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Old Kyiv MO Geometry
Kyiv City MO Juniors Round2 2010+ geometry
Kyiv City MO Juniors Round2 2010+ geometry
Part of
Old Kyiv MO Geometry
Subcontests
(40)
2022.9.4
1
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Bisectors meeting on bisector
Let
ω
\omega
ω
denote the circumscribed circle of triangle
A
B
C
ABC
A
BC
,
I
I
I
be its incenter, and
K
K
K
be any point on arc
A
C
AC
A
C
of
ω
\omega
ω
not containing
B
B
B
. Point
P
P
P
is symmetric to
I
I
I
with respect to point
K
K
K
. Point
T
T
T
on arc
A
C
AC
A
C
of
ω
\omega
ω
containing point
B
B
B
is such that
∠
K
C
T
=
∠
P
C
I
\angle KCT = \angle PCI
∠
K
CT
=
∠
PC
I
. Show that the bisectors of angles
A
K
C
AKC
A
K
C
and
A
T
C
ATC
A
TC
meet on line
C
I
CI
C
I
. (Proposed by Anton Trygub)
2022.8.4
1
Hide problems
Geometry about the right triangle from the right person
Points
D
,
E
,
F
D, E, F
D
,
E
,
F
are selected on sides
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
correspondingly of triangle
A
B
C
ABC
A
BC
with
∠
C
=
9
0
∘
\angle C = 90^\circ
∠
C
=
9
0
∘
such that
∠
D
A
B
=
∠
C
B
E
\angle DAB = \angle CBE
∠
D
A
B
=
∠
CBE
and
∠
B
E
C
=
∠
A
E
F
\angle BEC = \angle AEF
∠
BEC
=
∠
A
EF
. Show that
D
B
=
D
F
DB = DF
D
B
=
D
F
.(Proposed by Mykhailo Shtandenko)
2022.7.3
1
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Geometry for people who don't know what the sum of angles of triangle is
In triangle
A
B
C
ABC
A
BC
the median
B
M
BM
BM
is equal to half of the side
B
C
BC
BC
. Show that
∠
A
B
M
=
∠
B
C
A
+
∠
B
A
C
\angle ABM = \angle BCA + \angle BAC
∠
A
BM
=
∠
BC
A
+
∠
B
A
C
.(Proposed by Anton Trygub)
2021.7.4
1
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equilateral? equal extensions, 6 points concyclic (2021 Kyiv City MO Round2 7.4)
The sides of the triangle
A
B
C
ABC
A
BC
are extended in both directions and on these extensions
6
6
6
equal segments
A
A
1
,
A
A
2
,
B
B
1
,
B
B
2
,
C
C
1
,
C
C
2
AA_1 , AA_2, BB_1,BB_2, CC_1, CC_2
A
A
1
,
A
A
2
,
B
B
1
,
B
B
2
,
C
C
1
,
C
C
2
are drawn (fig.). It turned out that all
6
6
6
points
A
1
,
A
2
,
B
1
,
B
2
,
C
1
,
C
2
A_1,A_2,B_1,B_2,C_1, C_2
A
1
,
A
2
,
B
1
,
B
2
,
C
1
,
C
2
lie on the same circle, is
△
A
B
C
\vartriangle ABC
△
A
BC
necessarily equilateral?(Bogdan Rublev) https://cdn.artofproblemsolving.com/attachments/0/3/a499f6e6d978ce63d2ab40460dc73b62882863.png
2021.7.41
1
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perimeter of ABC>2 OC, C in <AOB =90^o (2021 Kyiv City MO Round2 7.4.1)
Point
C
C
C
lies inside the right angle
A
O
B
AOB
A
OB
. Prove that the perimeter of triangle
A
B
C
ABC
A
BC
is greater than
2
O
C
2 OC
2
OC
.
2021.9.2
1
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<AQT= 90^o wanted, altitudes, circumcircle (2021 Kyiv City MO Round2 9.2)
In an acute triangle
A
B
AB
A
B
the heights
B
E
BE
BE
and
C
F
CF
CF
intersect at the orthocenter
H
H
H
, and
M
M
M
is the midpoint of
B
C
BC
BC
. The line
E
F
EF
EF
intersects the lines
M
H
MH
M
H
and
B
C
BC
BC
at the points
P
P
P
and
T
T
T
, respectively.
