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Old Kyiv MO Geometry
Kyiv City MO Seniors 2003+ geometry
Kyiv City MO Seniors 2003+ geometry
Part of
Old Kyiv MO Geometry
Subcontests
(47)
2022.11.3
1
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Cute geometry with orthocenter and circumcenter
Let
H
H
H
and
O
O
O
be the orthocenter and the circumcenter of the triangle
A
B
C
ABC
A
BC
. Line
O
H
OH
O
H
intersects the sides
A
B
,
A
C
AB, AC
A
B
,
A
C
at points
X
,
Y
X, Y
X
,
Y
correspondingly, so that
H
H
H
belongs to the segment
O
X
OX
OX
. It turned out that
X
H
=
H
O
=
O
Y
XH = HO = OY
X
H
=
H
O
=
O
Y
. Find
∠
B
A
C
\angle BAC
∠
B
A
C
.(Proposed by Oleksii Masalitin)
2022.10.2
1
Hide problems
Legend returns with a new masterpiece
Diagonals of a cyclic quadrilateral
A
B
C
D
ABCD
A
BC
D
intersect at point
P
P
P
. The circumscribed circles of triangles
A
P
D
APD
A
P
D
and
B
P
C
BPC
BPC
intersect the line
A
B
AB
A
B
at points
E
,
F
E, F
E
,
F
correspondingly.
Q
Q
Q
and
R
R
R
are the projections of
P
P
P
onto the lines
F
C
,
D
E
FC, DE
FC
,
D
E
correspondingly. Show that
A
B
∥
Q
R
AB \parallel QR
A
B
∥
QR
.(Proposed by Mykhailo Shtandenko)
2004.11.4
1
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<EFA=< PQB_1 in rect. parallelepiped (Kyiv City Olympiad 2004 11.4)
Given a rectangular parallelepiped
A
B
C
D
A
1
B
1
C
1
D
1
ABCDA_1B_1C_1D_1
A
BC
D
A
1
B
1
C
1
D
1
. Let the points
E
E
E
and
F
F
F
be the feet of the perpendiculars drawn from point
A
A
A
on the lines
A
1
D
A_1D
A
1
D
and
A
1
C
A_1C
A
1
C
, respectively, and the points
P
P
P
and
Q
Q
Q
be the feet of the perpendiculars drawn from point
B
1
B_1
B
1
on the lines
A
1
C
1
A_1C_1
A
1
C
1
and
A
1
C
A_1C
A
1
C
, respectively. Prove that
∠
E
F
A
=
∠
P
Q
B
1
\angle EFA = \angle PQB_1
∠
EF
A
=
∠
PQ
B
1
2004.11.2
1
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angles wanted, _|_ bisectors, MN=BC (Kyiv City Olympiad 2004 11.2)
Given a triangle
A
B
C
ABC
A
BC
, in which
∠
B
>
9
0
o
\angle B> 90^o
∠
B
>
9
0
o
. Perpendicular bisector of the side
A
B
AB
A
B
intersects the side
A
C
AC
A
C
at the point
M
M
M
, and the perpendicular bisector of the side
A
C
AC
A
C
intersects the extension of the side
A
B
AB
A
B
beyond the vertex
B
B
B
at point
N
N
N
. It is known that the segments
M
N
MN
MN
and
B
C
BC
BC
are equal and intersect at right angles. Find the values of all angles of triangle
A
B
C
ABC
A
BC
.
2004.10.5
1
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angle wanted, centroid of AMN=orthocenter of ABC (Kyiv City Olympiad 2004 10.5 )
Let the points
M
M
M
and
N
N
N
in the triangle
A
B
C
ABC
A
BC
be the midpoints of the sides
B
C
BC
BC
and
A
C
AC
A
C
, respectively. It is known that the point of intersection of the altitudes of the triangle
A
B
C
ABC
A
BC
coincides with the point of intersection of the medians of the triangle
A
M
N
AMN
A
MN
. Find the value of the angle
A
B
C
ABC
A
BC
.
