Mixtilinear Incircle Proof

Date: 12/11/2000 at 13:13:32
From: Ailette
Subject: Geometry

I have had a bit of trouble with this question. Can you please help?

In triangle ABC, AB = AC. A circle is tangent internally to the 
circumcircle of triangle ABC and also to sides AB and AC at points P 
and Q, respectively. Prove that the midpoint of the segment PQ is the 
center of the incircle of triangle ABC.

Date: 12/13/2000 at 04:45:49
From: Doctor Floor
Subject: Re: Geometry

Hi, Ailette,

Thanks for writing.

Let's consider the following figure:



In this figure, D is the midpoint of BC, and D' is the point where AD 
meets the circumcircle for the second time (the first is A). Also, AB 
and AC are extended to B' and C' so that B'C' and BC are parallel, and 
D' is the midpoint of B'C'. I is the center of the inscribed circle, 
and I' is the center of the new circle (named by Leon Bankoff a 
"mixtilinear incircle"). Finally, X is the midpoint of PQ. Our task is 
to show that X = I.

Now note that the figures ABCDI and AB'C'D'I' are similar, so that

     AB/AB' = AC/AC' = AD/AD' = AI/AI'   ........................[1]

Observe that angles API' and AQI' are right, and that angles ABD' and 
ACD' are right as well because they are inscribed angles on the 
diameter of the circumcircle of ABC. Thus the angles PI'Q and BD'C 
must be congruent. But then the figures ABCD' and APQI' are similar 
figures, and by extension even ABCDD' and APQXI' are similar figures. 
That provides:

     AD/AD' = AX/AI'   ..........................................[2]

Combining [1] and [2] provides that X = I, as desired.

By the way, Leon Bankoff, and after him Paul Yiu (Florida Atlantic 
University; he published a year ago in the American Mathematical 
Monthly), have studied these mixtilinear incircles. One of Bankoff's 
results implies that in the above theorem it is not necessary that the 
triangle be isosceles. A scalene triangle will do.

If you have more questions, just write back.

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