338. ROU National Math Olympiad 2012.
by Virgil Nicula, Apr 3, 2012, 11:58 PM
See here (<== click) - Rou O.N.M. , Constanta.
PP1 (Grade IX).
and 
. Prove that 
Proof.
![$4\sum \frac {\left(x_k-x_{k+1}\right)\left[\left(x_k+x_{k+1}\right)^2-x_kx_{k+1}\right]}{x_k+x_{k+1}}=$](//latex.artofproblemsolving.com/e/9/a/e9ae6415ad864cb33020074391e1c38743b8097d.png)
.
Remark. I used the simple inequality
, where 
and 
.
PP2 (grade IX). Let
be a triangle with 
. Consider the altitude 
, the 
-angle bisector 
and the 
-angle bisector 
, where 
,
and 
. Denote the intersections 
, 
and the midpoints 
, 
of ![$[QD]$](//latex.artofproblemsolving.com/2/b/2/2b24b6b34a55e29ef31a8cc9a72f116d3896a69f.png)
, ![$[PE]$](//latex.artofproblemsolving.com/7/9/2/792976015f6d59bd3b89b6d2d53901be85cecd92.png)
respectively. Prove that 
.
Proof. Choose origin
and let 
. Thus, 
, 
and 
. Menelaus' theorem
to the transversals :
.
Therefore,
.
.
Therefore,
.
In conclusion,
.
PP3 (grade VII). Let
be an interior point of the square 
so that 
. Ascertain the length of ![$[PD]$](//latex.artofproblemsolving.com/3/b/3/3b3ab695b20d0a266cb5c4fad99fa342facb80ef.png)
and 
.
Proof 1. Let projections , of on 
, 
respectively. So 
and 
. Thus, 
.
Proof 2. Let
and 
i.e. . Let 
. Thus, 
. Therefore, 
and 
. Hence 
.
Remark.
. Since obtain that 
.
Generally. is interior w.r.t. -right and isosceles 
, where 
, 
, 
. In our particular case
obtain
. I denoted 
- the area of the triangle with lengths of the sides 
.
PP4 (grade VII). In the-right-angled triangle consider the points 
and 
so that 
. Prove that 
.
Proof 1 (synthetic). Let
be symmetric of 
w.r.t. . Thus, 
is -isosceles , i.e. 
.
So
, i.e. 
is -isosceles.
From
, 
obtain 
, i.e. 
.
Proof 2 (metric). Denote
. Observe that 
![$(BE+DC)^2=[(BE+b)-AD]^2\implies$](//latex.artofproblemsolving.com/1/5/3/153a553e4f937e2d5b8b196a8c118b25459d0352.png)
.
PP1 (Grade IX).
and 
. Prove that 
Proof.
![$4\sum \frac {\left(x_k-x_{k+1}\right)\left[\left(x_k+x_{k+1}\right)^2-x_kx_{k+1}\right]}{x_k+x_{k+1}}=$](http://latex.artofproblemsolving.com/e/9/a/e9ae6415ad864cb33020074391e1c38743b8097d.png)
.Remark. I used the simple inequality
, where 
and 
.PP2 (grade IX). Let
be a triangle with 
. Consider the altitude 
, the 
-angle bisector 
and the 
-angle bisector 
, where 
,
and 
. Denote the intersections 
, 
and the midpoints 
, 
of ![$[QD]$](http://latex.artofproblemsolving.com/2/b/2/2b24b6b34a55e29ef31a8cc9a72f116d3896a69f.png)
, ![$[PE]$](http://latex.artofproblemsolving.com/7/9/2/792976015f6d59bd3b89b6d2d53901be85cecd92.png)
respectively. Prove that 
.Proof. Choose origin
and let 
. Thus, 
, 
and 
. Menelaus' theoremto the transversals :
.Therefore,
.
.Therefore,
.In conclusion,
.PP3 (grade VII). Let
be an interior point of the square 
so that 
. Ascertain the length of ![$[PD]$](http://latex.artofproblemsolving.com/3/b/3/3b3ab695b20d0a266cb5c4fad99fa342facb80ef.png)
and 
.Proof 1. Let projections
, of on 
, 
respectively. So 
and 
. Thus, 
.Proof 2. Let
and 
i.e.
. Let 
. Thus, 
. Therefore, 
and 
. Hence 
.Remark.
. Since obtain that 
.Generally.
is interior w.r.t. -right and isosceles 
, where 
, 
, 
. In our particular caseobtain
. I denoted 
- the area of the triangle with lengths of the sides 
.PP4 (grade VII). In the
-right-angled triangle consider the points 
and 
so that 
. Prove that 
.Proof 1 (synthetic). Let
be symmetric of 
w.r.t. . Thus, 
is -isosceles , i.e. 
.So
, i.e. 
is -isosceles.From
, 
obtain 
, i.e. 
.Proof 2 (metric). Denote
. Observe that 
![$(BE+DC)^2=[(BE+b)-AD]^2\implies$](http://latex.artofproblemsolving.com/1/5/3/153a553e4f937e2d5b8b196a8c118b25459d0352.png)
.This post has been edited 57 times. Last edited by Virgil Nicula, Nov 18, 2015, 12:51 PM
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