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Kyiv City MO
Kyiv City MO - geometry
Kyiv City MO Juniors 2003+ geometry
Kyiv City MO Juniors 2003+ geometry
Part of
Kyiv City MO - geometry
Subcontests
(53)
2004.9.7
1
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triangle construction given 3 points (Kyiv City Olympiad 2004 9.7)
The board depicts the triangle
A
B
C
ABC
A
BC
, the altitude
A
H
AH
A
H
and the angle bisector
A
L
AL
A
L
which intersectthe inscribed circle in the triangle at the points
M
M
M
and
N
,
P
N, P
N
,
P
and
Q
Q
Q
, respectively. After that, the figure was erased, leaving only the points
H
,
M
H, M
H
,
M
and
Q
Q
Q
. Restore the triangle
A
B
C
ABC
A
BC
.(Bogdan Rublev)
2004.8.7
1
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angles in isosceles wanted, AM=2ВК, 24^o (Kyiv City Olympiad 2004 8.7 )
In an isosceles triangle
A
B
C
ABC
A
BC
with base
A
C
AC
A
C
, on side
B
C
BC
BC
is selected point
K
K
K
so that
∠
B
A
K
=
2
4
o
\angle BAK = 24^o
∠
B
A
K
=
2
4
o
. On the segment
A
K
AK
A
K
the point
M
M
M
is chosen so that
∠
A
B
M
=
9
0
o
\angle ABM = 90^o
∠
A
BM
=
9
0
o
,
A
M
=
2
B
K
AM=2BK
A
M
=
2
B
K
. Find the values of all angles of triangle
A
B
C
ABC
A
BC
.
2004.7.3
1
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<DOE=90^o if BD=AD, CB=CE, right triangle (Kyiv City Olympiad 2004 7.3)
Given a right triangle
A
B
C
ABC
A
BC
(
∠
A
<
4
5
o
\angle A <45^o
∠
A
<
4
5
o
,
∠
C
=
9
0
o
\angle C = 90^o
∠
C
=
9
0
o
), on the sides
A
C
AC
A
C
and
A
B
AB
A
B
which are selected points
D
,
E
D,E
D
,
E
respectively, such that
B
D
=
A
D
BD = AD
B
D
=
A
D
and
C
B
=
C
E
CB = CE
CB
=
CE
. Let the segments
B
D
BD
B
D
and
C
E
CE
CE
intersect at the point
O
O
O
. Prove that
∠
D
O
E
=
9
0
o
\angle DOE = 90^o
∠
D
OE
=
9
0
o
.
2003.9.4
1
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equal circumradii, diagonals of quadr. (Kyiv City Olympiad 2003 9.4)
The diagonals of a convex quadrilateral divide it into four triangles. The radii of the circles circumscribed around these triangles are equal. Can such a property have a quadrilateral other than: a) parallelogram, b) rhombus?(Sharygin Igor)
2003.8.5
1
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trapezoid construction (Kyiv City Olympiad 2003 8.5)
Three segments
2
2
2
cm,
5
5
5
cm and
12
12
12
cm long are constructed on the plane. Construct a trapezoid with bases of
2
2
2
cm and
5
5
5
cm, the sum of the sides of which is
12
12
12
cm, and one of the angles is
6
0
o
60^o
6
0
o
.(Bogdan Rublev)
2021.8.41
1
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AM=KB wanted, AK =AO, KM =MC , KM //AC (2021 Kyiv City MO 8.4.1)
On the sides
A
B
AB
A
B
and
B
C
BC
BC
of the triangle
A
B
C
ABC
A
BC
, the points
K
K
K
and
M
M
M
are chosen so that
K
M
∥
A
C
KM \parallel AC
K
M
∥
A
C
. The segments
A
M
AM
A
M
and
K
C
KC
K
C
intersect at the point
O
O
O
. It is known that
A
K
=
A
O
AK =AO
A
K
=
A
O
and
K
M
=
M
C
KM =MC
K
M
=
MC
. Prove that
A
M
=
K
B
AM=KB
A
M
=
K
B
.
