66. Remarkable perpendicularities in a triangle.
by Virgil Nicula, Jul 22, 2010, 3:09 PM
PP1. Let 
with incenter 
and circumcenter 
. Consider 
, 
so that 
. Denote the second
intersection
between circumcircles of 
, 
. Prove that 
, 
and 
(constant).
Proof. Denote
, i.e. 
, 
. Using power of 
, 
w.r.t. circumcircle 
obtain 
. Since quadrilaterals 
, 
are cyclically obtain 
.
Since
obtain (s.a.s.) 
, i.e. and 
, 
. If denote projections 
, 
of on this sidelines, then from upper equivalence obtain 
(i.e. belongs to 
-bisector) , 
and 
, 
. Therefore, 
![$x[(b-AV)-(c-AU)]$](//latex.artofproblemsolving.com/0/a/c/0ac85ebc404f665bdbf50e5da02434733a196c7d.png)
. From relations 
, 
obtain 
, i.e. 
.
Remark. Observe that
. Apply Ptolemy's theorem to 
, and obtain 
- constant for any 
.
PP2. Let
be a triangle with incenter and circumcenter . Consider , 
, 
so that . Denote the second
intersection between circumcircles of , . Prove that belongs to exterior -bisector, and 
(constant).
Proof. Prove easily that
and 
, where 
and 
, 
are the projections of to 
, 
respectively.
PP3. Let with the circumcircle 
, the incircle 
and 
. Let 
and 
such that 
. Prove that 
and 
.
Proof. Suppose w.l.o.g.
and 
. In this case 
, 
.
.
Denote the tangent points , 
of the incircle of with , respectively. Thus, 
and 
. Obtain that
. From the relations , obtain that .
Apply the generalized Pythagoras' theorem to the triangle 
![$bc\left[a^2-(a-b)^2-(a-c)^2+2(a-b)(a-c)\cos A\right]=$](//latex.artofproblemsolving.com/4/2/4/4242b0b8818eed69c05c9a6b3f199e0170f2133e.png)
![$bc\left\{a^2-\left[(a-b)-(a-c)\right]^2-2(a-b)(a-c)+2(a-b)(a-c)\cos A\right\}=$](//latex.artofproblemsolving.com/7/6/2/762218014c4a6a824c17e55e294c73b8f47c5a81.png)
![$bc\left[a^2-(b-c)^2-4(a-b)(a-c)\sin^2\frac A2\right]=$](//latex.artofproblemsolving.com/8/2/2/822c18bf7da47e2c44691015188dd0317b5b8a10.png)
. Therefore, 
. In conclusion, 
, i.e. 
. Since 
obtain from the relation 
that .
Remark. If denote the length
of , then 
.
PP4. Consider with 
. Let 
be two mobile points different from , such that 
. Show that the circumcircle of 
pass through a fixed point different from .
Proof. The common point (say
) is actually the reflection of point about the perpendicular to 
at the point 
given by 
, or 
. Let be the center of the circumcircle
of
(of radius 
) and the foot of the perpendicular from to . Then the given expression is equivalent to 
or (using the power of 
and 
around circle )
, or 
. That is, is a fixed, given point on . All circumcenters belong on the perpendicular to at . This perpendicular
bisects the common chords
of all circumcircles , and thus is the reflection of about this perpendicular. It is interesting to examine some limiting cases. If 
is on so that 
then 
is one limiting
case of with 
and 
. Indeed, in this case, 
and 
. On the other hand, in the limiting case of 
, we get 
.
To make sure that this is possible, we note that the previous quadratic in has a solution if 
which is allowed by the given condition that 
.
with incenter 
and circumcenter 
. Consider 
, 
so that 
. Denote the secondintersection
between circumcircles of 
, 
. Prove that 
, 
and 
(constant).Proof. Denote
, i.e. 
, 
. Using power of 
, 
w.r.t. circumcircle 
obtain 
. Since quadrilaterals 
, 
are cyclically obtain 
.Since
obtain (s.a.s.) 
, i.e. and 
, 
. If denote projections 
, 
of on this sidelines, then from upper equivalence obtain 
(i.e.
belongs to 
-bisector) , 
and 
, 
. Therefore, 
![$x[(b-AV)-(c-AU)]$](http://latex.artofproblemsolving.com/0/a/c/0ac85ebc404f665bdbf50e5da02434733a196c7d.png)
. From relations 
, 
obtain 
, i.e. 
.Remark. Observe that
. Apply Ptolemy's theorem to 
, and obtain 
- constant for any 
.PP2. Let
be a triangle with incenter and circumcenter . Consider , 
, 
so that . Denote the secondintersection
between circumcircles of , . Prove that belongs to exterior -bisector, and 
(constant).Proof. Prove easily that
and 
, where 
and 
, 
are the projections of to 
, 
respectively.PP3. Let
with the circumcircle 
, the incircle 
and 
. Let 
and 
such that 
. Prove that 
and 
.Proof. Suppose w.l.o.g.
and 
. In this case 
, 
.
.Denote the tangent points
, 
of the incircle of with , respectively. Thus, 
and 
. Obtain that
. From the relations , obtain that .
Apply the generalized Pythagoras' theorem to the triangle 
![$bc\left[a^2-(a-b)^2-(a-c)^2+2(a-b)(a-c)\cos A\right]=$](http://latex.artofproblemsolving.com/4/2/4/4242b0b8818eed69c05c9a6b3f199e0170f2133e.png)
![$bc\left\{a^2-\left[(a-b)-(a-c)\right]^2-2(a-b)(a-c)+2(a-b)(a-c)\cos A\right\}=$](http://latex.artofproblemsolving.com/7/6/2/762218014c4a6a824c17e55e294c73b8f47c5a81.png)
![$bc\left[a^2-(b-c)^2-4(a-b)(a-c)\sin^2\frac A2\right]=$](http://latex.artofproblemsolving.com/8/2/2/822c18bf7da47e2c44691015188dd0317b5b8a10.png)
. Therefore, 
. In conclusion, 
, i.e. 
. Since 
obtain from the relation 
that .Remark. If denote the length
of , then 
.PP4. Consider
with 
. Let 
be two mobile points different from , such that 
. Show that the circumcircle of 
pass through a fixed point different from .Proof. The common point (say
) is actually the reflection of point about the perpendicular to 
at the point 
given by 
, or 
. Let be the center of the circumcircleof
(of radius 
) and the foot of the perpendicular from to . Then the given expression is equivalent to 
or (using the power of 
and 
around circle )
, or 
. That is, is a fixed, given point on . All circumcenters belong on the perpendicular to at . This perpendicularbisects the common chords
of all circumcircles , and thus is the reflection of about this perpendicular. It is interesting to examine some limiting cases. If 
is on so that 
then 
is one limitingcase of
with 
and 
. Indeed, in this case, 
and 
. On the other hand, in the limiting case of 
, we get 
.To make sure that this is possible, we note that the previous quadratic in
has a solution if 
which is allowed by the given condition that 
.This post has been edited 19 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:08 PM
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