184. An interesting geometry problem from RMO 2010.
by Virgil Nicula, Dec 5, 2010, 4:23 PM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=380706
Proposed problem (RMO, 2010). Let
be a triangle with 
. Let 
and 
be the bisectors of 
and
respectively, where 
and 
. Let 
be the reflection of 
in the line 
. Prove that 
.
Proof 1 (synthetic). Consider the point
for which 
. The point 
exists because 
.
Observe that the quadrilaterals
and 
are cyclically and 
, i.e. the line bisects the segment ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
. Therefore,
the quadrilateral
is a kite with the symmetry axis 
. In conclusion, is the reflection of in the line , i.e. 
.
Proof 2 (proiective - Luisgeometria).
is the incenter of 
and 
cuts 
and at 
respectively. Since of 
,
obtain that
lie on a circle with center 
is the polar of 
WRT 
is perpendicular to through a point 
. Since cross ratio
is harmonic and 
, we deduce that 
and bisect 
internally and externally, then 
is cyclic. But, since lies on the diagonal 
of the rhombus 
formed by equilateral 
and 
obtain that
is an isosceles trapezoid with symmetry axis is the reflection of about , i.e. 
.
Proof 3 (metric). Let
for which 
and 
are cyclically. Observe that 
and 
.
Therefore,
and 
bisects ![$[AX]$](//latex.artofproblemsolving.com/3/d/1/3d107510e4dbf408903f44ba8d086233c93e9135.png)
Proof 4 (synthetic). Since
is cyclically. Denote the second intersection of the line with the
circumcircle of
. From the relations 
obtain that 
, i.e. 
is cyclically
is cyclically 
is kite is the reflection of in .
Proof 5 (synthetic). Denote
. Prove easily that 
is a cyclically 
and 
.
Since
![$90^{\circ}-\left[m(\angle ABI)+m(\angle IAB)\right]=$](//latex.artofproblemsolving.com/0/b/3/0b33295c9c3d16b093d26de4d6b5be0462810fbc.png)
obtain that
, i.e. 
is cyclically. Thus, 
the point is the reflection of onto which lies on .
Proposed problem (RMO, 2010). Let
be a triangle with 
. Let 
and 
be the bisectors of 
and
respectively, where 
and 
. Let 
be the reflection of 
in the line 
. Prove that 
.Proof 1 (synthetic). Consider the point
for which 
. The point 
exists because 
.Observe that the quadrilaterals
and 
are cyclically and 
, i.e. the line bisects the segment ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
. Therefore,the quadrilateral
is a kite with the symmetry axis 
. In conclusion, is the reflection of in the line , i.e. 
.Proof 2 (proiective - Luisgeometria).
is the incenter of 
and 
cuts 
and at 
respectively. Since of 
,obtain that
lie on a circle with center 
is the polar of 
WRT 
is perpendicular to through a point 
. Since cross ratio
is harmonic and 
, we deduce that 
and bisect 
internally and externally, then 
is cyclic. But, since lies on the diagonal 
of the rhombus 
formed by equilateral 
and 
obtain that
is an isosceles trapezoid with symmetry axis is the reflection of about , i.e. 
.Proof 3 (metric). Let
for which 
and 
are cyclically. Observe that 
and 
.Therefore,
and 
bisects ![$[AX]$](http://latex.artofproblemsolving.com/3/d/1/3d107510e4dbf408903f44ba8d086233c93e9135.png)
Proof 4 (synthetic). Since
is cyclically. Denote the second intersection of the line with thecircumcircle of
. From the relations 
obtain that 
, i.e. 
is cyclically 
is cyclically 
is kite is the reflection of in .Proof 5 (synthetic). Denote
. Prove easily that 
is a cyclically 
and 
.Since
![$90^{\circ}-\left[m(\angle ABI)+m(\angle IAB)\right]=$](http://latex.artofproblemsolving.com/0/b/3/0b33295c9c3d16b093d26de4d6b5be0462810fbc.png)
obtain that 
, i.e. 
is cyclically. Thus, 
the point is the reflection of onto which lies on .This post has been edited 33 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:28 PM
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