A
P
AP
A
P
intersects the cirumcscribed circle of
△
A
B
C
\vartriangle ABC
△
A
BC
for second time at the point
Q
Q
Q
. Prove that
∠
A
Q
T
=
9
0
o
\angle AQT= 90^o
∠
A
QT
=
9
0
o
.(Fedir Yudin)
2021.8.2
1
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3PI+IT=AC wanted, incenter in 90-60-30 (2021 Kyiv City MO Round2 8.2)
In a triangle
A
B
C
ABC
A
BC
,
∠
B
=
9
0
o
\angle B=90^o
∠
B
=
9
0
o
and
∠
A
=
6
0
o
\angle A=60^o
∠
A
=
6
0
o
,
I
I
I
is the point of intersection of its angle bisectors. A line passing through the point
I
I
I
parallel to the line
A
C
AC
A
C
, intersects the sides
A
B
AB
A
B
and
B
C
BC
BC
at the points
P
P
P
and
T
T
T
respectively. Prove that
3
P
I
+
I
T
=
A
C
3PI+IT=AC
3
P
I
+
I
T
=
A
C
.(Anton Trygub)
2020.9.2
1
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HX bisects AC, circumcircles, altitude (2020 Kyiv City MO Round2 9.2)
In the acute-angled triangle
A
B
C
ABC
A
BC
is drawn the altitude
C
H
CH
C
H
. A ray beginning at point
C
C
C
that lies inside the
∠
B
C
A
\angle BCA
∠
BC
A
and intersects for second time the circles circumscribed circles of
△
B
C
H
\vartriangle BCH
△
BC
H
and
△
A
B
C
\vartriangle ABC
△
A
BC
at points
X
X
X
and
Y
Y
Y
respectively. It turned out that
2
C
X
=
C
Y
2CX = CY
2
CX
=
C
Y
. Prove that the line
H
X
HX
H
X
bisects the segment
A
C
AC
A
C
.(Hilko Danilo)
2020.8.2
1
Hide problems
CA =CD wanted, <CBD = 90^o,<BCD =<CAD, AD=2BC (2020 Kyiv City MO Round2 8.2)
Given a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
, in which
∠
C
B
D
=
9
0
o
\angle CBD = 90^o
∠
CB
D
=
9
0
o
,
∠
B
C
D
=
∠
C
A
D
\angle BCD =\angle CAD
∠
BC
D
=
∠
C
A
D
and
A
D
=
2
B
C
AD= 2BC
A
D
=
2
BC
. Prove that
C
A
=
C
D
CA =CD
C
A
=
C
D
.(Anton Trygub)
2019.9.31
1
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r_1+r_2+r_3=r, indradii (2019 Kyiv City MO Round2 9.3.1)
A circle
k
k
k
of radius
r
r
r
is inscribed in
△
A
B
C
\vartriangle ABC
△
A
BC
, tangent to the circle
k
k
k
, which are parallel respectively to the sides
A
B
,
B
C
AB, BC
A
B
,
BC
and
C
A
CA
C
A
intersect the other sides of
△
A
B
C
\vartriangle ABC
△
A
BC
at points
M
,
N
;
P
,
Q
M, N; P, Q
M
,
N
;
P
,
Q
and
L
,
T
L, T
L
,
T
(
P
,
T
∈
A
B
P, T \in AB
P
,
T
∈
A
B
,
L
,
N
∈
B
C
L, N \in BC
L
,
N
∈
BC
and
M
,
Q
∈
A
C
M, Q\in AC
M
,
Q
∈
A
C
). Denote by
r
1
,
r
2
,
r
3
r_1,r_2,r_3
r
1
,
r
2
,
r
3
the radii of inscribed circles in triangles
M
N
C
,
P
Q
A
MNC, PQA
MNC
,
PQ
A
and
L
T
B
LTB
L
TB
. Prove that
r
1
+
r
2
+
r
3
=
r
r_1+r_2+r_3=r
r
1
+
r
2
+
r
3
=
r
.
