267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.
by Virgil Nicula, Apr 17, 2011, 3:36 PM
Germany, 2002. Let 
be an acute-angled triangle with the incenter 
and the circumcenter 
. Consider the feet
, 
and 
of the altitudes from 
, 
and 
. Denote the intersections 
, 
and 
of the angle bisectors from , and
with the opposite sidelines respectively. Show that
.
In conclusion,
Proof 1 (barycentrical coordinates). Prove easily that
.
It is well-known that
. Thus, 
and
.
Proof 2 (metric). Denote
. From the well-known properties 
and 
obtain that :
.
.
An easy extension. Let be an acute-angled triangle with the incenter and denote 
.
For an interior point
denote 
and for its isogonal point 
denote 
. Prove that
, where has the barycentrical coordinates 
w.r.t. 
.
Proof (barycentrical coordinates). It is well-known that
and 
.
Thus,
and 
.
Generalization. Let be a triangle. Consider , 
and its isogonals , respectively. For any 
define 
,
i.e.
is the cevian triangle of w.r.t. . Prove that 
,
where the points , have the barycentrical coordinates and 
respectively w.r.t. .
Proof. Observe that
and
.
Russia, 2003. In with circumcircle 
, 
denote 
and 
so that 
is tangent to 
. Prove 
.
Proof 1 (metric). From the relations
obtain 
and well-known property 
.
Proof 2 (synthetic). Prove easily that
.
From well-known property
obtain 
. If is the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, then 
and 
. Indeed,
.
=========================================================================================================================================
From the relation 
obtain that 
.
44 IMO 2003 Tokyo. Let be a point in the interior of the triangle . Let , , be the feet of the perpendiculars from to 
, 
, 
respectively.
Assume that
. Show that is the circumcenter of 
, where 
, 
, 
are the exincenters of .
Proof. Observe that
and 
. In the same
way we show that
and 
. Prove easily that , , 
are the projections of 
, 
, 
on , , respectively, i.e.
, 
, 
. Denote the projection 
. Observe that 
and
![$BF+BF'=FF'=\mathrm{pr}_{AB}[PI_a]=PI_a\cdot\sin\widehat{PI_aF'}=PI_a\cdot\sin B$](//latex.artofproblemsolving.com/7/e/1/7e12283e2ccda19d4cc352a7c2d5f5b914760401.png)
, where
is the length of the circumcenter for . In conclusion, 
, i.e. the point is the circumcenter of the triangle .
Otherwise. Prove easily that the triangles
, 
,
are isosceles , i.e. 
.Observe that in 
exists the relation 
.
Therefore,
. In conclusion, .
Otherwise. If
, 
, 
are the intersections of 
, 
, 
respectively with the circumcircle of , then 
,
where
and 
, 
, 
(homothety). Since 
obtain that the length
of the circumradius for
is equally to 
, where is the length of the circumradius for 
, i.e. the circumradius of .
Austrian-Polish MO 1992. Let![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
be a fixed diameter of the circle 
. Consider a fixed point 
.
Denote by
the line which is tangent to at 
. For any mobile chord ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
other than for which the
point
denote 
and 
. Prove that the product 
is constant.
Proof.
![\[\boxed{\ \begin{array}{c}
\frac {KA}{KB}\cdot PQ=\frac {CP}{CB}\cdot AQ+\frac {DQ}{DB}\cdot AP\\\\
\frac {KA}{KB}\cdot PQ=\left(\frac {AP}{AB}\right)^2\cdot AQ+\left(\frac {AQ}{AB}\right)^2\cdot AP\\\\
AB^2\cdot\frac {KA}{KB}\cdot PQ=AP\cdot AQ\cdot (AP+AQ)\\\\
AP\cdot AQ=AB^2\cdot\frac {KA}{KB}\\\\
AP\cdot AQ\ \mathrm{- constant.}\end{array}\ }\]](//latex.artofproblemsolving.com/c/d/7/cd75ee3a34eaa493e378b45508a97001cb949116.png)
Polish Mathematical Olympiad. Let a
- isosceles . For 
, 
let the reflection 
of 
w.r.t. 
. Prove that 
.
Proof. Prove easily that
, i.e. the quadrilateral 
is cyclically. Apply the Ptolemy's theorem
in the quad.rilateral
.
