55. Two important circles - mixtilinear incircle/excircle.
by Virgil Nicula, Jul 10, 2010, 12:24 AM
Proposed problem. Let 
be the incircle and the 
-exincircle of 
which touches 
at 
, 
respectively. Define and note the circles 
and
the circles which are tangent to 
in 
, 
respectively and to the circumcircle of (the first -interior, the second-exterior) at 
,
respectively.
Prove that
; 
; 
; 
; 
, 
and the bisector of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
are concurrently.
Remark. Let be a triangle with incenter 
and the circle 
internally tangent to , and to the circumcircle of given triangle.
Denote the contact points
, 
of circle with , respectively. Show that , , 
are collinear and 
.
PP1. Let
be a triangle -right-angled and 
be its circumcircle. Let 
be the circle touching the lines and , and the circle internally.
Further, let
be the circle touching the lines and and the circle externally. If 
, 
be the radii of , 
prove that ![$r_1 \cdot r_2 = 4\cdot [ABC]$](//latex.artofproblemsolving.com/d/3/0/d305df44a65706f774f19cdf0a7b6db11c773b68.png)
.
Proof.
. From upper problem obtain : ![$\frac{r_1}{r}=\frac{r_2}{r_a}=\frac{bc}{s(s-a)}\Longrightarrow r_1\cdot r_2=rr_a\cdot \left[ \frac{bc}{s(s-a)}\right]^2 =4S$](//latex.artofproblemsolving.com/0/7/f/07f56a355c19fd730fe1b92b3eee01976ad3b6cd.png)
.
PP2. Let , the circumcircle 
and the circle 
which touches , at ,
and which is internal tangent to
at . Let 
. Prove that 
.
Proof. It's known that the incenter of is the midpoint of 
(see here). Let 
, 
cut 
again at 
, 
. Prove easily
is the bisector of 
and 
is the bisector of 
, 
belong to . Apply the Pascal's
theorem to the cyclic hexagon
.
PP3 (Miguel Ocho Sanchez). The mixtlinear circle of touches , and the circumcircle
at , and respectively. Denote the projections 
, 
of on 
, 
respectively. Prove 
.
Proof. Is well-known that the incenter is the midpoint of ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
. Denote the midpoint 
of and the second intersection of 
with . From the previous
problem results that
. Observe that 
, what means that 
.
PP4 (Ruben Dario). Let with the incircle 
and the -excircle 
. Let 
and 
so that 
. Prove 
.
Proof 1. Denote
, i.e. 
. Prove easily that 
. Hence 
.
Since obtain that 
is cyclically
. Remark. The point is the tangent point of the mixtlinear -circle with the sideline , i.e. 
, where 
.
Proof 2 .
. Denote 
, where 
. Thus, 
.
PP4 (China). Let and the mixtilinear incircles 
which is tangent to , at , respectively 
which is tangent to , 
at
, respectively 
which is tangent to 
, 
at 
, respectively. Prove that ![$[RI_1Q]=[MI_2S]=[NI_3P]=\frac {r^2(2R+r)}s$](//latex.artofproblemsolving.com/d/9/1/d912e670b12e530b066783761c926a92fb4de6bf.png)
, where .
Proof.
and 
Thus, 
and ![$2[RAQ]=$](//latex.artofproblemsolving.com/9/1/0/9106fde3fe755263191a6cd442e05f2b27d58472.png)
![$\frac {bc\sin A\cdot (s-a)^2}{s^2}\ \stackrel{bc\sin A=2S=2sr}{\implies}\ \boxed{[RAQ]=\frac {r(s-a)^2}s}\ (2)\ .$](//latex.artofproblemsolving.com/2/e/1/2e14632221ab6cf46530a25228f48a03d9c23068.png)
Therefore, ![$2[RI_1Q]=2([RI_1A]+[QI_1A]-[RAQ])=$](//latex.artofproblemsolving.com/8/8/3/88367521c1ead16d2dd43279c91a6ca75fc1fff5.png)
![$r_1(AR+AQ)-2[RAQ]=$](//latex.artofproblemsolving.com/7/b/5/7b58f6834416f374d36d668e067b408c8affae6f.png)
![$\boxed{2[RI_1Q]=\frac r{s^2}\cdot \left[bc(b+c)-2s(s-a)^2\right]}\ (3)\ .$](//latex.artofproblemsolving.com/6/3/8/638b4612dae723ee61028b54d39b379ad078b365.png)
From 
![$bc[a+(b+c-a)]-2s(s-a)^2=$](//latex.artofproblemsolving.com/2/e/5/2e5826d49c3e9b758c5ac8768d432777c0c9765f.png)
![$abc+2(s-a)[\underline{bc-s(s-a)}]=$](//latex.artofproblemsolving.com/3/1/a/31a36714214fc645345804f7ab4be2eb175ccb9d.png)
obtain that
I used the well-known identity 
In conclusion, using the relation 
,
the relation
becomes ![$[RI_1Q]=\frac r{s^2}\cdot sr(2R+r)$](//latex.artofproblemsolving.com/c/8/e/c8e956aef129b971fa06cf12ae9d9b33fd169061.png)
, i.e. ![$[RI_1Q]=\frac {r^2(2R+r)}s$](//latex.artofproblemsolving.com/9/6/f/96f60cf38ff5fa113ef128ef653aefe97a903855.png)
what is symmetricaly w.r.t. . Hence ![$[RI_1Q]=[MI_2S]=[NI_3P]=\frac {r^2(2R+r)}s\ .$](//latex.artofproblemsolving.com/8/0/1/8013b876af2413149a43b13e0333a1a503514b06.png)
be the incircle and the 
-exincircle of 
which touches 
at 
, 
respectively. Define and note the circles 
and
the circles which are tangent to 
in 
, 
respectively and to the circumcircle of (the first -interior, the second-exterior) at 
,
respectively.Prove that
; 
; 
; 
; 
, 
and the bisector of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
are concurrently.Remark. Let
be a triangle with incenter 
and the circle 
internally tangent to , and to the circumcircle of given triangle.Denote the contact points
, 
of circle with , respectively. Show that , , 
are collinear and 
.PP1. Let
be a triangle -right-angled and 
be its circumcircle. Let 
be the circle touching the lines and , and the circle internally.Further, let
be the circle touching the lines and and the circle externally. If 
, 
be the radii of , 
prove that ![$r_1 \cdot r_2 = 4\cdot [ABC]$](http://latex.artofproblemsolving.com/d/3/0/d305df44a65706f774f19cdf0a7b6db11c773b68.png)
.Proof.