2021.8.4
1
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<ABM=<MBP wanted, AB //PC , PM_|_BM , median (2021 Kyiv City MO 8.4)
Let
B
M
BM
BM
be the median of the triangle
A
B
C
ABC
A
BC
, in which
A
B
>
B
C
AB> BC
A
B
>
BC
. Point
P
P
P
is chosen so that
A
B
∥
P
C
AB \parallel PC
A
B
∥
PC
and
P
M
⊥
B
M
PM \perp BM
PM
⊥
BM
. Prove that
∠
A
B
M
=
∠
M
B
P
\angle ABM = \angle MBP
∠
A
BM
=
∠
MBP
.(Mikhail Standenko)
2021.9.51
1
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OB_|_CD wanted, concurrent circles (2021 Kyiv City MO 9.5.1)
Two circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
intersect at points
A
A
A
and
B
B
B
. A line passing through point
B
B
B
intersects
ω
1
\omega_1
ω
1
for the second time at point
C
C
C
and
ω
2
\omega_2
ω
2
at point
D
D
D
. The line
A
C
AC
A
C
intersects circle
ω
2
\omega_2
ω
2
for the second time at point
F
F
F
, and the line
A
D
AD
A
D
intersects the circle
ω
1
\omega_1
ω
1
for the second time at point
E
E
E
. Let point
O
O
O
be the center of the circle circumscribed around
△
A
E
F
\vartriangle AEF
△
A
EF
. Prove that
O
B
⊥
C
D
OB \perp CD
OB
⊥
C
D
.
2021.9.5
1
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AB=BQ wanted, AB\\ PC , PM_|_ BM, median (2021 Kyiv City MO 9.5)
Let
B
M
BM
BM
be the median of the triangle
A
B
C
ABC
A
BC
, in which
A
B
>
B
C
AB> BC
A
B
>
BC
. Point
P
P
P
is chosen so that
A
B
∥
P
C
AB \parallel PC
A
B
∥
PC
and
P
M
⊥
B
M
PM \perp BM
PM
⊥
BM
. The point
Q
Q
Q
is chosen on the line
B
P
BP
BP
so that
∠
A
Q
C
=
9
0
o
\angle AQC = 90^o
∠
A
QC
=
9
0
o
, and the points
B
B
B
and
Q
Q
Q
lie on opposite sides of the line
A
C
AC
A
C
. Prove that
A
B
=
B
Q
AB = BQ
A
B
=
BQ
.(Mikhail Standenko)
2020.9.41
1
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angle wanted, cyclic ABCD, AB=BC=CD, AE//CD (2020 Kyiv City MO 9.4.1)
The points
A
,
B
,
C
,
D
A, B, C, D
A
,
B
,
C
,
D
are selected on the circle as followed so that
A
B
=
B
C
=
C
D
AB = BC = CD
A
B
=
BC
=
C
D
. Bisectors of
∠
A
B
D
\angle ABD
∠
A
B
D
and
∠
A
C
D
\angle ACD
∠
A
C
D
intersect at point
E
E
E
. Find
∠
A
B
C
\angle ABC
∠
A
BC
, if it is known that
A
E
∥
C
D
AE \parallel CD
A
E
∥
C
D
.
2020.9.4
1
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fixed point for circumcircle, equal angles (2020 Kyiv City MO 9.4)
Let the point
D
D
D
lie on the arc
A
C
AC
A
C
of the circumcircle of the triangle
A
B
C
ABC
A
BC
(
A
B
<
B
C
AB < BC
A
B
<
BC
), which does not contain the point
B
B
B
. On the side
A
C
AC
A
C
are selected an arbitrary point
X
X
X
and a point
X
′
X'
X
′
for which
∠
A
B
X
=
∠
C
B
X
′
\angle ABX= \angle CBX'
∠
A
BX
=
∠
CB
X
′
. Prove that regardless of the choice of the point
X
X
X
, the circle circumscribed around
△
D
X
X
′
\vartriangle DXX'
△
D
X
X
′
, passes through a fixed point, which is different from point
D
D
D
.(Nikolaev Arseniy)
2020.8.51
1
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concurrency in cyclic hexagon,AB=BC, CD=DE, EF=FA (2020 Kyiv City MO 8.5.1)
Let
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
be a hexagon inscribed in a circle in which
A
B
=
B
C
,
C
D
=
D
E
AB = BC, CD = DE
A
B
=
BC
,
C
D
=
D
E
and
E
F
=
F
A
EF = FA
EF
=
F
A
. Prove that the lines
A
D
,
B
E
AD, BE
A
D
,
BE
and
C
F
CF
CF
intersect at one point.
