Incenters, Orthocenters, and the Spieker Point

Date: 02/13/2000 at 11:00:16
From: Emily Wilson
Subject: Geometry

For community service I am working as an assistant teacher to a class 
of 9th grade geometry honors students in New York. The teacher has 
asked me to make an extra-credit packet with difficult proofs. I have 
found some interesting problems, but unfortunately the answers were 
not provided. My geometry is a little rusty as I took the course two 
years ago. I would appreciate your help on a problem.

Given:

M1, M2, M3 are the three midpoints of the sides of triangle ABC, so 
triangle M1M2M3 is the medial triangle of triangle ABC. Point I is the 
incenter of triangle ABC and point G is the centroid of triangle ABC. 
Point S is collinear with I and G as shown in the diagram. 2SG = IG.

Prove: S is the incenter of triangle M1M2M3

The book I got this from said a successful completion of the proof 
would give you that the points I, G and S are collinear in any 
triangle and that IG and GS are in a ratio of 2:1. It also said that 
point S is sometimes called the Spieker point.

Thanks,
Emily

Date: 02/13/2000 at 13:55:37
From: Doctor Floor
Subject: Re: Geometry

Hi Emily,

Thanks for your question.

If you are to present the problem, you should be very clear that S is 
on the other side of G from I.

If that is clear, there is not much to prove really. The result 
follows from the observation that G is the center of similitude of ABC 
and its medial triangle. Thus we conclude from AG:GM1 = 2:1 that if we 
take a point X in triangle ABC and the corresponding point Y in 
triangle M1M2M3, then G is between X and Y and XG:GY = 2:1. So when we 
take X is the incenter, then Y is the incenter of M1M2M3.

This point S - the Spieker point - is a remarkable point. It is the 
center of mass of the perimeter of triangle ABC. This can be seen in 
the following way:

Let a = |BC|, b = |AC| and c = |AB|. We can represent the masses of 
the three sides of ABC by placing weights of magnitudes a on M1, b on 
M2 and c on M3. Let Z be the point that divides segment M2M3 in ratio 
M2Z:ZM3 = c:b (so Z is the center of mass of the two weights on M2 and 
M3), then we can replace the weights of M2 and M3 by a weight of 
magnitude b+c on Z.

Now the center of mass of the weights on M1 and Z, and thus the center 
of mass of the perimeter, is the point T dividing segment M1Z in ratio 
M1T:TZ = a:b+c. From this we can see that area of triangle M2M3T is 
a/(a+b+c) times the area of triangle M1M2M3.

In the same way we can see that the area of triangles M1M3T and M1M2T 
are b/(a+b+c) and c/(a+b+c) times the area of triangle M1M2M3 
respectively.

Using that the sides of M1M2M3 are of half the lengths of the sides of 
ABC and that area(M2M3T):area(M1M3T):area(M1M2T) = a:b:c, we can 
conclude that the altitudes from T to the sides of triangle M1M2M3 are 
all three of equal length. And thus T is the incenter of M1M2M3, the 
Spieker point.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   

Date: 02/13/2000 at 14:55:07
From: Emily Wilson
Subject: Re: Geometry

I don't really remember what the center of similitude is and I'm 
pretty sure my students haven't gotten to that yet. What is it and is 
there another way to prove that S is the incenter of triangle M1M2M3?

Date: 02/15/2000 at 02:26:56
From: Doctor Floor
Subject: Re: Geometry

Hi, Emiliy,

Thanks for responding.

Take a plane geometric figure and a fixed point P. When, from this 
point P, you multiply the distances to the points of the figure with a 
fixed number (negative -> take the opposite side of P) you get a 
figure similar to the original figure.

When of two similar figures such a point exists, which is the case for 
a triangle and its medial triangle, then the point is called sometimes 
"Center of Similitude" or "Center of Multiplication."

I hope that clears things up.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   

Date: 02/13/2000 at 16:20:17
From: Emily Wilson
Subject: Re: Geometry

Hi, this is Emily again.

When you bisect the exterior angles of a triangle, they intersect at 
three points, called the excenters. The incenter of the triangle is 
the orthocenter of the triangle formed by the 3 excenters. How do I 
prove this? (What tells me that the angle bisectors of the smaller 
triangle are also the altitudes of the larger one?) Does it have 
something to do with drawing the circumcircle around the smaller 
triangle?

Date: 02/15/2000 at 03:28:00
From: Doctor Floor
Subject: Re: Geometry

Hi, Emily,

Thanks for your new question.

There are two observations you should make to prove this:

1. The internal and external bisectors from one vertex are 
perpendicular to each other. This can be seen in the following way: 
When the internal bisector bisects an angle of magnitude A degrees, 
then the external bisector bisects an angle of magnitude 180 - A 
degrees. So the bisected angles are A/2 and 90-A/2 degrees, which 
makes 90 degrees together.

2. The point of intersection of two external bisectors is on the 
internal bisector of the third angle (this can be shown by reasoning 
that the angle bisectors are lines of points equidistant from the 
sidelines of the triangle, and seeing in what regions the internal and 
external bisectors are).

When you draw this, you see that this gives that the incenter is the 
orthocenter of the excenters

If you need more help, just write back.

Kind regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/