34. IMO Shortlist 2002 geometry problem 7.
by Virgil Nicula, May 2, 2010, 8:31 PM
PP1 (IMO shortlist 2002). The incircle 
of an acute 
is tangent to side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
at 
. Let 
be the point for which 
. Denote the
midpoint
of ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
. The line 
cut again the circle at the point 
. Prove that the circle and the circumcircle of 
are tangent to each other at .
Proof 1. Suppose w.l.o.g.
. Let 
, 
, 
, 
and 
. Observe that 
and 
. Thus
. Since 
obtain 
, i.e. 
. Since the points 
, are harmonical conjugate w.r.t. the points 
, 
and 
is the midpoint
of
obtain from an well-known property that 
, i.e. 
what means that the line 
is tangent at the point to the circumcircle of 
.
Remark.
. If ![$[KK']$](//latex.artofproblemsolving.com/3/5/4/35471d26e6bb747eb0e97c0663c5a99bb6458f7d.png)
is diameter in the circle 
, then 
and 
.
Proof 2. Suppose w.l.o.g. that . Denote the excircle 
with 
, 
and 
, i.e. 
, 
. Prove easily that
Observe that 
, i.e.
So 
and the circumcircle of is tangent to in 
![$PK^2=[PK-(s-b)][PK+(s-c)]\iff$](//latex.artofproblemsolving.com/0/5/c/05c78ddba6a9c5d5444dcb50ccafdb7dd9ebb83a.png)
what is truly.
Remark.![$\left\{\begin{array}{ccccccccc}
MD\parallel IK & \implies & [MKL]=[IDL] & \implies & MD\cdot KL=IK\cdot DL & \implies & \frac {LD}{KL}=\frac {MD}{IK} & \implies & M\in IL\\\\
MD\parallel I_aL & \implies & [MKL]=[DKI_a] & \implies & MD\cdot KL=I_aL\cdot KD & \implies & \frac {KD}{KL}=\frac {MD}{I_aL} & \implies & M\in I_aK\end{array}\right\|$](//latex.artofproblemsolving.com/4/9/e/49eb91ee684118c15f27ce931b9d935ffea478a1.png)
.
of an acute 
is tangent to side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
at 
. Let 
be the point for which 
. Denote themidpoint
of ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
. The line 
cut again the circle at the point 
. Prove that the circle and the circumcircle of 
are tangent to each other at .Proof 1. Suppose w.l.o.g.
. Let 
, 
, 
, 
and 
. Observe that 
and 
. Thus
. Since 
obtain 
, i.e. 
. Since the points 
, are harmonical conjugate w.r.t. the points 
, 
and 
is the midpointof
obtain from an well-known property that 
, i.e. 
what means that the line 
is tangent at the point to the circumcircle of 
.Remark.
. If ![$[KK']$](http://latex.artofproblemsolving.com/3/5/4/35471d26e6bb747eb0e97c0663c5a99bb6458f7d.png)
is diameter in the circle 
, then 
and 
.Proof 2. Suppose w.l.o.g. that
. Denote the excircle 
with 
, 
and 
, i.e. 
, 
. Prove easily that
Observe that 
, i.e.
So 
and the circumcircle of is tangent to in 
![$PK^2=[PK-(s-b)][PK+(s-c)]\iff$](http://latex.artofproblemsolving.com/0/5/c/05c78ddba6a9c5d5444dcb50ccafdb7dd9ebb83a.png)
what is truly.Remark.
![$\left\{\begin{array}{ccccccccc}
MD\parallel IK & \implies & [MKL]=[IDL] & \implies & MD\cdot KL=IK\cdot DL & \implies & \frac {LD}{KL}=\frac {MD}{IK} & \implies & M\in IL\\\\
MD\parallel I_aL & \implies & [MKL]=[DKI_a] & \implies & MD\cdot KL=I_aL\cdot KD & \implies & \frac {KD}{KL}=\frac {MD}{I_aL} & \implies & M\in I_aK\end{array}\right\|$](http://latex.artofproblemsolving.com/4/9/e/49eb91ee684118c15f27ce931b9d935ffea478a1.png)
.This post has been edited 40 times. Last edited by Virgil Nicula, Apr 14, 2016, 12:57 PM
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