MathDB
Problems
Contests
National and Regional Contests
Ukraine Contests
Official Ukraine Selection Cycle
Kyiv City MO
Kyiv City MO - geometry
Kyiv City MO Seniors Round2 2010+ geometry
2020.10.2
2020.10.2
Part of
Kyiv City MO Seniors Round2 2010+ geometry
Problems
(1)
<PBM+<CBM=<PCA,<BM=90^o, <ABC+<APC=180^o (2020 Kyiv City MO Round2 10.2)
Source:
9/21/2020
Let
M
M
M
be the midpoint of the side
A
C
AC
A
C
of triangle
A
B
C
ABC
A
BC
. Inside
△
B
M
C
\vartriangle BMC
△
BMC
was found a point
P
P
P
such that
∠
B
M
P
=
9
0
o
\angle BMP = 90^o
∠
BMP
=
9
0
o
,
∠
A
B
C
+
∠
A
P
C
=
18
0
o
\angle ABC+ \angle APC =180^o
∠
A
BC
+
∠
A
PC
=
18
0
o
. Prove that
∠
P
B
M
+
∠
C
B
M
=
∠
P
C
A
\angle PBM + \angle CBM = \angle PCA
∠
PBM
+
∠
CBM
=
∠
PC
A
.(Anton Trygub)
geometry
angles
right ange