Posts Tagged ‘Euler Line’

Monday Math 157

August 4, 2014

Consider a triangle, which we label ∆ABC, with circumcenter O and circumradius R=AO=BO=CO. Let us label the midpoints of the sides as MA, MB and MC, so that MA is the midpoint of BC (the side opposite A), and similarly, so that $\overline{AM_{A}}$, $\overline{BM_{B}}$ and $\overline{CM_{CA}}$ are the medians. Then $\overline{OM_{A}}$, <$\overline{OM_{B}}$ and $\overline{OM_{C}}$ are segments of the perpendicular bisectors of the sides. Let us also label the feet of the altitudes from A, B and C as HA, HB and HC, respectively, and let H be the orthocenter (the intersection of the altitudes $\overline{AH_{A}}$, $\overline{BH_{B}}$ and $\overline{CH_{C}}$).

Let us construct the point D on the circumcircle diametrically opposed to A; that is to say, the point D such that AD is a diameter of the circumcenter. Then AD=2R, and O is the midpoint of AD.

Now, by Thales’ theorem, ∠ABD and ∠ACD are both right angles. Now, since CD and the altitude $\overline{BH_{B}}$ are both perpendicular to AC, they are parallel to each other. Similarly the altitude $\overline{CH_{C}}$ and segment BD are parallel, both being perpendicular to AB. Thus, the quadrilateral BDCH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, we see that the midpoint MA of BC is also the midpoint of HD.

Now, let PA be the midpoint of the segment AH. Then $\overline{P_{A}M_{A}}$ is a midline of the triangle ∆AHD, and by the triangle midline theorem, $\stackrel{\longleftrightarrow}{P_{A}M_{A}}\parallel\stackrel{\longleftrightarrow}{AD}$ and PAMAAD=R.

Now, let N be the intersection of $\overline{P_{A}M_{A}}$ and HO. By the midline-median bisection theorem proven in this post, we see that, as HO is the median of ∆AHD that crosses midline $\overline{P_{A}M_{A}}$, N is the midpoint of both $\overline{P_{A}M_{A}}$ and HO. Thus, NMA=NPAPAMAR.

Now, consider the quadrilateral HOMAHA. Since $\overline{HH_{A}}$ and $\overline{OM_{A}}$ are both perpendicular to BC, HOMAHA is a right trapezoid. Letting QA be the midpoint of $\overline{H_{A}M_{A}}$, we see then that NQ is the median (or midline) of trapezoid HOMAHA.

By the first of the three items proven here, we see that $\overline{NQ}\parallel\overline{HH_{A}}\parallel\overline{OM_{A}}$, and so $\overline{NQ}\perp\overline{H_{A}M_{A}}$. Thus, we see that $\stackrel{\longleftrightarrow}{NQ}$ is the perpendicular bisector of $\overline{H_{A}M_{A}}$, and so, by the perpendicular bisector theorem, NHA=NMA, and so
NHA=NMA=NPAR.

Constructing diameter BE of the circumcircle gives us parallelogram CEAH, by a similar argument as above. Letting PB be the midpoint of the segment BH, analogous reasoning to the above shows that NHB=NMB=NPBR as well. Lastly, diameter CF, and midpoint PC of CH gives, by similar proof, that NHC=NMC=NPCR. Thus, the nine points MA, MB, MC, HA, HB, HC, PA, PB and MC are all equidistant from N.

Therefore, we see that for any triangle, the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter all lie on a single circle, which, for this reason, is usually known as the nine-point circle, and its center N the nine-point center. We also see that the radius of the the nine-point circle is half the radius of the circumcircle, and the nine-point center is the midpoint of the segment connecting the circumcenter and orthocenter. This latter tells us that for any non-equilateral triangle, the nine-point center lies on the Euler line (for an equilateral triangle, it coincides with the circumcenter, orthocenter, and centroid).

Monday Math 154

July 14, 2014

Continuing the series on triangle centers, let us consider ∆ABC, with circumcenter O and centroid G. Construct the line segment OG, and extend it out from G to the point H such that GH=2OG.

Next, construct the median from vertex A to the midpoint M of side BC. Then G lies on AM, with AG=2GM, as we showed here. Recalling that the circumcenter is the intersection of the perpendicular bisectors of the sides, we see that OMBC.

Now, since AG=2GM, GH=2OG, and ∠AGH≅∠MGO, we see (by the SAS similarity condition) that ∆AGH~∆MGO. And since these triangles are similar, corresponding angles ∠HAG and ∠OMG are congruent. However, these are alternate interior angles for lines $\stackrel{\longleftrightarrow}{AH}$ and $\stackrel{\longleftrightarrow}{OM}$ cut by transversal $\stackrel{\longleftrightarrow}{AM}$, and therefore $\stackrel{\longleftrightarrow}{AH}\parallel\stackrel{\longleftrightarrow}{OM}$. And since OMBC, we see $\stackrel{\longleftrightarrow}{AH}\perp\overline{BC}$, and $\stackrel{\longleftrightarrow}{AH}$ is the triangle altitude from A to BC.

Analogous constructions show that H must also be on the triangle altitudes from B and C:

Thus, we see that H is the orthocenter of ∆ABC. So, we see that for any non-equilateral triangle, the circumcenter O, centroid G and orthocenter H are collinear, with GH=2OG; the line through these triangle centers is known as the Euler line of the triangle. (For an equilateral triangle, O, G and H are all the same point.)