2019.9.3
1
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fixed incenter, equilateral, <ABE+< ACF=60^o (2019 Kyiv City MO Round2 9.3)
The equilateral triangle
A
B
C
ABC
A
BC
is inscribed in the circle
w
w
w
. Points
F
F
F
and
E
E
E
on the sides
A
B
AB
A
B
and
A
C
AC
A
C
, respectively, are chosen such that
∠
A
B
E
+
∠
A
C
F
=
6
0
o
\angle ABE+ \angle ACF = 60^o
∠
A
BE
+
∠
A
CF
=
6
0
o
. The circumscribed circle of
△
A
F
E
\vartriangle AFE
△
A
FE
intersects the circle
w
w
w
at the point
D
D
D
for the second time. The rays
D
E
DE
D
E
and
D
F
DF
D
F
intersect the line
B
C
BC
BC
at the points
X
X
X
and
Y
Y
Y
, respectively. Prove that the center of the inscribed circle of
△
D
X
Y
\vartriangle DXY
△
D
X
Y
does not depend on the choice of points
F
F
F
and
E
E
E
.(Hilko Danilo)
2019.8.41
1
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AY = CZ wanted, circle and parallelogram (2019 Kyiv City MO Round2 8.4.1)
Through the vertices
A
,
B
A, B
A
,
B
of the parallelogram
A
B
C
D
ABCD
A
BC
D
passes a circle that intersects for the second time diagonals
B
D
BD
B
D
and
A
C
AC
A
C
at points
X
X
X
and
Y
Y
Y
, respectively. The circumsccribed circle of
△
A
D
X
\vartriangle ADX
△
A
D
X
intersects diagonal
A
C
AC
A
C
for the second time at the point
Z
Z
Z
. Prove that
A
Y
=
C
Z
AY = CZ
A
Y
=
CZ
.
2019.8.4
1
Hide problems
angle wanted, 75-60-45 triangle, BC=CT (2019 Kyiv City MO Round2 8.4)
In the triangle
A
B
C
ABC
A
BC
it is known that
∠
A
=
7
5
o
,
∠
C
=
4
5
o
\angle A = 75^o, \angle C = 45^o
∠
A
=
7
5
o
,
∠
C
=
4
5
o
. On the ray
B
C
BC
BC
beyond the point
C
C
C
the point
T
T
T
is taken so that
B
C
=
C
T
BC = CT
BC
=
CT
. Let
M
M
M
be the midpoint of the segment
A
T
AT
A
T
. Find the measure of the
∠
B
M
C
\angle BMC
∠
BMC
.(Anton Trygub)
2019.7.31
1
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O(0,0) construction given A (1, 2) ,B (3,1) (2019 Kyiv City MO Round2 7.3.1)
The teacher drew a coordinate plane on the board and marked some points on this plane. Unfortunately, Vasya's second-grader, who was on duty, erased almost the entire drawing, except for two points
A
(
1
,
2
)
A (1, 2)
A
(
1
,
2
)
and
B
(
3
,
1
)
B (3,1)
B
(
3
,
1
)
. Will the excellent Andriyko be able to follow these two points to construct the beginning of the coordinate system point
O
(
0
,
0
)
O (0, 0)
O
(
0
,
0
)
? Point A on the board located above and to the left of point
B
B
B
.
