Ken's POTW


Bisection in a Triangle
The circle inscribed in the triangle ABC touches the side AC in the point K. Let the center of the circle be O, and the midpoint of AC be D. Prove that the line connecting O with D bisects the BK segment.

Source: Submitted by reader Phillippe Fondanaiche, citing W. PERTSEL ( The problems of the All-Soviet-Union mathematical competitions 1961-1987, #112 )

Solutions were received from Phillipe and from Radu Ionescu. I didn't realize how difficult this problem would actually be, and I have not thoroughly checked these solutions - if you find any improvements, please let me know.
Radu's Solution: 
Let be:AB=c, BC=a, CA=b, r=circle radius, p=(a+b+c)/2 and S=area of ABC
BB' perpendicular to AC, B' on AC and M intersection of BB' and DO

AK=r/tg(A/2)=p-a; AB'=c cosA; BB'=2S/b=2pr/b

DK=b/2-p+a=(a-c)/2;  DB'b/2-c cosA=(b-2c cosA)/2=(b^2-2bc cos A)/2b=(c^2-a^2)/2b
MB'D and OKD are similary triangles==>MB'/r=DB'/KD==>MB'=r(c+a)/b

BM=BB'-MB'=r(2p/b-(a+c)/b)=r==>KO=BM but KO and BM are parallel==>BMKO
parallelogram==>
OD bisects the BK segment.
Phillipe's Solution: 
Preliminary comment: in the following demonstration, I have admitted some
properties concerning the inscribed and escribed circles and the angle
bisectors. These properties are described in
Eric's Treasure Trove of
Mathematics
http://www.astro.virginia.edu/~eww6n/books/RecreationalMathematics.html


Given a triangle BAC with the height BH, the angle bisector BI and the median
BD with H,I and D lying on AC.
Let BC=a  AC=b  BA=c  s=half of the perimeter=1/2*(a+b+c).
Let the inscribed circle,with O as incenter,tangent to the side AC at the
point K. It is easy to check that AK=s-a and CK =s-c.
Extending the side BA towards BAx and the side BC towards BCy, let the
escribed circle,with O' as center,tangent to the two lines Ax and Cy and to
the side AC at the point L. As above, it is easy to demonstrate the identity
CL=s-a so that K and L are symmetric with respect to D. So KD=LD.
As AO and AO' are respectively bisectors of the angles BAC and CAx, it is well
known that the four points B,O,I,O' which are one the same line, are harmonic
conjugates. As H,K and L are the perpendicular projections of B,O and O' on
AC, the four points H,K,I,L are also harmonic conjugates.
Therefore we have,for example,the following postion of the points H,K,I,D and
L:

----H----------------K---------I------D------------------L--------------   KL = 2KD

 HK / HL = IK / IL
 As HK = HI - IK ,we have HI / HL = IK / HL + IK / IL (1)
 As HK = HL - KL and IK = KL - IL,we have (HL - KL) / HL = (KL - IL) / IL
 Multilying the two menbers by IK / KL, we obtain the following identity:
 IK / KL - IK / HL = IK / IL - IK / KL
 that is to say: IK / HL + IK / IL = 2IK / KL = IK / KD (2)
 With the relations (1) and (2), we infer HI / HL = IK / KD or HI / IK = HL /
KD.
 As HI / KI = BH / OK, then BH / OK = HL / KD or BH / HL = OK / KD.
 So the rectangular triangles BHL and OKL are similar. Then OD is parallel to
BL 
 As OD bisects KL (see above KD=LD), then OM halves BK.
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