Saturday, February 5, 2011

Last Saturday I took Nicholas to a Metroplex Math Circle presentation about “Important points in a triangle: centroid, orthocenter, incenter, circumcenter, …” by a TI computer scientist - Branislav Kisačanin.

The presentation

The concept is simple and straight forward.

Centroid - the intersection of medians ( a line between vertex and middle point of the opposite side) of a triangle, orthocenter - the intersection of the altitudes (heights), incenter - the intersection of angle bisectors, and circumcenter - the intersection of perpendicular bisectors.

Then he raised the question: How do you prove that the medians, altitudes, angle bisectors and perpendicular bisectors are concurrent? The proof for medians -centroid, angle bisector - incenter, and angle bisectors - incenter, and perpendicular bisectors - circumcenters are not hard; but he used the seldom used Ceva theorem to prove the orthocenter.

I was literally amazed when he mentioned that centroid, circumcenter and orthocenter are collinear, and the line they are on is called Euler Line. And there is a nine point circle - which pass the middle point of each side, and foot of each altitude and mid-point between orthocenter and each vertex

The speaker let students participated in the process of his presentation; many students - either answered questions or went to the white board to give a proof. The interactive approach really helped keep students' focused and interested in the subject. One time when he asked an alternate proof to his proof that the medians divided a triangle into six smaller triangle which have equal area, no students could provide one, so he asked parents to help - I volunteered an alternate proof :). Near the end of his presentation, the speaker somehow stuck on a proof, he remembered what to do, but he could not make a logic derivation of the result he wanted. I volunteered again to help him figure things out, and I provided an alternate proof one more time!

Nicholas was very much interested in the presentation until the last quarter of the presentation, when methods used to solve some problems involved trigonometry.

The interesting facts and my proofs

Due to time limit, the speaker did not show the proof of Euler line and 9-point circle. Fascinated by these facts, I decided to try my capability of geometry proof. I was mentally as excited as when I was at the peak of my Ph.D. thesis research! I spent several hours on Sunday and proved that 1) altitudes are concurrent witout using Ceva's theorem; 2) Euler line: centroid, orthorcenter and circumcenter are collinear; and 3) nine point circle: the middle point of each side, and foot of each altitude and mid-point between orthocenter and each vertex are on the same circle.

1) Altitudes are concurrent

Assume altitude AA' and CC' intersect at O, and OA'=X, then we have

m/h=X/n, thus X = mn/h

Similarly assume that AA' and BB' intersects at O', O'A' =Y

n/h= Y/m, thus Y=mn/h = X. i.e. A'O=A'O'.So O and O' coincides, altitudes are concurrent.

Note: this is much easier and simpler than using Ceva's theorem

2) Euler Line

I used algebra to prove the Euler Line. A triangle is placed in a Cartesian coordinate as shown.

Write out the linear equation for two altitudes, one can solve for the orthocenter, which is at
(0, mn/h) . Similarly the centroid is at ((n-m)/3, h/3) and the circumcenter is located at
(0.5(n-m), (h^2-mn)/(2h)).

The line passing orthocenter and centroid is

y-mn/h = (h-3mn/h)/(n-m) x.

It is easily verified that the circumcenter is on this line. So orthorcenter, centroid and circumcenter are on the same line - Euler line.

Note: it is easy to determine that the radius of circumcircle is AB*AC/(2h)

3) 9 - point circle

Using the same coordinates to define a triangle as before, it is found the that the center of the circle that passes the mid-points of each side of the triangle is at ((n-m)/4, (h+mn/h)/4). The radius of this circle is sqrt{[(m-n)/4]^2+[(h+mn/h)/4]^2} = AB*AC/(4h).

It turns out that the distance between this center and the foot of a altitude (0,0) is the same as above. Similarly one can prove that the feet of other two altitudes are on the circle as well.

The remaining three points are mid-points between orthocenter and each vertex. One such point is on y axis, (0, 0.5(h+mn/h)). Its distance to the center of the circle is exactly the radius!

So all nine points are on the same circle - illustration of a nine point circle is shown below (chart from http://www.math.sunysb.edu/~scott/mat360.spr04/cindy/ninepoint.html)

9 - point circle has additional interesting properties: it is tangent to incircle, its center is on Euler line and is midway between Orthocenter and Circumcenter.

In additional to these interesting geometric relations, there are also many elegant and clean quantitative relations, e.g. the radius of 9-point circle is half of that of circumcircle (which has been shown above).

(to be continued ...)