167. Nice problems - OIM, 1995 and ... Toshio Seimyia.
by Virgil Nicula, Oct 26, 2010, 2:39 PM
OIM - 1997. Given are two parallel lines 
, 
, a circle 
which is tangent to these lines, a circle 
which is exterior tangent to 
and is
tangent to , a circle 
which is exterior tangent to and is tangent to so that the circles 
, 
are exterior tangent one to other. Denote
, 
, 
, 
, 
and 
. Prove that the point 
is the circumcenter of the triangle 
.
Proof 1 (synthetic).
Lemma 1 (PLC). Let , be the circles which are exterior tangent in 
and let 
be a exterior common tangent line of these in 
and 
. Let the diameters ![$[A_1B_1]$](//latex.artofproblemsolving.com/9/5/7/95797ca7706aab870b4043e5c14a8c81ca149120.png)
and
![$[A_2B_2]$](//latex.artofproblemsolving.com/4/9/a/49a944abec599ee77a5879345f6afd5d9afdc6c2.png)
of , respectively. Prove that 
(construct of a circle which pass through a given point 
, is tangent to a given line 
and is tangent to a given circle 
).
Lemma 2 (PCC). Let
, 
be two exterior circles and let be a exterior common tangent line of , in , . Denote 
.
Let a circle which is exterior tangent to , in 
, 
. Prove that 
and 
is cyclically (construct of a circle which pass
through a given point and is tangent to 
and 
). See the XCIII-message of my blog.
From lemma 1 results that
and 
. From lemma 2 obtain that the quadrilaterals 
and 
are cyclically. Thus,
is cyclically and 
the point 
has same power w.r.t. , the points , belong
to the radical axis of , the line 
is common tangent of , in . Prove analogously that the line 
is common tangent of , in 
.
In conclusion, the point is radical center of , , and 
, i.e. the point is circumcenter of 
.
Proof 2 (of one from my students). Denote
, 
and 
for which 
. Consider the line 
for which 
.
Denote the projections
, 
, 
on the line 
of 
, , respectively and 
, 
. Therefore, 
, 
,
, 
, 
, 
and 
. Thus, 
. Therefore, 
and 
. From where obtain
![$\frac {4r}{(r_1+r_2)^2}\cdot \left\{r^3+\left[\sqrt {r_1}(r+r_2)-r\sqrt {r_2}\right]^2\right\}=$](//latex.artofproblemsolving.com/4/b/d/4bdb94dcae8e485f5f4009ac21106044a6a0a29b.png)
![$\frac {4r}{(r_1+r_2)^2}\cdot \left[r^3+r_1(r+r_2)^2+r^2r_2-2r(r+r_2)\sqrt {r_1r_2}\right]$](//latex.artofproblemsolving.com/9/f/f/9ffec83d6e566d4a9715b1525fa240f715c0634e.png)
. Using the relation 
obtain
![$AD^2=\frac {4r}{(r_1+r_2)^2}\cdot \left[r^3+r_1(r+r_2)^2+r^2r_2-r^2(r+r_2)\right]=4rr_1=AF_1^2$](//latex.artofproblemsolving.com/8/1/d/81d77f90620b9f47b05c77308f45ecd7d34eb553.png)
. In conclusion, 
, i.e. the line is a common tangent
for , . Prove analogously that the line is a common tangent for , . Thus, is the radical center for the circles , ,
and in this case , i.e. the point is the circumcenter of the triangle .
Generalization. Let
be a triangle. Consider three circles 
which are situated inside of 
so that each from them is exterior tangent to others two
and is tangent to only two sides of given triangle. More exactly : is tangent to 
, 
in , 
respectively ; is tangent to , 
in 
, respectively ; 
is tangent to 
,
in 
, 
respectively. Denote 
, 
, 
and the diameters ![$[A_1A_1']$](//latex.artofproblemsolving.com/0/d/1/0d1b604ec2403a0ad32d6c3955aa39c3a9f714cb.png)
, ![$[A_2A_2']$](//latex.artofproblemsolving.com/c/5/d/c5d5de2b1eb8194c83c8b09da62213b141907107.png)
, ![$[B_1B_1']$](//latex.artofproblemsolving.com/3/4/1/341d212bddd2e37e0e01693f38022880511ef741.png)
, ![$[B_2B_2']$](//latex.artofproblemsolving.com/a/4/c/a4cedea307e1096e58ecc66740b5ef4c7c02b4e0.png)
, ![$[C_1C_1']$](//latex.artofproblemsolving.com/a/9/1/a91efdca18f2551a58910ffb2592823b1c383419.png)
, ![$[C_2C_2']$](//latex.artofproblemsolving.com/d/0/3/d0359ed55badb214ffbf4deaeead33a970d3e2f4.png)
in the corresponding circles.
Define the intersections
, 
, 
. Prove that :
, 
,
.
The quadrilaterals 
, 
, 
are cyclically.
The lines 
, 
, 
are concurrently in the circumcenter of the triangle 
.
Toshio Seimyia. Let be a triangle with the incircle 
, with the Lemoine's point 
and 
. Denote 
.
Prove that
.
Proof 1 (synthetic). Denote
, 
, 
. From the well-known property 
obtain 
.
Since , are conjugate harmonical w.r.t. , obtain that 
is the polar line of w.r.t. the circumcircle 
of 
,
i.e.
is tangent line to in . From another well-known property obtain that 
the point is the foot of the -symmedian in .
Proof 2 (metric). Recall the relations
, 
and 
, where 
. The line is -symmedian
. Therefore, 
. Since
, where 
is the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. Thus, 
, i.e. the relation .
In conclusion, is the -symmedian in .
