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KoMaL A Problems
KoMaL A Problems 2017/2018
A. 726
A. 726
Part of
KoMaL A Problems 2017/2018
Problems
(1)
A and P equidistant from incenter
Source: KöMaL A 726
6/12/2018
In triangle
A
B
C
ABC
A
BC
with incenter
I
I
I
, line
A
I
AI
A
I
intersects the circumcircle of
A
B
C
ABC
A
BC
at
S
≠
A
S\ne A
S
=
A
. Let the reflection of
I
I
I
with respect to
B
C
BC
BC
be
J
J
J
, and suppose that line
S
J
SJ
S
J
intersects the circumcircle of
A
B
C
ABC
A
BC
for the second time at point
P
≠
S
P\ne S
P
=
S
. Show that
A
I
=
P
I
.
AI=PI.
A
I
=
P
I
.
József Mészáros
geometry
incenter