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Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch BxMO/EGMO TST
2023 Dutch BxMO TST
4
4
Part of
2023 Dutch BxMO TST
Problems
(1)
Reciprocal geometric equation with excenter and circumcenter
Source: 2022 Turkey JBMO TST P7 + 2023 Dutch BxMO TST, Problem 4
3/15/2022
In a triangle
△
A
B
C
\triangle ABC
△
A
BC
with
∠
A
B
C
<
∠
B
C
A
\angle ABC < \angle BCA
∠
A
BC
<
∠
BC
A
, we define
K
K
K
as the excenter with respect to
A
A
A
. The lines
A
K
AK
A
K
and
B
C
BC
BC
intersect in a point
D
D
D
. Let
E
E
E
be the circumcenter of
△
B
K
C
\triangle BKC
△
B
K
C
. Prove that
1
∣
K
A
∣
=
1
∣
K
D
∣
+
1
∣
K
E
∣
.
\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.
∣
K
A
∣
1
=
∣
KD
∣
1
+
∣
K
E
∣
1
.
geometry
circumcircle