2003.11.3
1
Hide problems
radicals of distances and heights in tetrahedron inequality
Let
x
1
,
x
2
,
x
3
,
x
4
x_1, x_2, x_3, x_4
x
1
,
x
2
,
x
3
,
x
4
be the distances from an arbitrary point inside the tetrahedron to the planes of its faces, and let
h
1
,
h
2
,
h
3
,
h
4
h_1, h_2, h_3, h_4
h
1
,
h
2
,
h
3
,
h
4
be the corresponding heights of the tetrahedron. Prove that
h
1
+
h
2
+
h
3
+
h
4
≥
x
1
+
x
2
+
x
3
+
x
4
\sqrt{h_1+h_2+h_3+h_4} \ge \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}
h
1
+
h
2
+
h
3
+
h
4
≥
x
1
+
x
2
+
x
3
+
x
4
(Dmitry Nomirovsky)
2003.10.4
1
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angle bisector wanted, 2 right angles (Kyiv City Olympiad 2003 10.4)
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral. The bisector of the angle
A
C
D
ACD
A
C
D
intersects
B
D
BD
B
D
at point
E
E
E
. It is known that
∠
C
A
D
=
∠
B
C
E
=
9
0
o
\angle CAD = \angle BCE= 90^o
∠
C
A
D
=
∠
BCE
=
9
0
o
. Prove that the
A
C
AC
A
C
is the bisector of the angle
B
A
E
BAE
B
A
E
.(Nikolay Nikolay)
2021.10.3
1
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CX=CY wanted, intersecting circles, # (2021 Kyiv City MO 10.3 11.3)
Circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
with centers at points
O
1
O_1
O
1
and
O
2
O_2
O
2
intersect at points
A
A
A
and
B
B
B
. A point
C
C
C
is constructed such that
A
O
2
C
O
1
AO_2CO_1
A
O
2
C
O
1
is a parallelogram. An arbitrary line is drawn through point
A
A
A
, which intersects the circles
ω
1
\omega_1
ω
1
and
ω
2
\omega_2
ω
2
for the second time at points
X
X
X
and
Y
Y
Y
, respectively. Prove that
C
X
=
C
Y
CX = CY
CX
=
C
Y
.(Oleksii Masalitin)
2020.10.5.1
1
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<EFD=<ACD+<ECB wanted, points on semicircle (2020 Kyiv City MO 10.5.1 11.4.1)
Let
Γ
\Gamma
Γ
be a semicircle with diameter
A
B
AB
A
B
. On this diameter is selected a point
C
C
C
, and on the semicircle are selected points
D
D
D
and
E
E
E
so that
E
E
E
lies between
B
B
B
and
D
D
D
. It turned out that
∠
A
C
D
=
∠
E
C
B
\angle ACD = \angle ECB
∠
A
C
D
=
∠
ECB
. The intersection point of the tangents to
Γ
\Gamma
Γ
at points
D
D
D
and
E
E
E
is denoted by
F
F
F
. Prove that
∠
E
F
D
=
∠
A
C
D
+
∠
E
C
B
\angle EFD=\angle ACD+ \angle ECB
∠
EF
D
=
∠
A
C
D
+
∠
ECB
.
2020.10.5
1
Hide problems
Euler Line perpendicular BI criterion in isosceles (2020 Kyiv City MO 10.5 11.4)
Given an acute isosceles triangle
A
B
C
,
A
K
ABC, AK
A
BC
,
A
K
and
C
N
CN
CN
are its angle bisectors,
I
I
I
is their intersection point . Let point
X
X
X
be the other intersection point of the circles circumscribed around
△
A
B
C
\vartriangle ABC
△
A
BC
and
△
K
B
N
\vartriangle KBN
△
K
BN
. Let
M
M
M
be the midpoint of
A
C
AC
A
C
. Prove that the Euler line of
△
A
B
C
\vartriangle ABC
△
A
BC
is perpendicular to the line
B
I
BI
B
I
if and only if the points
X
,
I
X, I
X
,
I
and
M
M
M
lie on the same line. (Kivva Bogdan)
2019.11.2
1
Hide problems
\sqrt2 KM > AB, midpoint of broken BAC (2019 Kyiv City MO 11.2)
In an acute-angled triangle
A
B
C
ABC
A
BC
, in which
A
B
<
A
C
AB<AC
A
B
<
A
C
, the point
M
M
M
is the midpoint of the side
B
C
,
K
BC, K
BC
,
K
is the midpoint of the broken line segment
B
A
C
BAC
B
A
C
. Prove that
2
K
M
>
A
B
\sqrt2 KM > AB
2
K
M
>
A
B
. (George Naumenko)
2019.10.3
1
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integer side triangles, one inside another (2019 Kyiv City MO 10.3)
Call a right triangle
A
B
C
ABC
A
BC
special if the lengths of its sides
A
B
,
B
C
AB, BC
A
B
,
BC
and
C
A
CA
C
A
are integers, and on each of these sides has some point
X
X
X
(different from the vertices of
△
A
B
C
\vartriangle ABC
△
A
BC
), for which the lengths of the segments
A
X
,
B
X
AX, BX
A
X
,
BX
and
C
X
CX
CX
are integers numbers. Find at least one special triangle. (Maria Rozhkova)
2018.11.4.1
1
Hide problems
orthocenter AXY lies on BD, <ADC =< ACB (2018 Kyiv City MO 11.4.1 )
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
, the diagonal
A
C
AC
A
C
is the bisector
∠
B
A
D
\angle BAD
∠
B
A
D
and
∠
A
D
C
=
∠
A
C
B
\angle ADC = \angle ACB
∠
A
D
C
=
∠
A
CB
. The points
X
,
Y
X, \, \, Y
X
,
Y
are the feet of the perpendiculars drawn from the point
A
A
A
on the lines
B
C
,
C
D
BC, \, \, CD
BC
,
C
D
, respectively. Prove that the orthocenter
Δ
A
X
Y
\Delta AXY
Δ
A
X
Y
lies on the line
B
D
BD
B
D
.
