Emelyanov L., Emelyanova T.
Around of Feet of Bisectors
Introduction
Solutions
1. Let I 2 be the B-excenter (fig. 1a). Let us consider the circle with diameter II 2 . The vertices
A and C lie on this circle, therefore its center lies on the perpendicular bisector of AC which
intersects the diameter II 2 at the midarc B 0 of AC of  . Hence B 0 is equidistant from A ,
C , I and I 2 .
Let I 1 and I 3 be the A-excenter and C-excenter (fig. 1b). Let us consider the circle with diameter I 1 I 3 . The vertices A and C lie on this circle, therefore its center lies on the perpendicular
bisector of AC which intersects the diameter I 1 I 3 at the midarc B0 of AC of  containing
B. Hence B0 is equidistant from A , C , I 1 and I 3 .
2. Let C  be the touch point of the incircle and AB (fig. 2a). Power of I
BIC
B0CB0 are
and
IO 2  R 2   BI  IB0 . The triangles
BI / IC   B0 B0 / B0 C . From problem 1 it follows that
BI  B0 I  B0 B0  IC   2 R  r . Therefore IO 2  R 2  2 R  r , i.e. IO 2 
with respect to  is
similar, therefore
B0 C  B0 I , hence
R2  2R  r .
Let C  be the touch point of the excircle  2 and AB (fig. 2b). Power of I 2 with respect to  is
I 2 O 2  R 2  I 2 B  I 2 B0 . The triangles BI 2 C  and B0CB0 are similar, therefore
BI 2 / I 2 C   B0 B0 / B0 C . From problem 1 it follows that B0 C  B0 I 2 , hence
BI 2  B0 I 2  B0 B0  I 2 C   2 R  r . Therefore I 2 O 2  R 2  2 R  r , i.e. I 2 O 2  R 2  2 R  r .
3. Let   (O, R) be the circumcircle and   ( I , r ) be the incircle of some triangle. From problem 2 it follows that IO 2  R 2  2 R  r . Take an arbitrary point on  , denote it B and draw
the chords BA and BC tangent to  (fig. 3). From similarity of the triangles BIC and
B0CB0 it follows that B0 C / 2 R  r / BI , i.e. 2 R  r  BI  B0 C . From the Euler formula it follows that power of I with respect to  is  2 R  r   BI  IB0 . Therefore BI  B0 C  BI  IB0 ,
B0 IC  ICB0 , but B0 IC  B / 2  ICB ,
it means that in the triangle B0 CI
ICB 0  B / 2  ICA . We obtain that IBC  ICA . It means that the lines AC and
BC are symmetric with respect to CI , therefor AC is tangent to  .
4. Consider the circles   (O, R) and  2  ( I , r2 ) , which are the circumcircle and the excircle
of some triangle. From problem 2 it yields that I 2 O 2  R 2  2 R  r2 . Take any point B in 
and let the lines BA and BC be tangents to  2 (fig. 4). As the triangles BI 2 C  and B0CB0
are similar B0 C / 2 R  r2 / BI 2 , i.e. 2 R  r2  BI 2  B0 C , but by Euler formula the degree of
point I 2 with respect to  is equal to 2 R  r2  BI 2  I 2 B0 . So, BI 2  B0 C  BI 2  I 2 B0 . This
follows that the triangle B0 CI 2 is isoscelles and B0 I 2 C  I 2 CB0 , but
B0 IC  I 2 CB  BB0 C  A ,
I 2 CB0  A / 2 .
i.e
We
obnain
that
I 2 CA  A / 2  B0 CA  (A  B) / 2 . It means that the line I 2 C is the external bisector
of angle B , therefor AC is tangent to  2 .
5. Firstly prove that the orthocentric axe is the radical axe of the circumcircle and the nine point
circle. Consider two circles:  B with diameter AC and  B with diameter HB (fig. 5). The
sideline H1 H 3 of orthotriangle is its common chord so lies in its radical axe. Therefor
H 2 H 3  H 2 H1  H 2 A  H 2 C . Now consider the circumcircle  and the nine point circle  0 .
The degrees of point H 2 with respect to  and  0 are equal to H 2 A  H 2 C and
H 2 H 3  H 2 H1 respectively, i.e the degrees of the common point of respective sidelines of the
triangle and its orthotriangle with respect to  and  0 are equal. This follows that the orthocentric axe is the radical axe of the circumcircle and nine point circle so it is perpendicular to
the Euler line.