Remark.
, where 
and 
.
, 
, a circle 
which is tangent to these lines, a circle 
which is exterior tangent to 
and istangent to
, a circle 
which is exterior tangent to and is tangent to so that the circles 
, 
are exterior tangent one to other. Denote
, 
, 
, 
, 
and 
. Prove that the point 
is the circumcenter of the triangle 
.Proof 1 (synthetic).
Lemma 1 (PLC). Let
, be the circles which are exterior tangent in 
and let 
be a exterior common tangent line of these in 
and 
. Let the diameters ![$[A_1B_1]$](http://latex.artofproblemsolving.com/9/5/7/95797ca7706aab870b4043e5c14a8c81ca149120.png)
and![$[A_2B_2]$](http://latex.artofproblemsolving.com/4/9/a/49a944abec599ee77a5879345f6afd5d9afdc6c2.png)
of , respectively. Prove that 
(construct of a circle which pass through a given point 
, is tangent to a given line 
and is tangent to a given circle 
).Lemma 2 (PCC). Let
, 
be two exterior circles and let be a exterior common tangent line of , in , . Denote 
.Let a circle
which is exterior tangent to , in 
, 
. Prove that 
and 
is cyclically (construct of a circle which passthrough a given point
and is tangent to 
and 
). See the XCIII-message of my blog.From lemma 1 results that
and 
. From lemma 2 obtain that the quadrilaterals 
and 
are cyclically. Thus, is cyclically and 
the point 
has same power w.r.t. , the points , belongto the radical axis of
, the line 
is common tangent of , in . Prove analogously that the line 
is common tangent of , in 
.In conclusion, the point
is radical center of , , and 
, i.e. the point is circumcenter of 
. Proof 2 (of one from my students). Denote
, 
and 
for which 
. Consider the line 
for which 
.Denote the projections
, 
, 
on the line 
of 
, , respectively and 
, 
. Therefore, 
, 
,
, 
, 
, 
and 
. Thus, 
. Therefore, 
and 
. From where obtain
![$\frac {4r}{(r_1+r_2)^2}\cdot \left\{r^3+\left[\sqrt {r_1}(r+r_2)-r\sqrt {r_2}\right]^2\right\}=$](http://latex.artofproblemsolving.com/4/b/d/4bdb94dcae8e485f5f4009ac21106044a6a0a29b.png)
![$\frac {4r}{(r_1+r_2)^2}\cdot \left[r^3+r_1(r+r_2)^2+r^2r_2-2r(r+r_2)\sqrt {r_1r_2}\right]$](http://latex.artofproblemsolving.com/9/f/f/9ffec83d6e566d4a9715b1525fa240f715c0634e.png)
. Using the relation 
obtain![$AD^2=\frac {4r}{(r_1+r_2)^2}\cdot \left[r^3+r_1(r+r_2)^2+r^2r_2-r^2(r+r_2)\right]=4rr_1=AF_1^2$](http://latex.artofproblemsolving.com/8/1/d/81d77f90620b9f47b05c77308f45ecd7d34eb553.png)
. In conclusion, 
, i.e. the line is a common tangentfor
, . Prove analogously that the line is a common tangent for , . Thus, is the radical center for the circles , , and in this case
, i.e. the point is the circumcenter of the triangle .Generalization. Let
be a triangle. Consider three circles 
which are situated inside of 
so that each from them is exterior tangent to others twoand is tangent to only two sides of given triangle. More exactly :
is tangent to 
, 
in , 
respectively ; is tangent to , 
in 
, respectively ; 
is tangent to 
,
in 
, 
respectively. Denote 
, 
, 
and the diameters ![$[A_1A_1']$](http://latex.artofproblemsolving.com/0/d/1/0d1b604ec2403a0ad32d6c3955aa39c3a9f714cb.png)
, ![$[A_2A_2']$](http://latex.artofproblemsolving.com/c/5/d/c5d5de2b1eb8194c83c8b09da62213b141907107.png)
, ![$[B_1B_1']$](http://latex.artofproblemsolving.com/3/4/1/341d212bddd2e37e0e01693f38022880511ef741.png)
, ![$[B_2B_2']$](http://latex.artofproblemsolving.com/a/4/c/a4cedea307e1096e58ecc66740b5ef4c7c02b4e0.png)
, ![$[C_1C_1']$](http://latex.artofproblemsolving.com/a/9/1/a91efdca18f2551a58910ffb2592823b1c383419.png)
, ![$[C_2C_2']$](http://latex.artofproblemsolving.com/d/0/3/d0359ed55badb214ffbf4deaeead33a970d3e2f4.png)
in the corresponding circles.Define the intersections
, 
, 
. Prove that :
, 
,
.
The quadrilaterals 
, 
, 
are cyclically. The lines 
, 
, 
are concurrently in the circumcenter of the triangle 
.Toshio Seimyia. Let
be a triangle with the incircle 
, with the Lemoine's point 
and 
. Denote 
.Prove that
.Proof 1 (synthetic). Denote
, 
, 
. From the well-known property 
obtain 
.Since
, are conjugate harmonical w.r.t. , obtain that 
is the polar line of w.r.t. the circumcircle 
of 
,i.e.
is tangent line to in . From another well-known property obtain that 
the point is the foot of the -symmedian in .Proof 2 (metric). Recall the relations
, 
and 
, where 
. The line is -symmedian 
. Therefore, 
. Since
, where 
is the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. Thus, 
, i.e. the relation .In conclusion,
is the -symmedian in .Remark.
, where 
and 
.This post has been edited 55 times. Last edited by Virgil Nicula, Dec 1, 2015, 9:45 AM
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