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### Proof of the Feuerbach Theorem

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Date: 03/14/2000 at 13:52:26
From: Stefa Ben Ari
Subject: Feuerbach theorem

Please submit the proof of the Feuerbach theorem (the nine point
circle is tangent to the incircle and the circumcircle of a triangle.)

Stefa
```

```
Date: 03/15/2000 at 13:12:57
From: Doctor Floor
Subject: Re: Feuerbach theorem

Hi Stefa,

Before we start, let me say that we will make use of a formula on the
distance between the circumcenter O and incenter I of a triangle. When
we denote by r the radius of the incircle and by R the radius of a
circumcircle of a triangle, then we have:

IO^2 = R(R-2r)

A proof of this formula is given in the Dr. Math archives:

http://mathforum.org/dr.math/problems/gural.03.26.99.html

I will show you how the tangency of the incircle and the nine-point
circle can be proven with the use of complex numbers.

To do that, we consider a triangle with vertices A1, A2 and A3, which
we represent by complex numbers z1, z2 and z3, respectively. We can do
this in such a way that z1, z2 and z3 are positioned counterclockwise
on the unit circle with center 0, so R = |z1| = |z2| = |z3| = 1, and
the circumcenter O = 0.

For the measures of the angles of triangle A1A2A3 at A1, A2 and A3 we
will write a1, a2 and a3, respectively.

The centroid G of triangle A1A2A3 is G = (z1+z2+z3)/3. Now let us
consider the orthocenter H. From the fact that OG:GH = 1:2 we see that
H = z1+z2+z3. And since the nine-point center N is halfway between O
and H, we see that N = (z1+z2+z3)/2.

We can choose t1, t2 and t3 with

z1 = e^(t1i)
z2 = e^(t2i)
z3 = e^(t3i)

and

v1 = e^(i*t1/2)
v2 = e^(i*t2/2)
v3 = e^(i*t3/2)

in such a way that Q1 = v2v3 bisects the arc A2A3 including A1,
Q2 = v1v3 bisects the arc A1A3 including A2 and Q3 = v1v2 bisects arc
A1A2 including A3.

(Note that v1, v2 and v3 are numbers on the unit circle, and that
z1 = v1^2, z2 = v2^2 and z3 = v3^2).

The points P1 = -v2v3, P2 = -v1v3 and P3 = -v1v2 are the points
opposite to Q1, Q2 and Q3, and are the points where the internal angle
bisectors of triangle A1A2A3 meet the circumcircle/unit circle.

We can see that the incenter I of A1A2A3 is the orthocenter of P1P2P3
in the following way:

Let X be the intersection of A1P1 and P2P3. Note that A1P1 passes
through I. Angle A1P1P2 (denoted by <A1P1P2) equals

<A1A2P2 = a2/2

Also

<P3P2P1 = <P3P2A2 + <A2P2P1
= <P3A3A2 + <A2A1P1
= a3/2 + a1/2

This means that

<P1XP2 = 180 degrees - <A1P1P2 - <P3P2P1
= 180 degrees - (a1+a2+a3)/2
= 90 degrees

So IP1 is an altitude in P1P2P3, and by symmetry so are IP2 and IP3,
which proves that I is the orthocenter of P1P2P3.

By this, in the same way that we found H = z1+z2+z3, we see

I = -(v1v2+v1v3+v2v3)
= -v1v2v3(1/v1 + 1/v2 + 1/v3)
= -v1v2v3(v1*+v2*+v3*)
= -v1v2v3(v1+v2+v3)*

where I write z* for the complex conjugate of z. So, for points z on
the unit circle, we know that zz* = 1.

We find that

IO = |O-I|
= |-I|
= |v1v2v3(v1+v2+v3)*|
= |v1||v2||v3||(v1+v2+v3)*|
= 1*1*1*|v1+v2+v3|
= |v1+v2+v3|

and

IN = |N-I|
= |(z1+z2+z3)/2 - -(v1v2+v1v3+v2v3)|
= |(v1^2+v2^2+v3^2)/2 + (v1v2+v1v3+v2v3)|
= |0.5(v1+v2+v3)^2|
= 0.5*|v1+v2+v3|^2.

We recall the formula IO^2 = R(R-2r), which implies, together with
R = 1, that r = 0.5 - 0.5*IO^2. We derive from this:

r = 0.5 - 0.5*IO^2
= 0.5 - 0.5*|v1+v2+v3|^2
= 0.5 - IN

Now we see that IN = 0.5 - r, and since the radius of the nine-point
circle is half the radius of the circumcircle, and thus 0.5, we see
that the incircle is (internally) tangent to the nine point circle.

The result for the excircles can be found in a similar way. Let's say
we consider the A1-excircle, having radius r1 and center I1. Now we
have to use the formula I1O^2 = R(R+2r1), and we find I1 as the
orthocenter of triangle P1Q2Q3. I leave it to you to follow the
computations as above.

If you have more questions, just write back.

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

```
Date: 03/26/2000 at 14:11:36
From: Michael Ben Ari
Subject: Re: Feuerbach theorem

Dear Doctor Floor,

Thank you for the proof of the Feuerbach theorem received on
03/14/2000 (Einstein's Birthday and also PI (3.14) !

There is something I didn't understand in the proof:

>     v1 = e^(i*t1/2)
>     v2 = e^(i*t2/2)
>     v3 = e^(i*t3/2)
>
>in such a way that Q1 = v2v3 bisects the arc A2A3 including A1,
>Q2 = v1v3 bisects the arc A1A3 including A2 and Q3 = v1v2 bisects arc
>A1A2 including A3.

I understand that the coordinates of the point Q1 that bisects the arc
A2A1A3 are

[-cos(t2/2+t3/2),-sin(t2/2+t3/2)]

it means Q1 = -v2v3, and the coordinates of the point Q2 that bisects
the arc A1A2A3 are

[cos(t1/2+t3/2),sin(t1/2+t3/2)]

it means Q2 = v1v3 and the coordinates of the point Q3 that bisects
the arc A1A3A2 are

[-cos(t1/2+t2/2),-sin(t1 /2 +t2 /2)]

it means Q3 = -v1v2 and it doesn't fit your statement.

I can not find my mistake. Maybe I misunderstood the notation
e^(i*t1/2). I understand it as "cos(t1/2) + i sin(t1/2)."

Please give this part of the proof in more detail. The remainder of
the proof is clear to me.

Thank you,
Stefa Ben Ari
```

```
Date: 03/27/2000 at 06:38:09
From: Doctor Floor
Subject: Re: Feuerbach theorem

Dear Michael,

It seems that you have overlooked the fact that t1, t2, and t3 can be
chosen in two ways.

Suppose that z1 = e^(i*t1), and also z1 = e^(i*(t1+2*pi)).

These give v1 = e^(i*t1/2) and v1' = e^(i*t1/2 + pi) = -v1.

The part of the proof you are quoting is saying that you can choose
these t1, t2, and t3 in such a way that the resulting v1, v2, and v3,
and consequently Q1, Q2, and Q3, are as desired.

It is stated this way to be sure that the right square root of z1 is
chosen (there are always two square roots of course.)

I hope this clears it up!

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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