2019.7.3
1
Hide problems
|BC - BH|=HA wanted, <ABD= <DBC, AD= CD (2019 Kyiv City MO Round2 7.3)
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
it is known that
∠
A
B
D
=
∠
D
B
C
\angle ABD= \angle DBC
∠
A
B
D
=
∠
D
BC
and
A
D
=
C
D
AD= CD
A
D
=
C
D
. Let
D
H
DH
DH
be the altitude of
△
A
B
D
\vartriangle ABD
△
A
B
D
. Prove that
∣
B
C
−
B
H
∣
=
H
A
| BC - BH | = HA
∣
BC
−
B
H
∣
=
H
A
.(Hilko Danilo)
2018.9.1
1
Hide problems
cut a right triangle into 3 isosceles (2018 Kyiv City MO Round2 9.1)
Cut a right triangle with an angle of
3
0
o
30^o
3
0
o
into three isosceles non-acute triangles, among which there are no congruent ones.(Maria Rozhkova)
2018.8.31
1
Hide problems
MD = KD wanted, isosceles, AM = 2DC,<AMD =<KDC (2018 Kyiv City MO Round2 8.3.1)
On the sides
A
B
AB
A
B
,
B
C
BC
BC
and
C
A
CA
C
A
of the isosceles triangle
A
B
C
ABC
A
BC
with the vertex at the point
B
B
B
marked the points
M
M
M
,
D
D
D
and
K
K
K
respectively so that
A
M
=
2
D
C
AM = 2DC
A
M
=
2
D
C
and
∠
A
M
D
=
∠
K
D
C
\angle AMD = \angle KDC
∠
A
M
D
=
∠
KD
C
. Prove that
M
D
=
K
D
MD = KD
M
D
=
KD
.
2018.8.3
1
Hide problems
<CBA - <P_1T_1A=45^o,PT = BC,AP= CP_1, AT = BT_1 (2018 Kyiv City MO Round2 8.3)
In the triangle
A
B
C
ABC
A
BC
it is known that
∠
A
C
B
>
90
∘
\angle ACB> 90 {} ^ \circ
∠
A
CB
>
90
∘
,
∠
C
B
A
>
45
∘
\angle CBA> 45 {} ^ \circ
∠
CB
A
>
45
∘
. On the sides
A
C
AC
A
C
and
A
B
AB
A
B
, respectively, there are points
P
P
P
and
T
T
T
such that
A
B
C
ABC
A
BC
and
P
T
=
B
C
PT = BC
PT
=
BC
. The points
P
1
{{P} _ {1}}
P
1
and
T
1
{{T} _ {1}}
T
1
on the sides
A
C
AC
A
C
and
A
B
AB
A
B
are such that
A
P
=
C
P
1
AP = C {{P} _ {1}}
A
P
=
C
P
1
and
A
T
=
B
T
1
AT = B {{T} _ {1}}
A
T
=
B
T
1
. Prove that
∠
C
B
A
−
∠
P
1
T
1
A
=
45
∘
\angle CBA- \angle {{P} _ {1}} {{T} _ {1}} A = 45 {} ^ \circ
∠
CB
A
−
∠
P
1
T
1
A
=
45
∘
. (Anton Trygub)
2017.9.1
1
Hide problems
circumcenter and excenter symmetrc wrt side (2017 Kyiv City MO Round2 9.1)
Find the angles of the triangle
A
B
C
ABC
A
BC
, if we know that its center
O
O
O
of the circumscribed circle and the center
I
A
I_A
I
A
of the exscribed circle (tangent to
B
C
BC
BC
) are symmetric wrt
B
C
BC
BC
.(Bogdan Rublev)
2017.8.2
1
Hide problems
angle chasing, starting with right isosceles (2017 Kyiv City MO Round2 8.2)
Triangle
A
B
C
ABC
A
BC
is right-angled and isosceles with a right angle at the vertex
C
C
C
. On rays
C
B
CB
CB
on vertex
B
B
B
is selected point F, on rays
B
A
BA
B
A
on vertex
A
A
A
is selected point G so that
A
G
=
B
F
.
AG = BF.
A
G
=
BF
.