2020.8.5
1
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angle chasing, BK=BO, circumcircle, midpoint, bisector (2020 Kyiv City MO 8.4)
Given a triangle
A
B
C
,
O
ABC, O
A
BC
,
O
is the center of the circumcircle,
M
M
M
is the midpoint of
B
C
,
W
BC, W
BC
,
W
is the second intersection of the bisector of the angle
C
C
C
with this circle. A line parallel to
B
C
BC
BC
passing through
W
W
W
, intersects
A
B
AB
A
B
at the point
K
K
K
so that
B
K
=
B
O
BK = BO
B
K
=
BO
. Find the measure of angle
W
M
B
WMB
W
MB
.(Anton Trygub)
2020.7.41
1
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concurrency of perp. bisectors (2020 Kyiv City MO 7.4.1)
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
,
A
B
=
B
C
AB = BC
A
B
=
BC
. The point
E
E
E
lies on the line
A
B
AB
A
B
is such that
B
D
=
B
E
BD= BE
B
D
=
BE
and
A
D
⊥
D
E
AD \perp DE
A
D
⊥
D
E
. Prove that the perpendicular bisectors to segments
A
D
,
C
D
AD, CD
A
D
,
C
D
and
C
E
CE
CE
intersect at one point.
2020.7.4
1
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2 equal rectangles inscribed in square (2020 Kyiv City MO 7.4)
Given a square
A
B
C
D
ABCD
A
BC
D
with side
10
10
10
. On sides BC and
A
D
AD
A
D
of this square are selected respectively points
E
E
E
and
F
F
F
such that formed a rectangle
A
B
E
F
ABEF
A
BEF
. Rectangle
K
L
M
N
KLMN
K
L
MN
is located so that its the vertices
K
,
L
,
M
K, L, M
K
,
L
,
M
and
N
N
N
lie one on each segments
C
D
,
D
F
,
F
E
CD, DF, FE
C
D
,
D
F
,
FE
and
E
C
EC
EC
, respectively. It turned out that the rectangles
A
B
E
F
ABEF
A
BEF
and
K
L
M
N
KLMN
K
L
MN
are equal with
A
B
=
M
N
AB = MN
A
B
=
MN
. Find the length of segment
A
L
AL
A
L
.
2019.9.2
1
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medians AN _|_ CM, when BC=\sqrt2 AC, right triangle (2019 Kyiv City MO 9.2)
In a right triangle
A
B
C
ABC
A
BC
, the lengths of the legs satisfy the condition:
B
C
=
2
A
C
BC =\sqrt2 AC
BC
=
2
A
C
. Prove that the medians
A
N
AN
A
N
and
C
M
CM
CM
are perpendicular.(Hilko Danilo)
2019.8.3
1
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<B= 30^o wanted, 2AC=AB,< A = 2<B, CL = ML (2019 Kyiv City MO 8.3)
In the triangle
A
B
C
ABC
A
BC
it is known that
2
A
C
=
A
B
2AC=AB
2
A
C
=
A
B
and
∠
A
=
2
∠
B
\angle A = 2\angle B
∠
A
=
2∠
B
. In this triangle draw the angle bisector
A
L
AL
A
L
, and mark point
M
M
M
, the midpoint of the side
A
B
AB
A
B
. It turned out that
C
L
=
M
L
CL = ML
C
L
=
M
L
. Prove that
∠
B
=
3
0
o
\angle B= 30^o
∠
B
=
3
0
o
.(Hilko Danilo)
2018.9.51
1
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angle wanted, circle, midpoint, square related(2018 Kyiv City MO 9.5.1 )
Given a circle
Γ
\Gamma
Γ
with center at point
O
O
O
and diameter
A
B
AB
A
B
.
O
B
D
E
OBDE
OB
D
E
is square,
F
F
F
is the second intersection point of the line
A
D
AD
A
D
and the circle
Γ
\Gamma
Γ
,
C
C
C
is the midpoint of the segment
A
F
AF
A
F
. Find the value of the angle
O
C
B
OCB
OCB
.
2018.9.5
1
Hide problems
AK bisects BM, <MAC =< PCB, <MPA = <CPK (2018 Kyiv City MO 9.5 10.4.1)
Given a triangle
A
B
C
ABC
A
BC
, the perpendicular bisector of the side
A
C
AC
A
C
intersects the angle bisector of the triangle
A
K
AK
A
K
at the point
P
P
P
,
M
M
M
- such a point that
∠
M
A
C
=
∠
P
C
B
\angle MAC = \angle PCB
∠
M
A
C
=
∠
PCB
,
∠
M
P
A
=
∠
C
P
K
\angle MPA = \angle CPK
∠
MP
A
=
∠
CP
K
, and points
M
M
M
and
K
K
K
lie on opposite sides of the line
A
C
AC
A
C
. Prove that the line
A
K
AK
A
K
bisects the segment
B
M
BM
BM
.(Anton Trygub)
2018.8.3
1
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angle wanted, isosceles, altitudes, rectangle (2018 Kyiv City MO 8.3)
In the isosceles triangle
A
B
C
ABC
A
BC
with the vertex at the point
B
B
B
, the altitudes
B
H
BH
B
H
and
C
L
CL
C
L
are drawn. The point
D
D
D
is such that
B
D
C
H
BDCH
B
D
C
H
is a rectangle. Find the value of the angle
D
L
H
DLH
D
L
H
.(Bogdan Rublev)
2018.7.41
1
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angle wanted, midpoints, perpendiculars (2018 Kyiv City MO 7.4.1)
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
point
E
E
E
- the midpoint of the side
A
B
AB
A
B
, point
F
F
F
- the midpoint of the side
B
C
BC
BC
, point
G
G
G
- the midpoint
A
D
AD
A
D
. It turned out that the segment
G
E
GE
GE
is perpendicular to
A
B
AB
A
B
, and the segment
G
F
GF
GF
is perpendicular to the segment
B
C
BC
BC
. Find the value of the angle
G
C
D
GCD
GC
D
, if it is known that
∠
A
D
C
=
70
∘
\angle ADC = 70 {} ^ \circ
∠
A
D
C
=
70
∘
.
