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Kyiv City MO
Kyiv City MO - geometry
Kyiv City MO Juniors 2003+ geometry
2018.7.4
2018.7.4
Part of
Kyiv City MO Juniors 2003+ geometry
Problems
(1)
right angle wanted, <APC=180^o-<ABC,BC = AP, AK=KB+PC (2018 Kyiv City MO 7.4)
Source:
9/11/2020
Inside the triangle
A
B
C
ABC
A
BC
, the point
P
P
P
is selected so that
B
C
=
A
P
BC = AP
BC
=
A
P
and
∠
A
P
C
=
180
∘
−
∠
A
B
C
\angle APC = 180 {} ^ \circ - \angle ABC
∠
A
PC
=
180
∘
−
∠
A
BC
. On the side
A
B
AB
A
B
there is a point
K
K
K
, for which
A
K
=
K
B
+
P
C
AK = KB + PC
A
K
=
K
B
+
PC
. Prove that
∠
A
K
C
=
90
∘
\angle AKC = 90 {} ^ \circ
∠
A
K
C
=
90
∘
.(Danilo Hilko)
geometry
right angle
equal segments
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