Home > Appreciation, Geometry > A Mathematical Tribute to Richard Geller

## A Mathematical Tribute to Richard Geller

Richard Geller, a  longtime math teacher and math team coach at Stuyvesant High School, recently passed away.  I only knew Richard professionally, but it was easy to see that he was a good man and a good teacher.  His dedication to his students, his school, and the math team was always apparent.

At a math circle one evening, Richard shared with me a lovely solution to a challenging problem that I’ll never forget.  I share it here as a tribute to him.

There are lots of famous concurrencies in triangles.  The medians of a triangle all intersect at the centroid; the angle bisectors at the incenter; and the perpendicular bisectors at the circumcenter.  We say that each set of lines is concurrent.

A less intuitive concurrency is that of a triangle’s altitudes, which all intersect at the triangle’s orthocenter.  It’s harder to see because you often have to rethink your notion of  altitude to see them intersect.

Not only is it harder to see the concurrency of the altitudes, but it’s harder to prove it as well.  There are many well-known methods, like using Ceva’s Theorem or areas, but they are rather complicated.  To me, the orthocenter was never as accessible as the circumcenter, incenter, or centroid.  Until Richard showed me this proof.

Start with an ordinary triangle.  We want to show that the altitudes of this triangle all intersect at a single point.

First, we create a new triangle by rotating three copies of our original around the midpoints of each side.  What we are doing is creating a new triangle whose medial triangle is our original triangle.

Now the magic:  construct the perpendicular bisectors of the new triangle.

The amazing fact here is that the perpendicular bisectors of the new triangle are the altitudes of the original triangle!  As long as we know that the perpendicular bisectors of any triangle are concurrent (which is fairly easy to prove), we know that the altitudes of any triangle are concurrent, too!

Richard didn’t invent this theorem or this proof, but he taught it to me, and for that I’ll be forever grateful.  When I share it with students, I think of him.  And from now on, when I show students the Geller Technique, I’ll wrap it up with one of Richard’s favorite phrases:  Math is #1!

www.MrHonner.com

1. November 7, 2011 at 12:58 pm

Thanks for sharing this wonderful proof..its one that truly can be called “trivial” ! Not at all “trivial”, till you actually know it ….

2. November 7, 2011 at 6:15 pm

Like so much mathematics, it seems impossible until you see how easy it is.

I’ll always be grateful to my colleague for making this seem so easy.

3. November 7, 2011 at 6:59 pm

This year has been utterly saddening for Stuyvesant High School. They suffered the tragic death of Aileen Chen, who was a genius in every possible way, last June.

4. November 7, 2011 at 11:36 pm

An eloquent tribute to a fellow teacher. A good reminder to share our insights (when we have them) and remember those who are kind enough to pass them along.

• November 8, 2011 at 8:05 am

Well said, Dave. And thanks–I’m glad to know my modest tribute here resonates with others.

5. November 12, 2011 at 7:39 pm

I also thought this was a lovely piece (and a lovely proof). My (non-mathematical) tribute is here: http://jblblog.wordpress.com/2011/11/01/richard-geller/

• November 12, 2011 at 8:34 pm

Thanks for sharing. Reading your tribute, I could easily imagine Richard patrolling the halls, clipboard in hand, poking his head into rooms. “He had a strong desire to have things in their right place” brought a smile to my face.