2018.11.4
1
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tangent wanted, isosceles, 2AC = AB + BC, incenter (2018 Kyiv City MO 11.4)
Given an isosceles
A
B
C
ABC
A
BC
, which has
2
A
C
=
A
B
+
B
C
2AC = AB + BC
2
A
C
=
A
B
+
BC
. Denote
I
I
I
the center of the inscribed circle,
K
K
K
the midpoint of the arc
A
B
C
ABC
A
BC
of the circumscribed circle. Let
T
T
T
be such a point on the line
A
C
AC
A
C
that
∠
T
I
B
=
90
∘
\angle TIB = 90 {} ^ \circ
∠
T
I
B
=
90
∘
. Prove that the line
T
B
TB
TB
touches the circumscribed circle
Δ
K
B
I
\Delta KBI
Δ
K
B
I
. (Anton Trygub)
2018.10.4
1
Hide problems
angle wanted, altitudes of 2 trianlges related (2018 Kyiv City MO 10.4 )
In the acute-angled triangle
A
B
C
ABC
A
BC
, the altitudes
B
P
BP
BP
and
C
Q
CQ
CQ
were drawn, and the point
T
T
T
is the intersection point of the altitudes of
Δ
P
A
Q
\Delta PAQ
Δ
P
A
Q
. It turned out that
∠
C
T
B
=
90
∘
\angle CTB = 90 {} ^ \circ
∠
CTB
=
90
∘
. Find the measure of
∠
B
A
C
\angle BAC
∠
B
A
C
. (Mikhail Plotnikov)
2017.11.5.1
1
Hide problems
tangent circumcircles at given point wanted (2017 Kyiv City MO 11.5.1)
The bisector
A
D
AD
A
D
is drawn in the triangle
A
B
C
ABC
A
BC
. Circle
k
k
k
passes through the vertex
A
A
A
and touches the side
B
C
BC
BC
at point
D
D
D
. Prove that the circle circumscribed around
A
B
C
ABC
A
BC
touches the circle
k
k
k
at point
A
A
A
.
2017.11.5
1
Hide problems
concyclic wanted, isosceles, altitudes, circumcircles (2017 Kyiv City MO 11.5)
In the acute isosceles triangle
A
B
C
ABC
A
BC
the altitudes
B
B
1
BB_1
B
B
1
and
C
C
1
CC_1
C
C
1
are drawn, which intersect at the point
H
H
H
. Let
L
1
L_1
L
1
and
L
2
L_2
L
2
be the feet of the angle bisectors of the triangles
B
1
A
C
1
B_1AC_1
B
1
A
C
1
and
B
1
H
C
1
B_1HC_1
B
1
H
C
1
drawn from vertices
A
A
A
and
H
H
H
, respectively. The circumscribed circles of triangles
A
H
L
1
AHL_1
A
H
L
1
and
A
H
L
2
AHL_2
A
H
L
2
intersects the line
B
1
C
1
B_1C_1
B
1
C
1
for the second time at points
P
P
P
and
Q
Q
Q
, respectively. Prove that points
B
,
C
,
P
B, C, P
B
,
C
,
P
and
Q
Q
Q
lie on the same circle.(M. Plotnikov, D. Hilko)
2017.10.3
1
Hide problems
AB = AH wanted inside a square ABCD (2017 Kyiv City MO 10.3)
Given the square
A
B
C
D
ABCD
A
BC
D
. Let point
M
M
M
be the midpoint of the side
B
C
BC
BC
, and
H
H
H
be the foot of the perpendicular from vertex
C
C
C
on the segment
D
M
DM
D
M
. Prove that
A
B
=
A
H
AB = AH
A
B
=
A
H
.(Danilo Hilko)
2016.11.4.1
1
Hide problems
IT=IA wanted, incircle, circumcircle related (2016 Kyiv City MO 11.4.1)
In the triangle
A
B
C
ABC
A
BC
the angle bisector
A
D
AD
A
D
is drawn,
E
E
E
is the point of tangency of the inscribed circle to the side
B
C
BC
BC
,
I
I
I
is the center of the inscribed circle
Δ
A
B
C
\Delta ABC
Δ
A
BC
. The point
A
1
{{A} _ {1}}
A
1
on the circumscribed circle
Δ
A
B
C
\Delta ABC
Δ
A
BC
is such that
A
A
1
∣
∣
B
C
A {{A} _ {1}} || BC
A
A
1
∣∣
BC
. Denote by
T
T
T
- the second point of intersection of the line
E
A
1
E {{A} _ {1}}
E
A
1
and the circumscribed circle
Δ
A
E
D
\Delta AED
Δ
A
E
D
. Prove that
I
T
=
I
A
IT = IA
I
T
=
I
A
.