2018.7.4
1
Hide problems
right angle wanted, <APC=180^o-<ABC,BC = AP, AK=KB+PC (2018 Kyiv City MO 7.4)
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)
2017.9.51
1
Hide problems
tangential quadrilateral by medians, isosceles (2017 Kyiv City MO 9.5.1)
In the triangle
A
B
C
ABC
A
BC
, the medians
B
B
1
BB_1
B
B
1
and
C
C
1
CC_1
C
C
1
, which intersect at the point
M
M
M
, are drawn. Prove that a circle can be inscribed in the quadrilateral
A
C
1
M
B
1
AC_1MB_1
A
C
1
M
B
1
if and only if
A
B
=
A
C
AB = AC
A
B
=
A
C
.
2017.9.5
1
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concyclic wanted, incenter, excenter (2017 Kyiv City MO 9.5)
Let
I
I
I
be the center of the inscribed circle of
A
B
C
ABC
A
BC
and let
I
A
I_A
I
A
be the center of the exscribed circle touching the side
B
C
BC
BC
. Let
M
M
M
be the midpoint of the side
B
C
BC
BC
, and
N
N
N
be the midpoint of the arc
B
A
C
BAC
B
A
C
of the circumscribed circle of
A
B
C
ABC
A
BC
. The point
T
T
T
is symmetric to the point
N
N
N
wrt point
A
A
A
. Prove that the points
I
A
,
M
,
I
,
T
I_A,M,I,T
I
A
,
M
,
I
,
T
lie on the same circle.(Danilo Hilko)
2018.8.41
1
Hide problems
AB=CD if PS// RQ, angle bisectors in trapezoid (2017 Kyiv City MO 8.4.1)
In a trapezoid
A
B
C
D
ABCD
A
BC
D
with bases
A
D
AD
A
D
and
B
C
BC
BC
, the bisector of the angle
∠
D
A
B
\angle DAB
∠
D
A
B
intersects the bisectors of the angles
∠
A
B
C
\angle ABC
∠
A
BC
and
∠
C
D
A
\angle CDA
∠
C
D
A
at the points
P
P
P
and
S
S
S
, respectively, and the bisector of the angle
∠
B
C
D
\angle BCD
∠
BC
D
intersects the bisectors of the angles
∠
A
B
C
\angle ABC
∠
A
BC
and
∠
C
D
A
\angle CDA
∠
C
D
A
at the points
Q
Q
Q
and
R
R
R
, respectively. Prove that if
P
S
∥
R
Q
PS\parallel RQ
PS
∥
RQ
, then
A
B
=
C
D
AB = CD
A
B
=
C
D
.
2017.8.4
1
Hide problems
collinear wanted, circle intersecting a square (2017 Kyiv City MO 8.4)
On the sides
B
C
BC
BC
and
C
D
CD
C
D
of the square
A
B
C
D
ABCD
A
BC
D
, the points
M
M
M
and
N
N
N
are selected in such a way that
∠
M
A
N
=
4
5
o
\angle MAN= 45^o
∠
M
A
N
=
4
5
o
. Using the segment
M
N
MN
MN
, as the diameter, we constructed a circle
w
w
w
, which intersects the segments
A
M
AM
A
M
and
A
N
AN
A
N
at points
P
P
P
and
Q
Q
Q
, respectively. Prove that the points
B
,
P
B, P
B
,
P
and
Q
Q
Q
lie on the same line.
