OH, G is the point where AM and OH meet.
AH = 2OM 6.png
Since OG : GH = 1 : 2 and MG : GA = 1 : 2 (since the centroid divides each median in this ratio), the triangle OGM enlarges to triangle HGA, scale factor -2, centre G. OM is taken to AH by this enlargement so 2OM = AH.

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