Consider now the triangle I 1 I 2 I 3 formed by three excenters. Original triangle ABC is its orthotriangle, and the point I is its orthocenter. So the common points of external bisectors of
the triangle ABC with respecive sidelines lie in the orthocentric axe of the triangle I 1 I 2 I 3 i.e
in the line perpendicular to the Euler line of this triangle. But the Euler line of the triangle
I 1 I 2 I 3 pass through its orthocenter ( I ) and nine point center (O) , therefof it coincide with
the line IO .
6. Firstly consider next problem: given two circles 1  (O1 , R1 ) and  2  (O2 , R2 ) , their radical
axe and center line intersect in the point P (fig. 6). Find the lenght of the segment PO1 .
As the degrees of P with respect to both circles are equal, PO12  R12  PO22  R22 ,
PO22  PO12  R22  R12 , O1O2  (2 PO1  O1O2 )  R22  R12 . So it is easy to express PO1
through the radius of the circles and the distance O1O2..
Now take the circumcircle of the triangle ABC with radius R as  1 , and the circle ( I 1 I 2 I 3 )
with radius 2R as  2 . Then the distance d 1 from the circumcenter O to radical axe ( ) is
equal
to
d  d1  IO 
7.
R 2  Rr
.
Therefor
R 2  2 Rr
R 2  Rr
 R 2  2 Rr 
2
R  2 Rr
the
required
3Rr
R 2  2 Rr
distance
is
equal
to
.
The solution is analogously to the solution of the problem 5 with replacing of the triangle
I1 I 2 I 3 to the triangles II 2 I 3 , I 1 II 3 , I 1 I 2 I . The circumcircle () is the common nine-point
circle of аll these triangles and the lines I k O are the Euler lines of respective triangles. So
the internal bisecors axis of the triangle ABC are the the radical axis of  and the circumcircles of the respective triangles.
8.
Let  and  2 be the circumcircle and the excircle of the triangle ABC (fig. 8). Let D be
the touching point of its common external tangent with  . There are two limit “triangles”
in the family of triangles with  and  2 as the circumcircle and the excircle. Consider a
case when the secant AB becomes tangent. This will be the common external tangent to
 and  2 . Then the points A and B coincide in the point D , and the lines BC and AC
coincide in the tangent PD .
Consider now the circles  and  2 , such that the tangent to  2 in its common point P
r1
 t . As
passes through the point D . Then DK 2  (r2  R) 2  OI 22 , tg ( I 2 DK ) =
DK
2  r2  DK
2t
PDK  2  I 2 DK we obnain sin( PDK ) 
. The lenght of chord
2 
1 t
DK 2  r22
DP of circle  is equal to DP  2 R  sin(DPK ) . But as DP and DK are the tangents
2r2  DK
2
2
to  2 , DP  DK . So 2 R
2
2  DP , 4 R  r2  DK  r2 . Using the expression for
DK  r2
DK 2 we obtain the Euler formula I 2 O 2  R 2  2 Rr2 .
9.
As the circles  and  are fixed, and by problem 6 the distance from the center of  to
the external bisectors axe can be expressed through the radius of these circles, we obtain
that all feet of external bisectors lies in the fixed line.
Inversely. Let A2 be an arbitrary point of this line. Let B and C be the common points of
tangent to  passing through A2 with  . By Poncelet theorem the sideline BC generate
the triangle ABC which have A2 as the foot of the external bisector.
10.
As the circles  and  1 are fixed, and by problem 8 the internal bisectors axe  1 passes
through the touching points of of common external tangents of these circles,  1 is the fixed
line.
Inversely. Let B1 be an arbitrary point in the segment PQ . Let A and C be the common
points of tangent to  1 passing through B1 with  . By external Poncelet theorem the sideline AC generate the triangle ABC which have B1 as the foot of the internal bisector.
11.
Let be R  r2 . By problem 8 this is eqivalent that the common external tangents to  and
 2 are parallel to the line OI 2 . So DE is the diameter of the circumcenter i.e. O  A1C1 .
12. Let   be the circle with IB0 as a diameter. Let   be the circle with II 2 as a diameter
(fig.12). The line B0 B2 is the radical axe of   and  . The line CB2 is the radical axe of
  and  . So the line IB2 is the radical axe of   and   . Let K be the second common
point of these circles. As IKB0  IKI 2  90 , the point K lies in the line B0 I 2 , and therefor B0 I 2  B2 I .