2016.9.51
1
Hide problems
equal angles wanted and given, square (2016 Kyiv City MO 9.5.1 10.4.1)
On the sides
A
B
AB
A
B
and
A
D
AD
A
D
of the square
A
B
C
D
ABCD
A
BC
D
, the points
N
N
N
and
P
P
P
are selected, respectively, so that
P
N
=
N
C
PN = NC
PN
=
NC
, the point
Q
Q
Q
Is a point on the segment
A
N
AN
A
N
for which
∠
N
C
B
=
∠
Q
P
N
\angle NCB = \angle QPN
∠
NCB
=
∠
QPN
. Prove that
∠
B
C
Q
=
1
2
∠
P
Q
A
\angle BCQ = \tfrac {1} {2} \angle PQA
∠
BCQ
=
2
1
∠
PQ
A
.
2016.9.5
1
Hide problems
AC = AA_1 if <ABC = 45^o, AA_1= _|_ CC_1 (2016 Kyiv City MO 9.5)
On the sides
B
C
BC
BC
and
A
B
AB
A
B
of the triangle
A
B
C
ABC
A
BC
the points
A
1
{{A} _ {1}}
A
1
and
C
1
{{C} _ {1}}
C
1
are selected accordingly so that the segments
A
A
1
A {{A} _ {1}}
A
A
1
and
C
C
1
C {{C} _ {1}}
C
C
1
are equal and perpendicular. Prove that if
∠
A
B
C
=
45
∘
\angle ABC = 45 {} ^ \circ
∠
A
BC
=
45
∘
, then
A
C
=
A
A
1
AC = A {{A} _ {1}}
A
C
=
A
A
1
.(Gogolev Andrew)
2016.8.51
1
Hide problems
AX = AD if <BAX=<CDA,<ABC = <BCD, 2AB = CD (2016 Kyiv City MO 8.5.1)
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
, shown in fig. , the equations are true:
∠
A
B
C
=
∠
B
C
D
\angle ABC = \angle BCD
∠
A
BC
=
∠
BC
D
and
2
A
B
=
C
D
2AB = CD
2
A
B
=
C
D
. On the side
B
C
BC
BC
, a point
X
X
X
is selected such that
∠
B
A
X
=
∠
C
D
A
\angle BAX = \angle CDA
∠
B
A
X
=
∠
C
D
A
. Prove that
A
X
=
A
D
AX = AD
A
X
=
A
D
. https://cdn.artofproblemsolving.com/attachments/2/9/0884eb311d1e40300c1e5980fd53eaadfa7a25.png
2016.8.5
1
Hide problems
<ACB=60 iffAE + BD = AB, angle bisectors (2016 Kyiv City MO 8.5)
In the triangle
A
B
C
ABC
A
BC
the angle bisectors
A
D
AD
A
D
and
B
E
BE
BE
are drawn. Prove that
∠
A
C
B
=
60
∘
\angle ACB = 60 {} ^ \circ
∠
A
CB
=
60
∘
if and only if
A
E
+
B
D
=
A
B
AE + BD = AB
A
E
+
B
D
=
A
B
.(Hilko Danilo)
2015.9.3
1
Hide problems
square in a right trapezoid construction (2015 Kyiv City MO 9.3)
It is known that a square can be inscribed in a given right trapezoid so that each of its vertices lies on the corresponding side of the trapezoid (none of the vertices of the square coincides with the vertex of the trapezoid). Construct this inscribed square with a compass and a ruler.(Maria Rozhkova)
2015.8.3
1
Hide problems
angle chasing, isosceles and angle bisectors (2015 Kyiv City MO 8.3)
In the isosceles triangle
A
B
C
ABC
A
BC
,
(
A
B
=
B
C
)
(AB = BC)
(
A
B
=
BC
)
the bisector
A
D
AD
A
D
was drawn, and in the triangle
A
B
D
ABD
A
B
D
the bisector
D
E
DE
D
E
was drawn. Find the values of the angles of the triangle
A
B
C
ABC
A
BC
, if it is known that the bisectors of the angles
A
B
D
ABD
A
B
D
and
A
E
D
AED
A
E
D
intersect on the line
A
D
AD
A
D
.(Fedak Ivan)
2014.9.3
1
Hide problems
2 circles internally tangent to third one (2014 Kyiv City MO 9.3)
Two circles
c
1
,
c
2
{{c} _ {1}}, \, \, {{c} _ {2}}
c
1
,
c
2
pass through the center
O
O
O
of the circle
c
c
c
and touch it internally in points
A
A
A
and
B
B
B
, respectively. Prove that the line
A
B
AB
A
B
passes though a common point of circles
c
1
,
c
2
{{c} _ {1}}, \, \, {{c} _ {2}}
c
1
,
c
2
.