Remark. If
, then the quadrilateral 
is cyclically and 
.
be an acute-angled triangle with the incenter 
and the circumcenter 
. Consider the feet
, 
and 
of the altitudes from 
, 
and 
. Denote the intersections 
, 
and 
of the angle bisectors from , and with the opposite sidelines respectively. Show that
.In conclusion,
Proof 1 (barycentrical coordinates). Prove easily that
.It is well-known that
. Thus, 
and
.Proof 2 (metric). Denote
. From the well-known properties 
and 
obtain that :
.
.An easy extension. Let
be an acute-angled triangle with the incenter and denote 
.For an interior point
denote 
and for its isogonal point 
denote 
. Prove that
, where has the barycentrical coordinates 
w.r.t. 
.Proof (barycentrical coordinates). It is well-known that
and 
.Thus,
and 
.Generalization. Let
be a triangle. Consider , 
and its isogonals , respectively. For any 
define 
,i.e.
is the cevian triangle of w.r.t. . Prove that 
,where the points
, have the barycentrical coordinates and 
respectively w.r.t. .Proof. Observe that
and
.Russia, 2003. In
with circumcircle 
, 
denote 
and 
so that 
is tangent to 
. Prove 
.Proof 1 (metric). From the relations
obtain 
and well-known property 
.Proof 2 (synthetic). Prove easily that
.From well-known property
obtain 
. If is the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, then 
and 
. Indeed,
.=========================================================================================================================================
From the relation 
obtain that 
. 44 IMO 2003 Tokyo. Let
be a point in the interior of the triangle . Let , , be the feet of the perpendiculars from to 
, 
, 
respectively.Assume that
. Show that is the circumcenter of 
, where 
, 
, 
are the exincenters of .Proof. Observe that
and 
. In the sameway we show that
and 
. Prove easily that , , 
are the projections of 
, 
, 
on , , respectively, i.e.
, 
, 
. Denote the projection 
. Observe that 
and
![$BF+BF'=FF'=\mathrm{pr}_{AB}[PI_a]=PI_a\cdot\sin\widehat{PI_aF'}=PI_a\cdot\sin B$](http://latex.artofproblemsolving.com/7/e/1/7e12283e2ccda19d4cc352a7c2d5f5b914760401.png)
, where is the length of the circumcenter for . In conclusion, 
, i.e. the point is the circumcenter of the triangle .Otherwise. Prove easily that the triangles
, 
,
are isosceles , i.e. 
.Observe that in 
exists the relation 
.Therefore,
. In conclusion, .Otherwise. If
, 
, 
are the intersections of 
, 
, 
respectively with the circumcircle of , then 
,where
and 
, 
, 
(homothety). Since 
obtain that the lengthof the circumradius for
is equally to 
, where is the length of the circumradius for 
, i.e. the circumradius of .Austrian-Polish MO 1992. Let
![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
be a fixed diameter of the circle 
. Consider a fixed point 
.Denote by
the line which is tangent to at 
. For any mobile chord ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
other than for which thepoint
denote 
and 
. Prove that the product 
is constant.Proof.
![\[\boxed{\ \begin{array}{c}
\frac {KA}{KB}\cdot PQ=\frac {CP}{CB}\cdot AQ+\frac {DQ}{DB}\cdot AP\\\\
\frac {KA}{KB}\cdot PQ=\left(\frac {AP}{AB}\right)^2\cdot AQ+\left(\frac {AQ}{AB}\right)^2\cdot AP\\\\
AB^2\cdot\frac {KA}{KB}\cdot PQ=AP\cdot AQ\cdot (AP+AQ)\\\\
AP\cdot AQ=AB^2\cdot\frac {KA}{KB}\\\\
AP\cdot AQ\ \mathrm{- constant.}\end{array}\ }\]](http://latex.artofproblemsolving.com/c/d/7/cd75ee3a34eaa493e378b45508a97001cb949116.png)
Polish Mathematical Olympiad. Let a
- isosceles . For 
, 
let the reflection 
of 
w.r.t. 
. Prove that 
.Proof. Prove easily that
, i.e. the quadrilateral 
is cyclically. Apply the Ptolemy's theoremin the quad.rilateral
.Remark. If
, then the quadrilateral 
is cyclically and 
.This post has been edited 82 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:39 AM
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