223. An inequality with complex numbers-RMO shortlist 2010.
by Virgil Nicula, Feb 17, 2011, 12:22 PM
RMO shortlist 2010. Prove that 
, where 
(the set of all complex numbers).
Proposed problem. Prove that
from the plane 
exists the inequality 
.
Proof. Denote the midpoints
, 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, ![$[XA]$](//latex.artofproblemsolving.com/a/5/3/a5317c3e6beee34d801bcb11bbbba6b9ae3095c6.png)
respectively. Apply the theorem of the median in the triangles :
.
Remark. For
and 
, where 
, 
, 
belong to the complex plane and 
is the circumcircle of
obtain that 
.
Another proof. If denote
, 
and 
, then the our inequality is equivalently with 
, what is truly because 
.
Generalization. Prove that for any
and for any 
from the plane of 
exists the inequality
![$\boxed{\ n\left[n(1-m)a^2-m(1-m)b^2+mc^2\right]\ge m(1-n)\cdot\left[-m\cdot XA^2+n\cdot XB^2-n(1-m)\cdot XC^2\right]\ }$](//latex.artofproblemsolving.com/d/d/f/ddf6d465cec78f24884888c235e0426a706d95b5.png)
. See and here.
Particular cases.
For 
obtain that ![$(1-m)\left(a^2-b^2\right)+c^2\ge (1-m)\cdot\left[XB^2-XA^2-(1-m)\cdot XC^2\right]$](//latex.artofproblemsolving.com/9/9/0/99014b84439880db6666d750723c923968004a10.png)
.
For 
obtain above proposed problem, i.e. 
.
For 
and 
obtain that 
for any 
.
Proof. Denote
. Apply the Stewart's relation to the cevians in the mentioned triangles :
![$[AM]/\triangle ABC\ :\ (1-m)^2\cdot AM^2=(1-m)\left(c^2-mb^2\right)+ma^2\ ;$](//latex.artofproblemsolving.com/b/d/8/bd81d507b8b1b476030bd14a5fae16baf2d2c94b.png)
![$[XM]/\triangle BXC\ :\ (1-m)^2\cdot XM^2=(1-m)\cdot\left(XB^2-m\cdot XC^2\right)+ma^2\ ;$](//latex.artofproblemsolving.com/e/6/2/e62d6c6abcffead4f78ab4e69c23909b3ee1291d.png)
![$[MN]/\triangle AMX\ :\ (1-n)^2\cdot MN^2=$](//latex.artofproblemsolving.com/7/c/f/7cfd12532b5db0e15b88018467bd5ebfa634becc.png)
. By eliminating the lengths 
, 
obtain :
![$(1-n)\left[(1-m)\left(c^2-mb^2\right)+ma^2\right]-$](//latex.artofproblemsolving.com/3/c/8/3c89a433117de204bbc68006e91e4ed61266874f.png)
![$n(1-n)\cdot\left[(1-m)\cdot\left(XB^2-m\cdot XC^2\right)+ma^2\right]+$](//latex.artofproblemsolving.com/e/2/3/e2365a632f86ecfb2f95dccd4ee9e120658a2526.png)
. For and 
obtain required inequality.
, where 
(the set of all complex numbers).Proposed problem. Prove that
from the plane 
exists the inequality 
.Proof. Denote the midpoints
, 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, ![$[XA]$](http://latex.artofproblemsolving.com/a/5/3/a5317c3e6beee34d801bcb11bbbba6b9ae3095c6.png)
respectively. Apply the theorem of the median in the triangles :
.Remark. For
and 
, where 
, 
, 
belong to the complex plane and 
is the circumcircle of
obtain that 
.Another proof. If denote
, 
and 
, then the our inequality is equivalently with 
, what is truly because 
. Generalization. Prove that for any
and for any 
from the plane of 
exists the inequality![$\boxed{\ n\left[n(1-m)a^2-m(1-m)b^2+mc^2\right]\ge m(1-n)\cdot\left[-m\cdot XA^2+n\cdot XB^2-n(1-m)\cdot XC^2\right]\ }$](http://latex.artofproblemsolving.com/d/d/f/ddf6d465cec78f24884888c235e0426a706d95b5.png)
. See and here.Particular cases.
For 
obtain that ![$(1-m)\left(a^2-b^2\right)+c^2\ge (1-m)\cdot\left[XB^2-XA^2-(1-m)\cdot XC^2\right]$](http://latex.artofproblemsolving.com/9/9/0/99014b84439880db6666d750723c923968004a10.png)
.
For 
obtain above proposed problem, i.e. 
.
For 
and 
obtain that 
for any 
.Proof. Denote
. Apply the Stewart's relation to the cevians in the mentioned triangles :![$[AM]/\triangle ABC\ :\ (1-m)^2\cdot AM^2=(1-m)\left(c^2-mb^2\right)+ma^2\ ;$](http://latex.artofproblemsolving.com/b/d/8/bd81d507b8b1b476030bd14a5fae16baf2d2c94b.png)
![$[XM]/\triangle BXC\ :\ (1-m)^2\cdot XM^2=(1-m)\cdot\left(XB^2-m\cdot XC^2\right)+ma^2\ ;$](http://latex.artofproblemsolving.com/e/6/2/e62d6c6abcffead4f78ab4e69c23909b3ee1291d.png)
![$[MN]/\triangle AMX\ :\ (1-n)^2\cdot MN^2=$](http://latex.artofproblemsolving.com/7/c/f/7cfd12532b5db0e15b88018467bd5ebfa634becc.png)
. By eliminating the lengths 
, 
obtain :
![$(1-n)\left[(1-m)\left(c^2-mb^2\right)+ma^2\right]-$](http://latex.artofproblemsolving.com/3/c/8/3c89a433117de204bbc68006e91e4ed61266874f.png)
![$n(1-n)\cdot\left[(1-m)\cdot\left(XB^2-m\cdot XC^2\right)+ma^2\right]+$](http://latex.artofproblemsolving.com/e/2/3/e2365a632f86ecfb2f95dccd4ee9e120658a2526.png)
. For and 
obtain required inequality.This post has been edited 30 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:09 PM
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