2014.851
1
Hide problems
MF: BK wanted, AK = AF, median related (2014 Kyiv City MO 8.5.1)
On the side
A
B
AB
A
B
of the triangle
A
B
C
ABC
A
BC
mark the point
K
K
K
. The segment
C
K
CK
C
K
intersects the median
A
M
AM
A
M
at the point
F
F
F
. It is known that
A
K
=
A
F
AK = AF
A
K
=
A
F
. Find the ratio
M
F
:
B
K
MF: BK
MF
:
B
K
.
2014.85
1
Hide problems
perpendicular wanted, starting with equilateral (2014 Kyiv City MO 8.5)
Given an equilateral
Δ
A
B
C
\Delta ABC
Δ
A
BC
, in which
A
1
,
B
1
,
C
1
{{A} _ {1}}, {{B} _ {1}}, {{C} _ {1}}
A
1
,
B
1
,
C
1
are the midpoint of the sides
B
C
,
A
C
,
A
B
BC, \, \, AC, \, \, AB
BC
,
A
C
,
A
B
respectively. The line
l
l
l
passes through the vertex
A
A
A
, we denote by
P
,
Q
P, Q
P
,
Q
the projection of the points
B
,
C
B, C
B
,
C
on the line
l
l
l
, respectively (the line
l
l
l
and the point
Q
,
A
,
P
Q, \, \, A, \, \, P
Q
,
A
,
P
are located as shown in fig.). Denote by
T
T
T
the intersection point of the lines
B
1
P
{{B} _ {1}} P
B
1
P
and
C
1
Q
{{C} _ {1}} Q
C
1
Q
. Prove that the line
A
1
T
{{A} _ {1}} T
A
1
T
is perpendicular to the line
l
l
l
. https://cdn.artofproblemsolving.com/attachments/4/b/61f2f4ec9e6b290dfcd47e9351110bebd3bd43.png (Serdyuk Nazar)
2014.7.41
1
Hide problems
triangle inequality with integer AC in ABC, ACD (2014 Kyiv City MO 7.4.1)
The sides of triangles
A
B
C
ABC
A
BC
and
A
C
D
ACD
A
C
D
satisfy the following conditions:
A
B
=
A
D
=
3
AB = AD = 3
A
B
=
A
D
=
3
cm,
B
C
=
7
BC = 7
BC
=
7
cm,
D
C
=
11
DC = 11
D
C
=
11
cm. What values can the side length
A
C
AC
A
C
take if it is an integer number of centimeters, is the average in
Δ
A
C
D
\Delta ACD
Δ
A
C
D
and the largest in
Δ
A
B
C
\Delta ABC
Δ
A
BC
?
2014.7.4
1
Hide problems
BP=CP when AD=AB+CD, angle bisectors related in ABCD (2014 Kyiv City MO 7.4)
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
the condition
A
D
=
A
B
+
C
D
AD = AB + CD
A
D
=
A
B
+
C
D
is fulfilled. The bisectors of the angles
B
A
D
BAD
B
A
D
and
A
D
C
ADC
A
D
C
intersect at the point
P
P
P
, as shown in Fig. Prove that
B
P
=
C
P
BP = CP
BP
=
CP
. https://cdn.artofproblemsolving.com/attachments/3/1/67268635aaef9c6dc3363b00453b327cbc01f3.png(Maria Rozhkova)
2013.8.5
1
Hide problems
4 incircles in convex ABCD concurrent if ABCD is rhombus (2013 Kyiv City MO 8.5)
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral. Prove that the circles inscribed in the triangles
A
B
C
ABC
A
BC
,
B
C
D
BCD
BC
D
,
C
D
A
CDA
C
D
A
and
D
A
B
DAB
D
A
B
have a common point if and only if
A
B
C
D
ABCD
A
BC
D
is a rhombus.
2012.7.4
1
Hide problems
perimeter of triangles inequality, isosceles (2012 Kyiv City MO 7.4)
Given an isosceles triangle
A
B
C
ABC
A
BC
with a vertex at the point
B
B
B
. Based on
A
C
AC
A
C
, an arbitrary point
D
D
D
is selected, different from the vertices
A
A
A
and
C
C
C
. On the line
A
C
AC
A
C
select the point
E
E
E
outside the segment
A
C
AC
A
C
, for which
A
E
=
C
D
AE = CD
A
E
=
C
D
. Prove that the perimeter
Δ
B
D
E
\Delta BDE
Δ
B
D
E
is larger than the perimeter
Δ
A
B
C
\Delta ABC
Δ
A
BC
.
