Euler LineDate: 06/08/98 at 10:27:41 From: Adam Christen Subject: Euler Line Dr. Math, What is the Euler line? Date: 06/26/98 at 18:12:19 From: Doctor Floor Subject: Re: Euler Line Hi Adam, Thank you for sending your question to Dr. Math. If a triangle ABC is not equilateral, then the Euler line is the line that connects three important triangle centers (if the triangle is equilateral, these three are the same point): - the orthocenter H (point of intersection of the three lines from a vertex perpendicular to the opposite side); - the centroid G (point of intersection of the three lines connecting a vertex and the midpoint of the opposite side); - the circumcenter O (point of intersection of the three lines from a midpoint of a side, perpendicular to that side. The circumcenter is also the midpoint of the circle that passes through the three vertices A, B and C). It is supposed to be Euler who first proved that in any triangle ABC these three points are on a line (in an equilateral triangle there are more lines passing through the three centers that come together in one point). Here is how I would prove the three points to be on one line. For this I let A' be the midpoint of BC, B' midpoint of AC and C' the midpoint of AB. C / \ / \ G: centroid / H \ O: circumcenter B' A' D: point on AB such that CD / G \ is perpendicular to AB. The / O \ segment AD is called "altitude / \ from C" A--------D--C'-----------B H: orthocenter Take a careful look at triangles A'B'C' and ABC. You will see that ABC and A'B'C' are similar, and that A'B'C' is half the size of ABC. Also AB and A'B' are parallel, AC and A'C' are parallel, and BC and B'C' are parallel: ABC and A'B'C' are sometimes called parallel triangles, or homothetic triangles. Of course you see that AA', BB' and CC' meet in G. But then we know that ABC is the multiplication figure of A'B'C' with factor -2 over G. This means that if you take a point P of A'B'C', measure the distance PG, multiply PG by two, and go that distance on the opposite side from G, you get the corresponding point of ABC. Now A'O is perpendicular to BC - this way you construct O - but since BC and B'C' are parallel, A'O is also perpendicular to B'C'. In this way we see that O is the orthocenter of A'B'C'. Aha, the circumcenter of ABC is the orthocenter of A'B'C'! Let's think about what that means with respect to ABC being the multiplication figure of A'B'C' with factor -2 over G. Take for instance D' on A'B' in such a way that A'D' and A'B' are perpendicular. Where will D' go when you multiply with factor -2 over G? To the corresponding point in ABC, D of course! And segment A'D' goes to AD. Or: the altitude from A' in A'B'C' goes to the altitude from A in ABC. And the same goes for the altitudes from B' and from C'. Thus their common point of intersection, the orthocenter of A'B'C', goes to the common point of intersection of the altitudes in ABC, that is, the orthocenter H of ABC. Conclusion: when you multiply O over G with factor -2, you get the orthocenter H of ABC, and the three points must be on one line. Also you see that, because the factor is -2, the distance ratio HG:GO = 2:1. For more on triangle centers and the Euler Line, see "Incenter, Orthocenter, Circumcenter, Centroid" in the Dr. Math archives: http://mathforum.org/dr.math/problems/beck1.5.97.html Best regards, - Doctors Floor and Sarah, The Math Forum http://mathforum.org/dr.math/ |
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