169. Some metrical problems in a circle (III).
by Virgil Nicula, Oct 28, 2010, 10:59 AM
PP1. Let 
with 
circumcircle 
; feet 
, 
of bisectors from 
, 
; 
. Prove that 
(Bulgaria, 1997) .
Generalization. Let with the circumcircle , two points , and a point . Prove that 
.
Proof 1 (synthetic). Denote
so that 
separates , 
. Apply the Ptolemy's theorem to 
.
From an well-known property applied in the quadrilaterals
and 
obtain 
and 
.
Therefore,
.
Proof 2 (trigonometric). Choose the point so that the line separates , . Denote 
. Then 
. Observe that
and 
and the relation from conclusion is equivalently with
, what is truly.
Remarks. Denote
. Here are some interesting particular cases.
.
(Bulgaria, 1997).
.
(Nagel's point) 
, where 
.
(Lemoine's point) 
.
Remark. Let
be a triangle with the circumcircle . Denote the feet 
, 
of the bisectors from , respectively and 
.
If the side separates the points 
and then 
and 
and 
PP2. Consider an interior point
of for which 
and 
. Suppose 
and denote the incenters and 
for incircles of
and 
respectively and the length 
of circumradius of 
. Prove that 
(Toshio Seimyia).
Proof. Show easily that
. Observe that 
and 
.
Therefore,
. Apply the Cosinus' theorem in the triangles 
. Therefore, 
. Observe that
. Prove easily that 
. Thus, 
.
PP3. Let . For 
let the incircles 
and 
of 
and 
respectively. Prove that
(Toshio Seimyia).
Proof. Denote
and 
. Therefore, 
is cyclically 
and 
. Hence
and 
. Denote the distancies
and 
. Thus, 
and 
![$\frac {[ABD]}{cr_1}=\frac {[ADC]}{br_2}$](//latex.artofproblemsolving.com/c/3/1/c31227033cfc2ec3f683b3fd824deb6647589243.png)
. Remark. Let 
and 
. From the well-known relations
obtain that 
.
PP4. Let an
-right with the excircles 
, 
and 
so that 
, 
. Prove that 
.
Proof. Denote
so that 
. Prove easily that the nice identity 
. Therefore,
and 
Thus, 
and 
![$\left(1+k^2\right)\left[(s-b)^2+(s-c)^2\right]\ \stackrel{(*)}{=}\ \frac {a^2+(b+c)^2}{a^2}$](//latex.artofproblemsolving.com/4/f/8/4f898227bab9fec46d29c14e120a47408690939c.png)
. Remark. 
is a harmonic division, i.e. 
.
Hence and
is a harmonic division. Since 
obtain that 
is bisector of 
.
PP5 (Miguel O. Sanchez). Let an-right and 
. Prove that 
.
Proof.
.
.
PP6 (Ruben Dario). Let with 
and 
so that 
and 
. Let the incircles
and 
of 
and 
respectively, where 
and 
. Prove that 
.
Proof. Denote
. Observe that 
.
Thus,
. Observe that 
.
Thus,
. From 
obtain that 
and 
.
Therefore,
![$\frac {bc(s-a)}{a}\left[\frac {(s-a)+(s-b)}{s-b}+\frac {(s-a)+(s-c)}{s-c}-2\right]=$](//latex.artofproblemsolving.com/c/3/7/c37fd19c3f3a17b2a20a5f3e6036aa5bc84e2168.png)
Remark. I used the well-known relations in any
.
with 
circumcircle 
; feet 
, 
of bisectors from 
, 
; 
. Prove that 
(Bulgaria, 1997) .Generalization. Let
with the circumcircle , two points , and a point . Prove that 
.Proof 1 (synthetic). Denote
so that 
separates , 
. Apply the Ptolemy's theorem to 
.From an well-known property applied in the quadrilaterals
and 
obtain 
and 
.Therefore,
.Proof 2 (trigonometric). Choose the point
so that the line separates , . Denote 
. Then 
. Observe that
and 
and the relation from conclusion is equivalently with
, what is truly.Remarks. Denote
. Here are some interesting particular cases.
.
(Bulgaria, 1997).
.
(Nagel's point) 
, where 
.
(Lemoine's point) 
.Remark. Let
be a triangle with the circumcircle . Denote the feet 
, 
of the bisectors from , respectively and 
.If the side
separates the points 
and then 
and 
and 
PP2. Consider an interior point
of for which 
and 
. Suppose 
and denote the incenters and 
for incircles of
and 
respectively and the length 
of circumradius of 
. Prove that 
(Toshio Seimyia). Proof. Show easily that
. Observe that 
and 
.Therefore,
. Apply the Cosinus' theorem in the triangles 
. Therefore, 
. Observe that
. Prove easily that 
. Thus, 
.PP3. Let
. For 
let the incircles 
and 
of 
and 
respectively. Prove that
(Toshio Seimyia). Proof. Denote
and 
. Therefore, 
is cyclically 
and 
. Hence
and 
. Denote the distancies
and 
. Thus, 
and 
![$\frac {[ABD]}{cr_1}=\frac {[ADC]}{br_2}$](http://latex.artofproblemsolving.com/c/3/1/c31227033cfc2ec3f683b3fd824deb6647589243.png)
. Remark. Let 
and 
. From the well-known relations
obtain that 
. PP4. Let an
-right with the excircles 
, 
and 
so that 
, 
. Prove that 
. Proof. Denote
so that 
. Prove easily that the nice identity 
. Therefore,
and 
Thus, 
and 
![$\left(1+k^2\right)\left[(s-b)^2+(s-c)^2\right]\ \stackrel{(*)}{=}\ \frac {a^2+(b+c)^2}{a^2}$](http://latex.artofproblemsolving.com/4/f/8/4f898227bab9fec46d29c14e120a47408690939c.png)
. Remark. 
is a harmonic division, i.e. 
.Hence and
is a harmonic division. Since 
obtain that 
is bisector of 
.PP5 (Miguel O. Sanchez). Let an
-right and 
. Prove that 
.Proof.
.
.PP6 (Ruben Dario). Let
with 
and 
so that 
and 
. Let the incircles
and 
of 
and 
respectively, where 
and 
. Prove that 
. Proof. Denote
. Observe that 
.Thus,
. Observe that 
.Thus,
. From 
obtain that 
and 
.Therefore,
![$\frac {bc(s-a)}{a}\left[\frac {(s-a)+(s-b)}{s-b}+\frac {(s-a)+(s-c)}{s-c}-2\right]=$](http://latex.artofproblemsolving.com/c/3/7/c37fd19c3f3a17b2a20a5f3e6036aa5bc84e2168.png)
Remark. I used the well-known relations in any
.This post has been edited 113 times. Last edited by Virgil Nicula, Dec 1, 2015, 9:27 AM
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