218. Some problems from the Suiss IMO Selection Team 2006.
by Virgil Nicula, Jan 28, 2011, 3:54 PM
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89094&ml=1
Proposed problem 1. In
denote 
. Suppose 
. Prove that 
.
Proof. If
, then is isosceles. If 
, then denote the symmetrical point 
of 
w.r.t. 
. Thus, 
and
. From 
the quadrilateral 
is inscribed 
.
Proposed problem 5. ]Let
be inside to and 
, . Let 
and 
be the circumscribed circles of 
and 
respectively.
The line
intersects the circles and in the interior points , 
respectively. Denote Let 
and 
. Show that 
.
Proof. Denote
. Apply Menelaus' theorem to 
and 
. Obtain that that 
and 
.
Proving that the two LHS's in above expressions are equal is equivalent with proving
which (in turn) is equivalent to 
.
This is just another way of saying that the powers of
w.r.t. , are equal which is clear because belongs to the radical axis 
of two circles.
Proposed problem 7. The roots of the equation
are 
and 
. Find the value of 
.
Proof. Observe that
, 
, 
and 
. Denote 
.
Prove easily that
, 
and 
.
Therefore,
and 
are roots of the equation: 
. Because 
, obtain that 
.
Remark. 
. Since 
obtain that .
Proposed problem 9. Let be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
and let 
be the midpoint
of the side![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Let be a point on the side 
and 
a point on the side 
such that 
and the points , , are on the same line.
Prove that the line
is perpendicular to the common chord of the circumcircles of and 
.
Lemma. Let be an acute triangle. Define: the circumcircle 
and the orthocentre of the triangle ;
the midpoint of the side ; the intersection 
between 
and the bisector of the angle 
; 
and 
so that
and 
. Then the point belongs to the circumcircle 
of the triangle 
.
Proof of the proposed problem. Denote:
; the middlepoint 
of the segment ![$[AH]$](//latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
; 
. From the above lemma results
. But 
(the point is the middlepoint of the segment ![$[HA']$](//latex.artofproblemsolving.com/b/d/d/bdd6284b56db3edb70c48d53b2cef3bbf7dfdb34.png)
), 
and 
. Thus the points , 
, 
are the middlepoints of the segments
, 
, 
respectively and 
and 
. Therefore, 
.
Proposed problem 1. In
denote 
. Suppose 
. Prove that 
.Proof. If
, then is isosceles. If 
, then denote the symmetrical point 
of 
w.r.t. 
. Thus, 
and
. From 
the quadrilateral 
is inscribed 
.Proposed problem 5. ]Let
be inside to and 
, . Let 
and 
be the circumscribed circles of 
and 
respectively.The line
intersects the circles and in the interior points , 
respectively. Denote Let 
and 
. Show that 
.Proof. Denote
. Apply Menelaus' theorem to 
and 
. Obtain that that 
and 
.Proving that the two LHS's in above expressions are equal is equivalent with proving
which (in turn) is equivalent to 
.This is just another way of saying that the powers of
w.r.t. , are equal which is clear because belongs to the radical axis 
of two circles.Proposed problem 7. The roots of the equation
are 
and 
. Find the value of 
.Proof. Observe that
, 
, 
and 
. Denote 
.Prove easily that
, 
and 
.Therefore,
and 
are roots of the equation: 
. Because 
, obtain that 
.
Remark. 
. Since 
obtain that .Proposed problem 9. Let
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
and let 
be the midpointof the side
![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Let be a point on the side 
and 
a point on the side 
such that 
and the points , , are on the same line.Prove that the line
is perpendicular to the common chord of the circumcircles of and 
.Lemma. Let
be an acute triangle. Define: the circumcircle 
and the orthocentre of the triangle ;the midpoint
of the side ; the intersection 
between 
and the bisector of the angle 
; 
and 
so that
and 
. Then the point belongs to the circumcircle 
of the triangle 
.
and 
, 
. Thus, 
. Denote the point 
,
. Thus, 
and 
,Proof of the proposed problem. Denote:
; the middlepoint 
of the segment ![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
; 
. From the above lemma results
. But 
(the point is the middlepoint of the segment ![$[HA']$](http://latex.artofproblemsolving.com/b/d/d/bdd6284b56db3edb70c48d53b2cef3bbf7dfdb34.png)
), 
and 
. Thus the points , 
, 
are the middlepoints of the segments
, 
, 
respectively and 
and 
. Therefore, 
.This post has been edited 34 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:43 PM
August 2018
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