2016.11.4
1
Hide problems
circumcircle of KTH_1 tangent to BC (2016 Kyiv City MO 11.4)
The median
A
M
AM
A
M
is drawn in the acute-angled triangle
A
B
C
ABC
A
BC
with different sides. Its extension intersects the circumscribed circle
w
w
w
of this triangle at the point
P
P
P
. Let
A
H
1
A {{H} _ {1}}
A
H
1
be the altitude
Δ
A
B
C
\Delta ABC
Δ
A
BC
,
H
H
H
be the point of intersection of its altitudes. The rays
M
H
MH
M
H
and
P
H
1
P {{H} _ {1}}
P
H
1
intersect the circle
w
w
w
at the points
K
K
K
and
T
T
T
, respectively. Prove that the circumscribed circle of
Δ
K
T
H
1
\Delta KT {{H} _ {1}}
Δ
K
T
H
1
touches the segment
B
C
BC
BC
. (Hilko Danilo)
2016.10.4
1
Hide problems
locus is a line, point in circle fixed (2016 Kyiv City MO 10.4)
On the circle with diameter
A
B
AB
A
B
, the point
M
M
M
was selected and fixed. Then the point
Q
i
{{Q} _ {i}}
Q
i
is selected, for which the chord
M
Q
i
M {{Q} _ {i}}
M
Q
i
intersects
A
B
AB
A
B
at the point
K
i
{{K} _ {i}}
K
i
and thus
∠
M
K
i
B
<
90
∘
\angle M {{K} _ {i}} B <90 {} ^ \circ
∠
M
K
i
B
<
90
∘
. A chord that is perpendicular to
A
B
AB
A
B
and passes through the point
K
i
{{K} _ {i}}
K
i
intersects the line
B
Q
i
B {{Q} _ {i}}
B
Q
i
at the point
P
i
{{P } _ {i}}
P
i
. Prove that the points
P
i
{{P} _ {i}}
P
i
in all possible choices of the point
Q
i
{{Q} _ {i}}
Q
i
lie on the same line.(Igor Nagel)
2015.11.4.1
1
Hide problems
MB = MC wanted, angle bisector, perpendiculars (2015 Kyiv City MO 11.4.1)
On the bisector of the angle
B
A
C
BAC
B
A
C
of the triangle
A
B
C
ABC
A
BC
we choose the points
B
1
,
C
1
{{B} _ {1}}, \, \, {{C} _ {1}}
B
1
,
C
1
for which
B
B
1
⊥
A
B
B {{B} _ {1 }}\perp AB
B
B
1
⊥
A
B
,
C
C
1
⊥
A
C
C {{C} _ {1}} \perp AC
C
C
1
⊥
A
C
. The point
M
M
M
is the midpoint of the segment
B
1
C
1
{{B} _ {1}} {{C} _ {1}}
B
1
C
1
. Prove that
M
B
=
M
C
MB = MC
MB
=
MC
.
2015.11.4
1
Hide problems
concurrent wanted, parallelogram, circumcircle (2015 Kyiv City MO 11.4)
In the acute-angled triangle
A
B
C
ABC
A
BC
, the sides
A
B
AB
A
B
and
B
C
BC
BC
have different lengths, and the extension of the median
B
M
BM
BM
intersects the circumscribed circle at the point
N
N
N
. On this circle we note such a point
D
D
D
that
∠
B
D
H
=
90
∘
\angle BDH = 90 {} ^ \circ
∠
B
DH
=
90
∘
, where
H
H
H
is the point of intersection of the altitudes of the triangle
A
B
C
ABC
A
BC
. The point
K
K
K
is chosen so that
A
N
C
K
ANCK
A
NC
K
is a parallelogram. Prove that the lines
A
C
AC
A
C
,
K
H
KH
KH
and
B
D
BD
B
D
intersect at one point.(Igor Nagel)
2015.10.5.1
1
Hide problems
AX:BX wanted, CX//DA,DX//CB.BY//CD,CY//BA (2015 Kyiv City MO 10.5.1)
The points
X
,
Y
X, \, \, Y
X
,
Y
are selected on the sides
A
B
AB
A
B
and
A
D
AD
A
D
of the convex quadrilateral
A
B
C
D
ABCD
A
BC
D
, respectively. Find the ratio
A
X
:
B
X
AX \, \,: \, \, BX
A
X
:
BX
if you know that
C
X
∣
∣
D
A
CX || DA
CX
∣∣
D
A
,
D
X
∣
∣
C
B
DX || CB
D
X
∣∣
CB
,
B
Y
∣
∣
C
D
BY || CD
B
Y
∣∣
C
D
and
C
Y
∣
∣
B
A
CY || BA
C
Y
∣∣
B
A
.
2015.10.5
1
Hide problems
PM = PN wanted, intersecting circles (2015 Kyiv City MO 10.5)
Circles
w
1
{{w} _ {1}}
w
1
and
w
2
{{w} _ {2}}
w
2
with centers at points
O
1
{{O} _ {1}}
O
1
and
O
2
{{ O} _ {2}}
O
2
intersect at points
A
A
A
and
B
B
B
, respectively. Around the triangle
O
1
O
2
B
{{O} _ {1}} {{O} _ {2}} B
O
1
O
2
B
circumscribe a circle
w
w
w
centered at the point
O
O
O
, which intersects the circles
w
1
{{w } _ {1}}
w
1
and
w
2
{{w} _ {2}}
w
2
for the second time at points
K
K
K
and
L
L
L
, respectively. The line
O
A
OA
O
A
intersects the circles
w
1
{{w} _ {1}}
w
1
and
w
2
{{w} _ {2}}
w
2
at the points
M
M
M
and
N
N
N
, respectively. The lines
M
K
MK
M
K
and
N
L
NL
N
L
intersect at the point
P
P
P
. Prove that the point
P
P
P
lies on the circle
w
w
w
and
P
M
=
P
N
PM = PN
PM
=
PN
.(Vadym Mitrofanov)
2014.11.4.1.