2012.8.3
1
Hide problems
equal angles wanted, intersecting circles (2012 Kyiv City MO 8.3)
On the circle
γ
\gamma
γ
the points
A
A
A
and
B
B
B
are selected. The circle
ω
\omega
ω
touches the segment
A
B
AB
A
B
at the point
K
K
K
and intersects the circle
γ
\gamma
γ
at the points
M
M
M
and
N
N
N
. The points lie on the circle
γ
\gamma
γ
in the following order:
A
,
M
,
N
,
B
A, \, \, M, \, \, N, \, \, B
A
,
M
,
N
,
B
. Prove that
∠
A
M
K
=
∠
K
N
B
\angle AMK = \angle KNB
∠
A
M
K
=
∠
K
NB
.(Yuri Biletsky)
2011.9.41
1
Hide problems
AD = CD wanted, tangents to circumcircle (2011 Kyiv City MO 9.4.1)
The triangle
A
B
C
ABC
A
BC
is inscribed in a circle. At points
A
A
A
and
B
B
B
are tangents to this circle, which intersect at point
T
T
T
. A line drawn through the point
T
T
T
parallel to the side
A
C
AC
A
C
intersects the side
B
C
BC
BC
at the point
D
D
D
. Prove that
A
D
=
C
D
AD = CD
A
D
=
C
D
.
2011.8.41
1
Hide problems
<ABM =<LAC if <ANC = <ALB, medians (2011 Kyiv City MO 8.4.1)
The medians
A
L
,
B
M
AL, BM
A
L
,
BM
, and
C
N
CN
CN
are drawn in the triangle
A
B
C
ABC
A
BC
. Prove that
∠
A
N
C
=
∠
A
L
B
\angle ANC = \angle ALB
∠
A
NC
=
∠
A
L
B
if and only if
∠
A
B
M
=
∠
L
A
C
\angle ABM =\angle LAC
∠
A
BM
=
∠
L
A
C
.(Veklich Bogdan)
2011.89.4
1
Hide problems
<DKL=<CLK if <BMN=<MNC, midpoints of cyclic (2011 Kyiv City MO 9.4)
Let
A
B
C
D
ABCD
A
BC
D
be an inscribed quadrilateral. Denote the midpoints of the sides
A
B
,
B
C
,
C
D
AB, BC, CD
A
B
,
BC
,
C
D
and
D
A
DA
D
A
through
M
,
L
,
N
M, L, N
M
,
L
,
N
and
K
K
K
, respectively. It turned out that
∠
B
M
N
=
∠
M
N
C
\angle BM N = \angle MNC
∠
BMN
=
∠
MNC
. Prove that: i)
∠
D
K
L
=
∠
C
L
K
\angle DKL = \angle CLK
∠
DK
L
=
∠
C
L
K
. ii) in the quadrilateral
A
B
C
D
ABCD
A
BC
D
there is a pair of parallel sides.
2010.9.4
1
Hide problems
TH bisects BC, circumcircle of acute related (2010 Kyiv City MO 9.4 )
In an acute-angled triangle
A
B
C
ABC
A
BC
, the point
O
O
O
is the center of the circumcircle,
C
H
CH
C
H
is the height of the triangle, and the point
T
T
T
is the foot of the perpendicular dropped from the vertex
C
C
C
on the line
A
O
AO
A
O
. Prove that the line
T
H
TH
T
H
passes through the midpoint of the side
B
C
BC
BC
.
2010.8.5
1
Hide problems
(MB- MS)(NC-NS) <= 0, midpoints (Kyiv City Olympiad 2010 8.5)
In an acute-angled triangle
A
B
C
ABC
A
BC
, the points
M
M
M
and
N
N
N
are the midpoints of the sides
A
B
AB
A
B
and
A
C
AC
A
C
, respectively. For an arbitrary point
S
S
S
lying on the side of
B
C
BC
BC
prove that the condition holds
(
M
B
−
M
S
)
(
N
C
−
N
S
)
≤
0
(MB- MS)(NC-NS) \le 0
(
MB
−
MS
)
(
NC
−
NS
)
≤
0
2010.89.4
1
Hide problems
angle chasing, OD = BD = 1/3 BC (Kyiv City Olympiad 2010 8.4 9.6)
Point
O
O
O
is the center of the circumcircle of the acute triangle
A
B
C
ABC
A
BC
. The line
A
O
AO
A
O
intersects the side
B
C
BC
BC
at point
D
D
D
so that
O
D
=
B
D
=
1
/
3
B
C
OD = BD = 1/3 BC
O
D
=
B
D
=
1/3
BC
. Find the angles of the triangle
A
B
C
ABC
A
BC
. Justify the answer.