The ray
G
D
GD
G
D
is drawn so that it intersects with ray
A
C
AC
A
C
at point
D
D
D
with
∠
F
G
D
=
4
5
o
\angle FGD = 45^o
∠
FG
D
=
4
5
o
. Find
∠
F
D
G
\angle FDG
∠
F
D
G
.(Bogdan Rublev)
2017.7.41
1
Hide problems
isosceles criterion involving equal segments (2017 Kyiv City MO Round2 7.4.1)
Let
A
C
AC
A
C
be the largest side of the triangle
A
B
C
ABC
A
BC
. The point M is selected on the ray
A
C
AC
A
C
ray, and point
N
N
N
on ray
C
A
CA
C
A
such that
C
N
=
C
B
CN = CB
CN
=
CB
and
A
M
=
A
B
AM = AB
A
M
=
A
B
. a) Prove that
△
A
B
C
\vartriangle ABC
△
A
BC
is isosceles if we know that
B
M
=
B
N
BM = BN
BM
=
BN
. b) Will the statement remain true if
A
C
AC
A
C
is not necessarily the largest side of triangle
A
B
C
ABC
A
BC
?
2017.7.4
1
Hide problems
perimeter ineq. in rectangle, AM=MN=ND=BP=PQ=QC (2017 Kyiv City MO Round2 7.4)
On the sides
A
D
AD
A
D
and
B
C
BC
BC
of a rectangle
A
B
C
D
ABCD
A
BC
D
select points
M
,
N
M, N
M
,
N
and
P
,
Q
P, Q
P
,
Q
respectively such that
A
M
=
M
N
=
N
D
=
B
P
=
P
Q
=
Q
C
AM = MN = ND = BP = PQ = QC
A
M
=
MN
=
N
D
=
BP
=
PQ
=
QC
. On segment
Q
C
QC
QC
selected point
X
X
X
, different from the ends of the segment. Prove that the perimeter of
△
A
N
X
\vartriangle ANX
△
A
NX
is more than the perimeter of
△
M
D
X
\vartriangle MDX
△
M
D
X
.
2016.9.2
1
Hide problems
BN + MC = AW wanted, 3 circumcircles (2016 Kyiv City MO Round2 9.2 )
The bisector of the angle
B
A
C
BAC
B
A
C
of the acute triangle
A
B
C
ABC
A
BC
(
A
C
≠
A
B
AC \ne AB
A
C
=
A
B
) intersects its circumscribed circle for the second time at the point
W
W
W
. Let
O
O
O
be the center of the circumscribed circle
Δ
A
B
C
\Delta ABC
Δ
A
BC
. The line
A
W
AW
A
W
intersects for the second time the circumcribed circles of triangles
O
W
B
OWB
O
W
B
and
O
W
C
OWC
O
W
C
at the points
N
N
N
and
M
M
M
, respectively. Prove that
B
N
+
M
C
=
A
W
BN + MC = AW
BN
+
MC
=
A
W
.(Mitrofanov V., Hilko D.)
2016.8.1
1
Hide problems
ratio of radii, right triangle related (2016 Kyiv City MO Round2 8.1)
In a right triangle, the point
O
O
O
is the center of the circumcircle. Another circle of smaller radius centered at the point
O
O
O
touches the larger leg and the altitude drawn from the top of the right angle. Find the acute angles of a right triangle and the ratio of the radii of the circumscribed and smaller circles.
2016.7.3
1
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angle wanted, concurrency given (2016 Kyiv City MO Round2 7.3 )
In an acute triangle
A
B
C
ABC
A
BC
, the bisector
A
L
AL
A
L
, the altitude
B
H
BH
B
H
, and the perpendicular bisector of the side
A
B
AB
A
B
intersect at one point. Find the value of the angle
B
A
C
BAC
B
A
C
.