1
Hide problems
circles intersection is midpoint of chord, 3 circles (2014 Kyiv City MO 11.4.1)
Construct for the triangle
A
B
C
ABC
A
BC
a circle
S
S
S
passing through the point
B
B
B
and touching the line
C
A
CA
C
A
at the point
A
A
A
, a circle
T
T
T
passing through the point
C
C
C
and touches the line
B
A
BA
B
A
at the point
A
A
A
. The second intersection point of the circles
S
S
S
and
T
T
T
is denoted by
D
D
D
. The intersection point of the line
A
D
AD
A
D
and the circumscribed circle
Δ
A
B
C
\Delta ABC
Δ
A
BC
is denoted by
E
E
E
. Prove that
D
D
D
is the midpoint of the segment
A
E
AE
A
E
.
2014.11.4
1
Hide problems
incircles of ABM,CBM are tangent , KA=AC=CN (2014 Kyiv City MO 11.4)
In the triangle
A
B
C
ABC
A
BC
, for which
A
C
<
A
B
<
B
C
AC <AB <BC
A
C
<
A
B
<
BC
, on the sides
A
B
AB
A
B
and
B
C
BC
BC
the points
K
K
K
and
N
N
N
were chosen, respectively, that
K
A
=
A
C
=
C
N
KA = AC = CN
K
A
=
A
C
=
CN
. The lines
A
N
AN
A
N
and
C
K
CK
C
K
intersect at the point
O
O
O
. From the point
O
O
O
held the segment
O
M
⊥
A
C
OM \perp AC
OM
⊥
A
C
(
M
∈
A
C
M \in AC
M
∈
A
C
) . Prove that the circles inscribed in triangles
A
B
M
ABM
A
BM
and
C
B
M
CBM
CBM
are tangent.(Igor Nagel)
2014.10.4.1
1
Hide problems
angle chasing, AC=1/2(AB + BC) bisector, midpoints (2014 Kyiv City MO 10.4.1)
In the triangle
A
B
C
ABC
A
BC
the side
A
C
=
1
2
(
A
B
+
B
C
)
AC = \tfrac {1} {2} (AB + BC)
A
C
=
2
1
(
A
B
+
BC
)
,
B
L
BL
B
L
is the bisector
∠
A
B
C
\angle ABC
∠
A
BC
,
K
,
M
K, \, \, M
K
,
M
- the midpoints of the sides
A
B
AB
A
B
and
B
C
BC
BC
, respectively. Find the value
∠
K
L
M
\angle KLM
∠
K
L
M
if
∠
A
B
C
=
β
\angle ABC = \beta
∠
A
BC
=
β
2014.10.4
1
Hide problems
equal segments from projections, altitudes related (2014 Kyiv City MO 10.4)
The altitueds
A
A
1
A {{A} _ {1}}
A
A
1
,
B
B
1
B {{B} _ {1}}
B
B
1
and
C
C
1
C {C} _ 1
C
C
1
are drawn in the acute triangle
A
B
C
ABC
A
BC
. . The perpendicular
A
K
AK
A
K
is drawn from the vertex
A
A
A
on the line
A
1
B
1
{{A} _ {1}} {{B} _ {1}}
A
1
B
1
, and the perpendicular
B
L
BL
B
L
is drawn from the vertex
B
B
B
on the line
C
1
B
1
{{C} _ {1}} {{B} _ {1}}
C
1
B
1
. Prove that
A
1
K
=
B
1
L
{{A} _ {1}} K = {{B} _ {1}} L
A
1
K
=
B
1
L
.(Maria Rozhkova)
2013.11.3
1
Hide problems
KN bisects CM, diameter , perpendiculars related (2013 Kyiv City MO 11.3)
The segment
A
B
AB
A
B
is the diameter of the circle. The points
M
M
M
and
C
C
C
belong to this circle and are located in different half-planes relative to the line
A
B
AB
A
B
. From the point
M
M
M
the perpendiculars
M
N
MN
MN
and
M
K
MK
M
K
are drawn on the lines
A
B
AB
A
B
and
A
C
AC
A
C
, respectively. Prove that the line
K
N
KN
K
N
intersects the segment
C
M
CM
CM
in its midpoint.(Igor Nagel)
2013.10.4
1
Hide problems
circle bisects segment, externally tangent circles (2013 Kyiv City MO 9.5 10.4)
The two circles
w
1
,
w
2
{{w} _ {1}}, \, \, {{w} _ {2}}
w
1
,
w
2
touch externally at the point
Q
Q
Q
. The common external tangent of these circles is tangent to
w
1
{{w} _ {1}}
w
1
at the point
B
B
B
,
B
A
BA
B
A
is the diameter of this circle. A tangent to the circle
w
2
{{w} _ {2}}
w
2
is drawn through the point
A
A
A
, which touches this circle at the point
C
C
C
, such that the points
B
B
B
and
C
C
C
lie in one half-plane relative to the line
A
Q
AQ
A
Q
. Prove that the circle
w
1
{{w} _ {1}}
w
1
bisects the segment
C
C
C
.(Igor Nagel)
2012.11.3
1
Hide problems
BM = KQ wanted, MK || AB, 2 circles related (2012 Kyiv City MO 11.3)
Inside the triangle
A
B
C
ABC
A
BC
choose the point
M
M
M
, and on the side
B
C
BC
BC
- the point
K
K
K
in such a way that
M
K
∣
∣
A
B
