290. Characterize "OP perpendicular on OA" in ABC a.s.o.
by Virgil Nicula, Jun 24, 2011, 1:46 PM
PP1. Let 
be a triangle with the circumcircle 
. Consider an interior point 
with the barycentrical coordinates
w.r.t. 
. Prove that 
. Study some particular cases.
Proof. Let
so that 
, 
, diameter ![$[AA']$](//latex.artofproblemsolving.com/f/8/3/f83bdc8e170d3bc867097c603fa03cf6edefbb4b.png)
of 
and 
. Since 
, 
belong to the circle with the diameter
![$[A'P]$](//latex.artofproblemsolving.com/8/2/d/82ddd69d01ce497d9384bb9971474a585439fc03.png)
obtain 
. Since 
and 
, the relation 
becomes
. In conclusion, 
.
Some particular cases.
.
.
.
PP2. Let be a triangle with the circumcircle . Consider the midpoint 
of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and a point
for which 
. Prove that 
.
Particular case For
obtain 
, where 
is centroid of .
Proof 1
and 
. Denote the second intersection 
of 
with . Observe that
. Thus, 
![$\frac {m_a}{\rho}\left[\frac {(\rho -1)m_a}{\rho}+\frac {a^2}{4m_a}\right]=$](//latex.artofproblemsolving.com/d/a/1/da157c95ffa5b0ad7d95d4d5bd09301ca82f4307.png)
![$\frac {1}{4\rho ^2}\cdot\left[2(\rho -1)\left(b^2+c^2\right)+\rho a^2\right]$](//latex.artofproblemsolving.com/b/d/0/bd087e2cfe984890deca5b95ecd646948314fab3.png)
. In conclusion, 
![$2R^2=\frac {1}{4\rho^2}\cdot\left[2(\rho -1)\left(b^2+c^2\right)+a^2\right]+\frac {m_a^2}{\rho ^2}$](//latex.artofproblemsolving.com/1/d/8/1d8d6a19505432579e5c9d4c614c9a511e336d71.png)
.
Proof 2 (Yetti). Denote the symmedian
and the altitude 
, where 
. Thus, the line 
is
antiparallel to
w.r.t. 
and from the w.k. relations 
and
obtain that 
.
Remark. Can obtain directly this problem from the general case PP1 for
.
An easy extension. Let be a triangle with the circumcircle . Consider a point ![$M\in [BC]$](//latex.artofproblemsolving.com/6/f/1/6f11082ff5ed53afefce1cb91c180d8a7342b4c1.png)
so that 
and a point for which . Prove that 
.
Proof. Denote the isogonal of and the altitude , where . Since the line is antiparallel to w.r.t.
obtain that and from the w.k. relations 
and
obtain that
.
PP3. Let be an 
-right triangle with incenter 
. Denote 
and the midpoint of ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Prove that 
.
Proof 1 (synthetic). Denote the projections
, 
on of 
, respectively. Observe that 
and 
. Therefore, 
and
, i.e. is the circumcenter of 
. In conclusion, 
is middleline in the right trapezoid 
.
Remark. Prove easily that the point is the 
-exincenter of 
and is the 
-exincenter of 
. Thus 
and 
.
Proof 2 (metric). I"ll use the equivalence
and the well-known relations 
.
Observe that
. Apply the theorem of the median to :
. On other hand, 
.
From the last relations
and 
obtain that 
, i.e. .
Proof 3 (metric). Denote the point
for which 
. I"ll show that 
. Using an well-known property 
.
Remark. Prove easily that in any triangle exists the relation ![$MB^2-MC^2=a(c-b)\cdot\left[1+\frac {bc\cdot \cos A}{(a+b)(a+c)}\right]=a(c-b)\cdot\frac {(a+b+c)^2}{2(a+b)(a+c)}$](//latex.artofproblemsolving.com/a/c/1/ac11139b06549c525b544bd00656fe5fe7266090.png)
.
Observe that
.
PP4.Let
be a convex quadrilateral. Denote 
, the centroids , 
of 
and
respectively, the orthocenters , 
of 
and 
respectively. Prove that 
.