2015.9.4
1
Hide problems
collinearities wanted, 3 circles related (2015 Kyiv City MO Round2 9.4 10.2)
Circles
w
1
{{w} _ {1}}
w
1
and
w
2
{{w} _ {2}}
w
2
with centers
O
1
{{O} _ {1}}
O
1
and
O
2
{{O} _ {2}}
O
2
intersect at points
A
A
A
and
B
B
B
, respectively. The line
O
1
O
2
{{O} _ {1}} {{O} _ {2}}
O
1
O
2
intersects
w
1
{{w} _ {1}}
w
1
at the point
Q
Q
Q
, which does not lie inside the circle
w
2
{{w} _ {2}}
w
2
, and
w
2
{{w} _ {2}}
w
2
at the point
X
X
X
lying inside the circle
w
1
{{w} _ {1} }
w
1
. Around the triangle
O
1
A
X
{{O} _ {1}} AX
O
1
A
X
circumscribe a circle
w
3
{{w} _ {3}}
w
3
intersecting the circle
w
1
{{w} _ {1}}
w
1
for the second time in point
T
T
T
. The line
Q
T
QT
QT
intersects the circle
w
3
{{w} _ {3}}
w
3
at the point
K
K
K
, and the line
Q
B
QB
QB
intersects
w
2
{{w} _ {2}}
w
2
the second time at the point
H
H
H
. Prove that a) points
T
,
X
,
B
T, \, \, X, \, \, B
T
,
X
,
B
lie on one line; b) points
K
,
X
,
H
K, \, \, X, \, \, H
K
,
X
,
H
lie on one line.(Vadym Mitrofanov)
2015.8.41
1
Hide problems
equilateral inside inscribed in a triangle (2015 Kyiv City MO Round2 8.4.1)
On the sides
A
B
,
B
C
,
C
A
AB, \, \, BC, \, \, CA
A
B
,
BC
,
C
A
of the triangle
A
B
C
ABC
A
BC
the points
C
1
,
A
1
,
B
1
{{C} _ {1}}, \, \, {{A} _ { 1}},\, \, {{B} _ {1}}
C
1
,
A
1
,
B
1
are selected respectively, that are different from the vertices. It turned out that
Δ
A
1
B
1
C
1
\Delta {{A} _ {1}} {{B} _ {1}} {{C} _ {1}}
Δ
A
1
B
1
C
1
is equilateral,
∠
B
C
1
A
1
=
∠
C
1
B
1
A
\angle B{{C}_{1}}{{A}_{1}}=\angle {{C}_{1}}{{B}_{1}}A
∠
B
C
1
A
1
=
∠
C
1
B
1
A
and
∠
B
A
1
C
1
=
∠
A
1
B
1
C
\angle B{{A}_{1}}{{C}_{1}}=\angle {{A}_{1}}{{B}_{1}}C
∠
B
A
1
C
1
=
∠
A
1
B
1
C
. Is
Δ
A
B
C
\Delta ABC
Δ
A
BC
equilateral?
2015.7.41
1
Hide problems
DM = MB wanted, AO: OB = CO: OD = 1: 2 (2015 Kyiv City MO Round2 7.4.1)
The equal segments
A
B
AB
A
B
and
C
D
CD
C
D
intersect at the point
O
O
O
and divide it by the relation
A
O
:
O
B
=
C
O
:
O
D
=
1
:
2
AO: OB = CO: OD = 1: 2
A
O
:
OB
=
CO
:
O
D
=
1
:
2
. The lines
A
D
AD
A
D
and
B
C
BC
BC
intersect at the point
M
M
M
. Prove that
D
M
=
M
B
DM = MB
D
M
=
MB
.
2015.789.4
1
Hide problems
AM = LC wanted, BL = AB, <AML = <BCA (2015 Kyiv City MO Round2 7.4 8.4 9.4.1)
In the acute triangle
A
B
C
ABC
A
BC
the side
B
C
>
A
B
BC> AB
BC
>
A
B
, and the angle bisector
B
L
=
A
B
BL = AB
B
L
=
A
B
. On the segment
B
L
BL
B
L
there is a point
M
M
M
, for which
∠
A
M
L
=
∠
B
C
A
\angle AML = \angle BCA
∠
A
M
L
=
∠
BC
A
. Prove that
A
M
=
L
C
AM = LC
A
M
=
L
C
.