MK || AB
M
K
∣∣
A
B
. The circle passing through the points
M
,
K
,
C
,
M, \, \, K, \, \, C,
M
,
K
,
C
,
crosses the side
A
C
AC
A
C
for the second time at the point
N
N
N
, a circle passing through the points
M
,
N
,
A
,
M, \, \, N, \, \, A,
M
,
N
,
A
,
crosses the side
A
B
AB
A
B
for the second time at the point
Q
Q
Q
. Prove that
B
M
=
K
Q
BM = KQ
BM
=
K
Q
.(Nagel Igor)
2012.10.4
1
Hide problems
(con)cyclic wanted, other circumcircle related (2012 Kyiv City MO 9.5 10.4)
The triangle
A
B
C
ABC
A
BC
with
A
B
>
A
C
AB> AC
A
B
>
A
C
is inscribed in a circle, the angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
intersects the side
B
C
BC
BC
of the triangle at the point
K
K
K
, and the circumscribed circle at the point
M
M
M
. The midline of
Δ
A
B
C
\Delta ABC
Δ
A
BC
, which is parallel to the side
A
B
AB
A
B
, intersects
A
M
AM
A
M
at the point
O
O
O
, the line
C
O
CO
CO
intersects the line
A
B
AB
A
B
at the point
N
N
N
. Prove that a circle can be circumscribed around the quadrilateral
B
N
K
M
BNKM
BN
K
M
.(Nagel Igor)
2011.11.4.1
1
Hide problems
tangent circles inside a parallelogram (2011 Kyiv City MO 11.4.1)
Inside the parallelogram
A
B
C
D
ABCD
A
BC
D
are the circles
γ
1
\gamma_1
γ
1
and
γ
2
\gamma_2
γ
2
, which are externally tangent at the point
K
K
K
. The circle
γ
1
\gamma_1
γ
1
touches the sides
A
D
AD
A
D
and
A
B
AB
A
B
of the parallelogram, and the circle
γ
2
\gamma_2
γ
2
touches the sides
C
D
CD
C
D
and
C
B
CB
CB
. Prove that the point
K
K
K
lies on the diagonal
A
C
AC
A
C
of the paralelogram.
2011.11.4
1
Hide problems
equal circumcircles, parallelogram, cyclic (2011 Kyiv City MO 11.4)
On the diagonals
A
C
AC
A
C
and
B
D
BD
B
D
of the inscribed quadrilateral A
B
C
D
BCD
BC
D
, the points
X
X
X
and
Y
Y
Y
are marked, respectively, so that the quadrilateral
A
B
X
Y
ABXY
A
BX
Y
is a parallelogram. Prove that the circumscribed circles of triangles
B
X
D
BXD
BX
D
and
C
Y
A
CYA
C
Y
A
have equal radii.(Vyacheslav Yasinsky)
2011.10.3
1
Hide problems
triangle ruler construction, equal area trapezoid (2011 Kyiv City MO 10.3)
A trapezoid
A
B
C
D
ABCD
A
BC
D
with bases
B
C
=
a
BC = a
BC
=
a
and
A
D
=
2
a
AD = 2a
A
D
=
2
a
is drawn on the plane. Using only with a ruler, construct a triangle whose area is equal to the area of the trapezoid. With the help of a ruler you can draw straight lines through two known points.(Rozhkova Maria)
2010.11.3
1
Hide problems
equal quadrilaterals, cyclic, _|_ diag., orthocenters (2010 Kyiv City MO 11.3)
The quadrilateral
A
B
C
D
ABCD
A
BC
D
is inscribed in a circle and has perpendicular diagonals. Points
K
,
L
,
M
,
Q
K,L,M,Q
K
,
L
,
M
,
Q
are the points of intersection of the altitudes of the triangles
A
B
D
,
A
C
D
,
B
C
D
,
A
B
C
ABD, ACD, BCD, ABC
A
B
D
,
A
C
D
,
BC
D
,
A
BC
, respectively. Prove that the quadrilateral
K
L
M
Q
KLMQ
K
L
MQ
is equal to the quadrilateral
A
B
C
D
ABCD
A
BC
D
.(Rozhkova Maria)
2010.10.3
1
Hide problems
equal sum of areas under homothety of a square (2010 Kyiv City MO 10.3)
A point
O
O
O
is chosen inside the square
A
B
C
D
ABCD
A
BC
D
. The square
A
′
B
′
C
′
D
′
A'B'C'D'
A
′
B
′
C
′
D
′
is the image of the square
A
B
C
D
ABCD
A
BC
D
under the homothety with center at point
O
O
O
and coefficient
k
>
1
k> 1
k
>
1
(points
A
′
,
B
′
,
C
′
,
D
′
A', B', C', D'
A
′
,
B
′
,
C
′
,
D
′
are images of points
A
,
B
,
C
,
D
A, B, C, D
A
,
B
,
C
,
D
respectively). Prove that the sum of the areas of the quadrilaterals
A
′
A
B
B
′
A'ABB'
A
′
A
B
B
′
and
C
′
C
D
D
′
C'CDD'
C
′
C
D
D
′
is equal to the sum of the areas quadrilaterals
B
′
B
C
C
′
B'BCC'
B
′
BC
C
′
and
D
′
D
A
A
′
D'DAA'
D
′
D
A
A
′
.