Proof. Soon !
be a triangle with the circumcircle 
. Consider an interior point 
with the barycentrical coordinates
w.r.t. 
. Prove that 
. Study some particular cases. Proof. Let
so that 
, 
, diameter ![$[AA']$](http://latex.artofproblemsolving.com/f/8/3/f83bdc8e170d3bc867097c603fa03cf6edefbb4b.png)
of 
and 
. Since 
, 
belong to the circle with the diameter![$[A'P]$](http://latex.artofproblemsolving.com/8/2/d/82ddd69d01ce497d9384bb9971474a585439fc03.png)
obtain 
. Since 
and 
, the relation 
becomes
. In conclusion, 
.Some particular cases.
.
.
.PP2. Let
be a triangle with the circumcircle . Consider the midpoint 
of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and a point
for which 
. Prove that 
.Particular case For
obtain 
, where 
is centroid of .Proof 1
and 
. Denote the second intersection 
of 
with . Observe that
. Thus, 
![$\frac {m_a}{\rho}\left[\frac {(\rho -1)m_a}{\rho}+\frac {a^2}{4m_a}\right]=$](http://latex.artofproblemsolving.com/d/a/1/da157c95ffa5b0ad7d95d4d5bd09301ca82f4307.png)
![$\frac {1}{4\rho ^2}\cdot\left[2(\rho -1)\left(b^2+c^2\right)+\rho a^2\right]$](http://latex.artofproblemsolving.com/b/d/0/bd087e2cfe984890deca5b95ecd646948314fab3.png)
. In conclusion, 
![$2R^2=\frac {1}{4\rho^2}\cdot\left[2(\rho -1)\left(b^2+c^2\right)+a^2\right]+\frac {m_a^2}{\rho ^2}$](http://latex.artofproblemsolving.com/1/d/8/1d8d6a19505432579e5c9d4c614c9a511e336d71.png)
.Proof 2 (Yetti). Denote the symmedian
and the altitude 
, where 
. Thus, the line 
isantiparallel to
w.r.t. 
and from the w.k. relations 
and
obtain that 
.Remark. Can obtain directly this problem from the general case PP1 for
.An easy extension. Let
be a triangle with the circumcircle . Consider a point ![$M\in [BC]$](http://latex.artofproblemsolving.com/6/f/1/6f11082ff5ed53afefce1cb91c180d8a7342b4c1.png)
so that 
and a point
for which . Prove that 
.Proof. Denote the isogonal
of and the altitude , where . Since the line is antiparallel to w.r.t. obtain that
and from the w.k. relations 
and obtain that
.PP3. Let
be an 
-right triangle with incenter 
. Denote 
and the midpoint of ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Prove that 
. Proof 1 (synthetic). Denote the projections
, 
on of 
, respectively. Observe that 
and 
. Therefore, 
and
, i.e. is the circumcenter of 
. In conclusion, 
is middleline in the right trapezoid 
.Remark. Prove easily that the point
is the 
-exincenter of 
and is the 
-exincenter of 
. Thus 
and 
.Proof 2 (metric). I"ll use the equivalence
and the well-known relations 
.Observe that
. Apply the theorem of the median to :
. On other hand, 
.From the last relations
and 
obtain that 
, i.e. .Proof 3 (metric). Denote the point
for which 
. I"ll show that 
. Using an well-known property 
.Remark. Prove easily that in any triangle
exists the relation ![$MB^2-MC^2=a(c-b)\cdot\left[1+\frac {bc\cdot \cos A}{(a+b)(a+c)}\right]=a(c-b)\cdot\frac {(a+b+c)^2}{2(a+b)(a+c)}$](http://latex.artofproblemsolving.com/a/c/1/ac11139b06549c525b544bd00656fe5fe7266090.png)
.Observe that
.PP4.Let
be a convex quadrilateral. Denote 
, the centroids , 
of 
and
respectively, the orthocenters , 
of 
and 
respectively. Prove that 
.Proof. Soon !
This post has been edited 72 times. Last edited by Virgil Nicula, Nov 21, 2015, 2:04 PM
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