2014.89.3
1
Hide problems
DF construction,<DEF=90^o, DE bisects BF (2014 Kyiv City MO Round2 8.3 9.3)
Given a triangle
A
B
C
ABC
A
BC
, on the side
B
C
BC
BC
which marked the point
E
E
E
such that
B
E
≥
C
E
BE \ge CE
BE
≥
CE
. Construct on the sides
A
B
AB
A
B
and
A
C
AC
A
C
the points
D
D
D
and
F
F
F
, respectively, such that
∠
D
E
F
=
90
∘
\angle DEF = 90 {} ^ \circ
∠
D
EF
=
90
∘
and the segment
B
F
BF
BF
is bisected by the segment
D
E
DE
D
E
.(Black Maxim)
2014.7.4
1
Hide problems
ratio chasing, median <ABM=140^o, <CBM=70^o (2014 Kyiv City MO Round2 7.4)
The median
B
M
BM
BM
is drawn in the triangle
A
B
C
ABC
A
BC
. It is known that
∠
A
B
M
=
40
∘
\angle ABM = 40 {} ^ \circ
∠
A
BM
=
40
∘
and
∠
C
B
M
=
70
∘
\angle CBM = 70 {} ^ \circ
∠
CBM
=
70
∘
Find the ratio
A
B
:
B
M
AB: BM
A
B
:
BM
.
2013.9.5
1
Hide problems
2 circles and a line concurrent, incenters (2013 Kyiv City MO Round2 9.5 10.3)
Given a triangle
A
B
C
ABC
A
BC
,
A
D
AD
A
D
is its angle bisector. Let
E
,
F
E, F
E
,
F
be the centers of the circles inscribed in the triangles
A
D
C
ADC
A
D
C
and
A
D
B
ADB
A
D
B
, respectively. Denote by
ω
\omega
ω
, the circle circumscribed around the triangle
D
E
F
DEF
D
EF
, and by
Q
Q
Q
, the intersection point of
B
E
BE
BE
and
C
F
CF
CF
, and
H
,
J
,
K
,
M
H, J, K, M
H
,
J
,
K
,
M
, respectively the second intersection point of the lines
C
E
,
C
F
,
B
E
,
B
F
CE, CF, BE, BF
CE
,
CF
,
BE
,
BF
with circle
ω
\omega
ω
. Let
ω
1
,
ω
2
\omega_1, \omega_2
ω
1
,
ω
2
the circles be circumscribed around the triangles
H
Q
J
HQJ
H
Q
J
and
K
Q
M
KQM
K
QM
Prove that the intersection point of the circles
ω
1
,
ω
2
\omega_1, \omega_2
ω
1
,
ω
2
different from
Q
Q
Q
lies on the line
A
D
AD
A
D
.(Kivva Bogdan)
2013.8.3
1
Hide problems
BC //MN wanted, 3 angles 45^o, projections (2013 Kyiv City MO Round2 8.3)
Inside
∠
B
A
C
=
45
∘
\angle BAC = 45 {} ^ \circ
∠
B
A
C
=
45
∘
the point
P
P
P
is selected that the conditions
∠
A
P
B
=
∠
A
P
C
=
45
∘
\angle APB = \angle APC = 45 {} ^ \circ
∠
A
PB
=
∠
A
PC
=
45
∘
are fulfilled. Let the points
M
M
M
and
N
N
N
be the projections of the point
P
P
P
on the lines
A
B
AB
A
B
and
A
C
AC
A
C
, respectively. Prove that
B
C
∥
M
N
BC\parallel MN
BC
∥
MN
.(Serdyuk Nazar)
2013.7.3
1
Hide problems
angle chasing in a square, <BMA=<NMD=60^o (2013 Kyiv City MO Round2 7.3)
In the square
A
B
C
D
ABCD
A
BC
D
on the sides
A
D
AD
A
D
and
D
C
DC
D
C
, the points
M
M
M
and
N
N
N
are selected so that
∠
B
M
A
=
∠
N
M
D
=
60
∘
\angle BMA = \angle NMD = 60 { } ^ \circ
∠
BM
A
=
∠
NM
D
=
60
∘
. Find the value of the angle
M
B
N
MBN
MBN
.