2009.10.4
1
Hide problems
rhombus of side \sqrt {DF x EK} wanted (Kyiv City Olympiad 2009 10.4 11.3)
In the triangle
A
B
C
ABC
A
BC
the angle bisectors
A
L
AL
A
L
and
B
T
BT
BT
are drawn, which intersect at the point
I
I
I
, and their extensions intersect the circle circumscribed around the triangle
A
B
C
ABC
A
BC
at the points
E
E
E
and
D
D
D
respectively. The segment
D
E
DE
D
E
intersects the sides
A
C
AC
A
C
and
B
C
BC
BC
at the points
F
F
F
and
K
K
K
, respectively. Prove that: a) quadrilateral
I
K
C
F
IKCF
I
K
CF
is rhombus; b) the side of this rhombus is
D
F
⋅
E
K
\sqrt {DF \cdot EK}
D
F
⋅
E
K
.(Rozhkova Maria)
2008.11.4
1
Hide problems
SA=SB=SC wanted in a tetrahedron (Kyiv City Olympiad 2008 11.4)
In the tetrahedron
S
A
B
C
SABC
S
A
BC
at the height
S
H
SH
S
H
the following point
O
O
O
is chosen, such that:
∠
A
O
S
+
α
=
∠
B
O
S
+
β
=
∠
C
O
S
+
γ
=
18
0
o
,
\angle AOS + \alpha = \angle BOS + \beta = \angle COS + \gamma = 180^o,
∠
A
OS
+
α
=
∠
BOS
+
β
=
∠
COS
+
γ
=
18
0
o
,
where
α
,
β
,
γ
\alpha, \beta, \gamma
α
,
β
,
γ
are dihedral angles at the edges
B
C
,
A
C
,
A
B
BC, AC, AB
BC
,
A
C
,
A
B
, respectively, at this point
H
H
H
lies inside the base
A
B
C
ABC
A
BC
. Let
A
1
,
B
1
,
C
1
{{A} _ {1}}, \, {{B} _ {1}}, \, {{C} _ {1}}
A
1
,
B
1
,
C
1
be the points of intersection of lines and planes:
A
1
=
A
O
∩
S
B
C
{{A} _ {1}} = AO \cap SBC
A
1
=
A
O
∩
SBC
,
B
1
=
B
O
∩
S
A
C
{{B} _ {1}} = BO \cap SAC
B
1
=
BO
∩
S
A
C
,
C
1
=
C
O
∩
S
B
A
{{C} _ {1}} = CO \cap SBA
C
1
=
CO
∩
SB
A
. Prove that if the planes
A
B
C
ABC
A
BC
and
A
1
B
1
C
1
{{A} _ {1}} {{B} _ {1}} {{C} _ {1}}
A
1
B
1
C
1
are parallel, then
S
A
=
S
B
=
S
C
SA = SB = SC
S
A
=
SB
=
SC
.(Alexey Klurman)
2008.10.4
1
Hide problems
midpoint wanted, concurrent cevians given (Kyiv City Olympiad 2008 10.4)
Given a triangle
A
B
C
ABC
A
BC
,
A
A
1
A {{A} _ {1}}
A
A
1
,
B
B
1
B {{B} _ {1}}
B
B
1
,
C
C
1
C {{C} _ {1}}
C
C
1
- its chevians intersecting at one point.