2009.89.5
1
Hide problems
orthocenter, perpendicular chords, AP = 2PB (Kyiv City Olympiad 2009 8.5 9.3)
A chord
A
B
AB
A
B
is drawn in the circle, on which the point
P
P
P
is selected in such a way that
A
P
=
2
P
B
AP = 2PB
A
P
=
2
PB
. The chord
D
E
DE
D
E
is perpendicular to the chord
A
B
AB
A
B
and passes through the point
P
P
P
. Prove that the midpoint of the segment
A
P
AP
A
P
is the orthocener of the triangle
A
E
D
AED
A
E
D
.
2008.9.5
1
Hide problems
<BFL=?AF=LC<1/2AC, AB^2+BC^2=AL^2+LC^2 (Kyiv City Olympiad 2008 9.5)
In the triangle
A
B
C
ABC
A
BC
on the side
A
C
AC
A
C
the points
F
F
F
and
L
L
L
are selected so that
A
F
=
L
C
<
1
2
A
C
AF = LC <\frac{1}{2} AC
A
F
=
L
C
<
2
1
A
C
. Find the angle
∠
F
B
L
\angle FBL
∠
FB
L
if
A
B
2
+
B
C
2
=
A
L
2
+
L
C
2
A {{B} ^ {2}} + B {{C} ^ {2}} = A {{L} ^ {2}} + L {{C } ^ {2}}
A
B
2
+
B
C
2
=
A
L
2
+
L
C
2
(Zhidkov Sergey)
2008.8.4
1
Hide problems
angles wanted, 2 triangles, trapezoid (Kyiv City Olympiad 2008 8.4)
There are two triangles
A
B
C
ABC
A
BC
and
B
K
L
BKL
B
K
L
on the plane so that the segment
A
K
AK
A
K
is divided into three equal parts by the point of intersection of the medians
△
A
B
C
\vartriangle ABC
△
A
BC
and the point of intersection of the bisectors
△
B
K
L
\vartriangle BKL
△
B
K
L
(
A
K
AK
A
K
- median
△
A
B
C
\vartriangle ABC
△
A
BC
,
K
A
KA
K
A
- bisector
△
B
K
L
\vartriangle BKL
△
B
K
L
) and quadrilateral
K
A
L
C
KALC
K
A
L
C
is trapezoid. Find the angles of the triangle
B
K
L
BKL
B
K
L
.(Bogdan Rublev)
2007.9.3
1
Hide problems
triangle construction, circumcircle, incircle (Kyiv City Olympiad 2007 9.3)
On a straight line
4
4
4
points are successively set ,
A
,
P
,
Q
,
W
A, P, Q,W
A
,
P
,
Q
,
W
, which are the points of intersection of the bisector
A
L
AL
A
L
of the triangle
A
B
C
ABC
A
BC
with the circumscribed and inscribed circle. Knowing only these points, construct a triangle
A
B
C
ABC
A
BC
.
2006.8.3
1
Hide problems
KL=LB, perpendiculars in right triangle (Kyiv City Olympiad 2006 8.3)
On the legs
A
C
,
B
C
AC, BC
A
C
,
BC
of a right triangle
△
A
B
C
\vartriangle ABC
△
A
BC
select points
M
M
M
and
N
N
N
, respectively, so that
∠
M
B
C
=
∠
N
A
C
\angle MBC = \angle NAC
∠
MBC
=
∠
N
A
C
. The perpendiculars from points
M
M
M
and
C
C
C
on the line
A
N
AN
A
N
intersect
A
B
AB
A
B
at points
K
K
K
and
L
L
L
, respectively. Prove that
K
L
=
L
B
KL=LB
K
L
=
L
B
.(O. Clurman)
2006.9.4
1
Hide problems
concurrency in a parallelogram (Kyiv City Olympiad 2006 9.4)
On the sides
A
B
AB
A
B
and
C
D
CD
C
D
of the parallelogram
A
B
C
D
ABCD
A
BC
D
mark points
E
E
E
and
F
F
F
, respectively. On the diagonals
A
C
AC
A
C
and
B
D
BD
B
D
chose the points
M
M
M
and
N
N
N
so that
E
M
∥
B
D
EM\parallel BD
EM
∥
B
D
and
F
N
∥
A
C
FN\parallel AC
FN
∥
A
C
. Prove that the lines
A
F
,
D
E
AF, DE
A
F
,
D
E
and
M
N
MN
MN
intersect at one point.(B. Rublev)
2005.89.5
1
Hide problems
<FXE inside regular hecagon (Kyiv City Olympiad 2005 8.5 9.5)
Let
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
be a regular hexagon. On the line
A
F
AF
A
F
mark the point
X
X
X
so that
∠
D
C
X
=
4
5
o
\angle DCX = 45^o
∠
D
CX
=
4
5
o
. Find the value of the angle
F
X
E
FXE
FXE
. (Vyacheslav Yasinsky)