2012.9.4
1
Hide problems
min angle, BC=//4OH, circumcenter, orthocenter (2012 Kyiv City MO Round2 9.4)
In an acute-angled triangle
A
B
C
ABC
A
BC
, the point
O
O
O
is the center of the circumcircle, and the point
H
H
H
is the orthocenter. It is known that the lines
O
H
OH
O
H
and
B
C
BC
BC
are parallel, and
B
C
=
4
O
H
BC = 4OH
BC
=
4
O
H
. Find the value of the smallest angle of triangle
A
B
C
ABC
A
BC
.(Black Maxim)
2012.8.5
1
Hide problems
2 equilateral on sides of ABC, A is incenter (2012 Kyiv City MO Round2 8.5)
In the triangle
A
B
C
ABC
A
BC
on the sides
A
B
AB
A
B
and
A
C
AC
A
C
outward constructed equilateral triangles
A
B
D
ABD
A
B
D
and
A
C
E
ACE
A
CE
. The segments
C
D
CD
C
D
and
B
E
BE
BE
intersect at point
F
F
F
. It turns out that point
A
A
A
is the center of the circle inscribed in triangle
D
E
F
DEF
D
EF
. Find the angle
B
A
C
BAC
B
A
C
.(Rozhkova Maria)
2012.7.3
1
Hide problems
computational with trisection of median (2012 Kyiv City MO Round2 7.3)
In the triangle
A
B
C
ABC
A
BC
the median
B
D
BD
B
D
is drawn, which is divided into three equal parts by the points
E
E
E
and
F
F
F
(
B
E
=
E
F
=
F
D
BE = EF = FD
BE
=
EF
=
F
D
). It is known that
A
D
=
A
F
AD = AF
A
D
=
A
F
and
A
B
=
1
AB = 1
A
B
=
1
. Find the length of the segment
C
E
CE
CE
.
2010.89.3
1
Hide problems
angle chasing, incenters, 30^o in acute (2010 Kyiv City MO Round2 8.3 9.3)
In the acute-angled triangle
A
B
C
ABC
A
BC
the angle
∠
B
=
3
0
o
\angle B = 30^o
∠
B
=
3
0
o
, point
H
H
H
is the intersection point of its altitudes. Denote by
O
1
,
O
2
O_1, O_2
O
1
,
O
2
the centers of circles inscribed in triangles
A
B
H
,
C
B
H
ABH ,CBH
A
B
H
,
CB
H
respectively. Find the degree of the angle between the lines
A
O
2
AO_2
A
O
2
and
C
O
1
CO_1
C
O
1
.
2011.9.4
1
Hide problems
_|_ wanted, // diameters of tangent circles (2011 Kyiv City MO Round2 9.4 10.4)
Let two circles be externally tangent at point
C
C
C
, with parallel diameters
A
1
A
2
,
B
1
B
2
A_1A_2, B_1B_2
A
1
A
2
,
B
1
B
2
(i.e. the quadrilateral
A
1
B
1
B
2
A
2
A_1B_1B_2A_2
A
1
B
1
B
2
A
2
is a trapezoid with bases
A
1
A
2
A_1A_2
A
1
A
2
and
B
1
B
2
B_1B_2
B
1
B
2
or parallelogram). Circle with the center on the common internal tangent to these two circles, passes through the intersection point of lines
A
1
B
2
A_1B_2
A
1
B
2
and
A
2
B
1
A_2B_1
A
2
B
1
as well intersects those lines at points
M
,
N
M, N
M
,
N
. Prove that the line
M
N
MN
MN
is perpendicular to the parallel diameters
A
1
A
2
,
B
1
B
2
A_1A_2, B_1B_2
A
1
A
2
,
B
1
B
2
.(Yuri Biletsky)
2011.8.3
1
Hide problems
right angle wanted inside a square, AM = BN (2011 Kyiv City MO Round2 8.3)
On the sides
A
D
,
B
C
AD , BC
A
D
,
BC
of the square
A
B
C
D
ABCD
A
BC
D
the points
M
,
N
M, N
M
,
N
are selected
N
N
N
, respectively, such that
A
M
=
B
N
AM = BN
A
M
=
BN
. Point
X
X
X
is the foot of the perpendicular from point
D
D
D
on the line
A
N
AN
A
N
. Prove that the angle
M
X
C
MXC
MXC
is right.(Mirchev Borislav)