. From upper problem obtain : ![$\frac{r_1}{r}=\frac{r_2}{r_a}=\frac{bc}{s(s-a)}\Longrightarrow r_1\cdot r_2=rr_a\cdot \left[ \frac{bc}{s(s-a)}\right]^2 =4S$](http://latex.artofproblemsolving.com/0/7/f/07f56a355c19fd730fe1b92b3eee01976ad3b6cd.png)
.PP2. Let
, the circumcircle 
and the circle 
which touches , at , and which is internal tangent to
at . Let 
. Prove that 
.Proof. It's known that the incenter
of is the midpoint of 
(see here). Let 
, 
cut 
again at 
, 
. Prove easily
is the bisector of 
and 
is the bisector of 
, 
belong to . Apply the Pascal'stheorem to the cyclic hexagon
.PP3 (Miguel Ocho Sanchez). The mixtlinear circle
of touches , and the circumcircle at
, and respectively. Denote the projections 
, 
of on 
, 
respectively. Prove 
. Proof. Is well-known that the incenter
is the midpoint of ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
. Denote the midpoint 
of and the second intersection of 
with . From the previousproblem results that
. Observe that 
, what means that 
.PP4 (Ruben Dario). Let
with the incircle 
and the -excircle 
. Let 
and 
so that 
. Prove 
.Proof 1. Denote
, i.e. 
. Prove easily that 
. Hence 
.Since
obtain that 
is cyclically 
. Remark. The point is the tangent point of the mixtlinear -circle with the sideline , i.e. 
, where 
.Proof 2 .
. Denote 
, where 
. Thus, 
.PP4 (China). Let
and the mixtilinear incircles 
which is tangent to , at , respectively 
which is tangent to , 
at , respectively 
which is tangent to 
, 
at 
, respectively. Prove that ![$[RI_1Q]=[MI_2S]=[NI_3P]=\frac {r^2(2R+r)}s$](http://latex.artofproblemsolving.com/d/9/1/d912e670b12e530b066783761c926a92fb4de6bf.png)
, where . Proof.
and 
Thus, 
and ![$2[RAQ]=$](http://latex.artofproblemsolving.com/9/1/0/9106fde3fe755263191a6cd442e05f2b27d58472.png)
![$\frac {bc\sin A\cdot (s-a)^2}{s^2}\ \stackrel{bc\sin A=2S=2sr}{\implies}\ \boxed{[RAQ]=\frac {r(s-a)^2}s}\ (2)\ .$](http://latex.artofproblemsolving.com/2/e/1/2e14632221ab6cf46530a25228f48a03d9c23068.png)
Therefore, ![$2[RI_1Q]=2([RI_1A]+[QI_1A]-[RAQ])=$](http://latex.artofproblemsolving.com/8/8/3/88367521c1ead16d2dd43279c91a6ca75fc1fff5.png)
![$r_1(AR+AQ)-2[RAQ]=$](http://latex.artofproblemsolving.com/7/b/5/7b58f6834416f374d36d668e067b408c8affae6f.png)
![$\boxed{2[RI_1Q]=\frac r{s^2}\cdot \left[bc(b+c)-2s(s-a)^2\right]}\ (3)\ .$](http://latex.artofproblemsolving.com/6/3/8/638b4612dae723ee61028b54d39b379ad078b365.png)
From 
![$bc[a+(b+c-a)]-2s(s-a)^2=$](http://latex.artofproblemsolving.com/2/e/5/2e5826d49c3e9b758c5ac8768d432777c0c9765f.png)
![$abc+2(s-a)[\underline{bc-s(s-a)}]=$](http://latex.artofproblemsolving.com/3/1/a/31a36714214fc645345804f7ab4be2eb175ccb9d.png)
obtain that
I used the well-known identity 
In conclusion, using the relation 
,the relation
becomes ![$[RI_1Q]=\frac r{s^2}\cdot sr(2R+r)$](http://latex.artofproblemsolving.com/c/8/e/c8e956aef129b971fa06cf12ae9d9b33fd169061.png)
, i.e. ![$[RI_1Q]=\frac {r^2(2R+r)}s$](http://latex.artofproblemsolving.com/9/6/f/96f60cf38ff5fa113ef128ef653aefe97a903855.png)
what is symmetricaly w.r.t. . Hence ![$[RI_1Q]=[MI_2S]=[NI_3P]=\frac {r^2(2R+r)}s\ .$](http://latex.artofproblemsolving.com/8/0/1/8013b876af2413149a43b13e0333a1a503514b06.png)
This post has been edited 80 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:36 PM
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