155. A constant ratio in a triangle ABC.
by Virgil Nicula, Oct 14, 2010, 10:42 AM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=371730
Proof. Denote 
. Observe that 
. Therefore,
![$IE+IF=\frac as\cdot \left[c\cdot\cos\frac B2\cdot\sin\left(y+\frac B2\right)+b\cdot\cos\frac C2\cdot\sin\left(x+\frac C2\right)\right]=$](//latex.artofproblemsolving.com/5/e/4/5e44b76525acfe5f86e2c922d06cbebe62df549e.png)
![$\frac {a}{2s}\cdot\left\{c\cdot\left[\sin(B+y)+\sin y\right]+b\cdot\left[\sin(C+x)+\sin x\right]\right\}\implies$](//latex.artofproblemsolving.com/3/b/3/3b376723fe00329e7b8104dfb07f67127594cdc6.png)
. Denote
Apply the Ptolemy's relation in the cyclical quadrilateral
. Thus, 
. Since 
obtain
![$(b+c)\sin(C+x)=\left(\frac {bc}{ar}\cdot\lambda -1\right)\cdot2R\cdot [\sin B\sin x+\sin C\sin (A-x))]$](//latex.artofproblemsolving.com/e/1/b/e1b6d4fb4cc21cf8590916212ad72825c01e0bdc.png)
![$R(bc\lambda -ar)\cdot [\cos(B-x)+\cos (C-A+x)]=$](//latex.artofproblemsolving.com/1/b/d/1bd82100ed816cbaeac7b56b7b893fd653fbec38.png)
, i.e. 
.
Remarks.
If 
, then 
.
If 
, then 
and 
.
If 
, where 
is the orthocenter, then 
and .
.
(N.M.O, Iran - 1998).
For any 
for which 
separates 
, 
we have 
.
For 
and the projections 
, 
, 
of 
on the tangent lines to 
in , 
, 
respectively obtain 
,
, 
and 
. Since 
obtain
with equality for 
. Is interesting and the identity 
.
Otherwise (syntetical), Denote the midpoint
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the projections 
, 
of , on respectively and 
. From an well-known property 
obtain that the quadrilateral
is a parallelogram. Since 
obtain 
. Since 
obtain 
. Since 
obtain 
. Since 
obtain 
from where obtain 
, i.e. 
.
Otherwise. Prove easily that
, i.e. 
. Thus, 
![$IX=R\cdot\left[1+\cos (B-C)-(\cos B+\cos C)\right]$](//latex.artofproblemsolving.com/8/e/5/8e5bb1d80596f21c76120800b4b7ddbd67bb1484.png)
![$IX=R\left[1+\cos (B-C)+\cos A-1-\frac rR\right]$](//latex.artofproblemsolving.com/7/a/0/7a0626fe0931f1bc4f506384cc1475a9b7aed7a6.png)
![$IX=-r+R\left[\cos (B-C)+\cos A\right]$](//latex.artofproblemsolving.com/f/5/d/f5d13d8bebe6c22b6cd812e717d9892857398d85.png)
.
Quote:
Proposed problem. Let 
be a triangle with the circumcircle and the incenter . Consider for which the line
separates , . Denote the projections 
, 
of on the line 
, 
respectively. Prove that 
.
Remark. For
obtain the proposed problem from N.M.O. IRAN, 1998.
be a triangle with the circumcircle 
and the incenter 
. Consider 
for which the line 
separates
, 
. Denote the projections 
, 
of on the line 
, 
respectively. Prove that 
.Remark. For
obtain the proposed problem from N.M.O. IRAN, 1998. 
. Observe that 
. Therefore,
![$IE+IF=\frac as\cdot \left[c\cdot\cos\frac B2\cdot\sin\left(y+\frac B2\right)+b\cdot\cos\frac C2\cdot\sin\left(x+\frac C2\right)\right]=$](http://latex.artofproblemsolving.com/5/e/4/5e44b76525acfe5f86e2c922d06cbebe62df549e.png)
![$\frac {a}{2s}\cdot\left\{c\cdot\left[\sin(B+y)+\sin y\right]+b\cdot\left[\sin(C+x)+\sin x\right]\right\}\implies$](http://latex.artofproblemsolving.com/3/b/3/3b376723fe00329e7b8104dfb07f67127594cdc6.png)
. DenoteApply the Ptolemy's relation in the cyclical quadrilateral
. Thus, 
. Since 
obtain ![$(b+c)\sin(C+x)=\left(\frac {bc}{ar}\cdot\lambda -1\right)\cdot2R\cdot [\sin B\sin x+\sin C\sin (A-x))]$](http://latex.artofproblemsolving.com/e/1/b/e1b6d4fb4cc21cf8590916212ad72825c01e0bdc.png)
![$R(bc\lambda -ar)\cdot [\cos(B-x)+\cos (C-A+x)]=$](http://latex.artofproblemsolving.com/1/b/d/1bd82100ed816cbaeac7b56b7b893fd653fbec38.png)
, i.e. 
.Remarks.
If 
, then 
. If 
, then 
and 
. If 
, where 
is the orthocenter, then 
and .
.
(N.M.O, Iran - 1998). For any for which separates , we have 
. For 
and the projections 
, 
, 
of on the tangent lines to in , 
, 
respectively obtain 
,
, 
and 
. Since 
obtain
with equality for 
. Is interesting and the identity 
.Otherwise (syntetical), Denote the midpoint
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the projections 
, 
of , on respectively and 
. From an well-known property 
obtain that the quadrilateral
is a parallelogram. Since 
obtain 
. Since 
obtain 
. Since 
obtain 
. Since 
obtain 
from where obtain 
, i.e. 
.Otherwise. Prove easily that
, i.e. 
. Thus, 
![$IX=R\cdot\left[1+\cos (B-C)-(\cos B+\cos C)\right]$](http://latex.artofproblemsolving.com/8/e/5/8e5bb1d80596f21c76120800b4b7ddbd67bb1484.png)
![$IX=R\left[1+\cos (B-C)+\cos A-1-\frac rR\right]$](http://latex.artofproblemsolving.com/7/a/0/7a0626fe0931f1bc4f506384cc1475a9b7aed7a6.png)
![$IX=-r+R\left[\cos (B-C)+\cos A\right]$](http://latex.artofproblemsolving.com/f/5/d/f5d13d8bebe6c22b6cd812e717d9892857398d85.png)
.This post has been edited 55 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:12 PM
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