A
0
,
C
0
{{A} _ {0}}, {{C} _ {0}}
A
0
,
C
0
- the midpoint of the sides
B
C
BC
BC
and
A
B
AB
A
B
respectively. Lines
B
1
C
1
{{B} _ {1}} {{C} _ {1}}
B
1
C
1
,
B
1
A
1
{{B} _ {1}} {{A} _ {1}}
B
1
A
1
and
B
1
B
{ {B} _ {1}} B
B
1
B
intersect the line
A
0
C
0
{{A} _ {0}} {{C} _ {0}}
A
0
C
0
at points
C
2
{{C} _ {2}}
C
2
,
A
2
{{A} _ {2}}
A
2
and
B
2
{{B} _ {2}}
B
2
, respectively. Prove that the point
B
2
{{B} _ {2}}
B
2
is the midpoint of the segment
A
2
C
2
{{A} _ {2}} {{C} _ {2}}
A
2
C
2
. (Eugene Bilokopitov)
2007.11.5
1
Hide problems
fixed point for the circumcircle, equal angles (Kyiv City Olympiad 2007 11.5)
The points
A
A
A
and
P
P
P
are marked on the plane. Consider all such points
B
,
C
B, C
B
,
C
of this plane that
∠
A
B
P
=
∠
M
A
B
\angle ABP = \angle MAB
∠
A
BP
=
∠
M
A
B
and
∠
A
C
P
=
∠
M
A
C
\angle ACP = \angle MAC
∠
A
CP
=
∠
M
A
C
, where
M
M
M
is the midpoint of the segment
B
C
BC
BC
. Prove that all the circumscribed circles around the triangle
A
B
C
ABC
A
BC
for different points
B
B
B
and
C
C
C
pass through some fixed point other than the point
A
A
A
.(Alexei Klurman)
2007.10.3
1
Hide problems
locus wanted, circumcircle, incircle (Kyiv City Olympiad 2007 10.3)
The points
P
,
Q
P, Q
P
,
Q
are given on the plane, which are the points of intersection of the angle bisector
A
L
AL
A
L
of some triangle
A
B
C
ABC
A
BC
with an inscribed circle, and the point
W
W
W
is the intersection of the angle bisector
A
L
AL
A
L
with a circumscribed circle other than the vertex
A
A
A
. a) Find the geometric locus of the possible location of the vertex
A
A
A
of the triangle
A
B
C
ABC
A
BC
. b) Find the geometric locus of the possible location of the vertex
B
B
B
of the triangle
A
B
C
ABC
A
BC
.
2006.11.3
1
Hide problems
<BAC =?, if WO = WH, arc midpoint of circumcircle (Kyiv City Olympiad 2006 11.3)
Let
O
O
O
be the center of the circle
ω
\omega
ω
circumscribed around the acute-angled triangle
△
A
B
C
\vartriangle ABC
△
A
BC
, and
W
W
W
be the midpoint of the arc
B
C
BC
BC
of the circle
ω
\omega
ω
, which does not contain the point
A
A
A
, and
H
H
H
be the point of intersection of the heights of the triangle
△
A
B
C
\vartriangle ABC
△
A
BC
. Find the angle
∠
B
A
C
\angle BAC
∠
B
A
C
, if
W
O
=
W
H
WO = WH
W
O
=
W
H
.(O. Clurman)
2006.10.4
1
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ABC isosceles if MN // PQ, incircle (Kyiv City Olympiad 2006 10.4)
A circle
ω
\omega
ω
is inscribed in the acute-angled triangle
△
A
B
C
\vartriangle ABC
△
A
BC
, which touches the side
B
C
BC
BC
at the point
K
K
K
. On the lines
A
B
AB
A
B
and
A
C
AC
A
C
, the points
P
P
P
and
Q
Q
Q
, respectively, are chosen so that
P
K
⊥
A
C
PK \perp AC
P
K
⊥
A
C
and
Q
K
⊥
A
B
QK \perp AB
Q
K
⊥
A
B
. Denote by
M
M
M
and
N
N
N
the points of intersection of
K
P
KP
K
P
and
K
Q
KQ
K
Q
with the circle
ω
\omega
ω
. Prove that if
M
N
∥
P
Q
MN \parallel PQ
MN
∥
PQ
, then
△
A
B
C
\vartriangle ABC
△
A
BC
is isosceles.(S. Slobodyanyuk)
2005.11.2
1
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isosceles criterion with touchpoints with circle (Kyiv City Olympiad 2005 11.2)
A circle touches the sides
A
C
AC
A
C
and
A
B
AB
A
B
of the triangle
A
B
C
ABC
A
BC
at the points
B
1
{{B}_ {1}}
B
1
and
C
1
{{C}_ {1}}
C
1
respectively. The segments
B
B
1
B {{B} _ {1}}
B
B
1
and
C
C
1
C {{C} _ {1}}
C
C
1
are equal. Prove that the triangle
A
B
C
ABC
A
BC
is isosceles. (Timoshkevich Taras)
2005.10.4
1
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NK bisects BC, CM=MN, <MNB=<CBM, right triangle (Kyiv City Olympiad 2005 10.4)
In a right triangle
A
B
C
ABC
A
BC
with a right angle
∠
C
\angle C
∠
C
, n the sides
A
C
AC
A
C
and
A
B
AB
A
B
, the points
M
M
M
and
N
N
N
are selected, respectively, that
C
M
=
M
N
CM = MN
CM
=
MN
and
∠
M
N
B
=
∠
C
B
M
\angle MNB = \angle CBM
∠
MNB
=
∠
CBM
. Let the point
K
K
K
be the projection of the point
C
C
C
on the segment
M
B
MB
MB
. Prove that the line
N
K
NK
N
K
passes through the midpoint of the segment
B
C
BC
BC
. (Alex Klurman)