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Random Geometry Problems from Ukrainian Contests
Ukrainian From Tasks to Tasks - geometry
Ukrainian From Tasks to Tasks - geometry
Part of
Random Geometry Problems from Ukrainian Contests
Subcontests
(22)
2016.3
1
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5 concurrent angle bisectors wanted, 5 given (2016 From Tasks to Tasks, p3 Ukr)
In fig. the bisectors of the angles
∠
D
A
C
\angle DAC
∠
D
A
C
,
∠
E
B
D
\angle EBD
∠
EB
D
,
∠
A
C
E
\angle ACE
∠
A
CE
,
∠
B
D
A
\angle BDA
∠
B
D
A
and
∠
C
E
B
\angle CEB
∠
CEB
intersect at one point. Prove that the bisectors of the angles
∠
T
P
Q
\angle TPQ
∠
TPQ
,
∠
P
Q
R
\angle PQR
∠
PQR
,
∠
Q
R
S
\angle QRS
∠
QRS
,
∠
R
S
T
\angle RST
∠
RST
and
∠
S
T
P
\angle STP
∠
STP
also intersect at one point. https://cdn.artofproblemsolving.com/attachments/6/e/870e4f20bc7fdcb37534f04541c45b1cd5034a.png
2016.13
1
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PQ _|_BC wanted, 45^o given, isosceles (2016 From Tasks to Tasks, p13 Ukr)
Let
A
B
C
ABC
A
BC
be an isosceles acute triangle (
A
B
=
B
C
AB = BC
A
B
=
BC
). On the side
B
C
BC
BC
we mark a point
P
P
P
, such that
∠
P
A
C
=
4
5
o
\angle PAC = 45^o
∠
P
A
C
=
4
5
o
, and
Q
Q
Q
is the point of intersection of the perpendicular bisector of the segment
A
P
AP
A
P
with the side
A
B
AB
A
B
. Prove that
P
Q
⊥
B
C
PQ \perp BC
PQ
⊥
BC
.
2016.8
1
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area chasing, # related (2016 From Tasks to Tasks, p8 Ukr)
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral. It is known that
S
A
B
D
=
7
S_{ABD} = 7
S
A
B
D
=
7
,
S
B
C
D
=
5
S_{BCD}= 5
S
BC
D
=
5
and
S
A
B
C
=
3
S_{ABC}= 3
S
A
BC
=
3
. Inside the quadrilateral mark the point
X
X
X
so that
A
B
C
X
ABCX
A
BCX
is a parallelogram. Find
S
A
D
X
S_{ADX}
S
A
D
X
and
S
B
D
X
S_{BDX}
S
B
D
X
.
2015.14
1
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concyclic, BM=BC,AN=AC, BP=BN, AQ=AM (2015 From Tasks to Tasks, p14 Ukr)
On the side
A
B
AB
A
B
of the triangle
A
B
C
ABC
A
BC
mark the points
M
M
M
and
N
N
N
, such that
B
M
=
B
C
BM = BC
BM
=
BC
and
A
N
=
A
C
AN = AC
A
N
=
A
C
. Then on the sides
B
C
BC
BC
and
A
C
AC
A
C
mark the points
P
P
P
and
Q
Q
Q
, respectively, such that
B
P
=
B
N
BP = BN
BP
=
BN
and
A
Q
=
A
M
AQ = AM
A
Q
=
A
M
. Prove that the points
C
,
Q
,
M
,
N
C, Q, M, N
C
,
Q
,
M
,
N
and
P
P
P
lie on the same circle.
2015.10
1
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bisector + median + altitude = perimeter (2015 From Tasks to Tasks, p10 Ukr)
Can the sum of the lengths of the median, angle bisector and altitude of a triangle be equal to its perimeter if a) these segments are drawn from three different vertices? b) these segments are drawn from one vertex?
2015.5
1
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restore coordinate system given 2 points (2015 From Tasks to Tasks, p5 Ukr)
A coordinate system was constructed on the board, points
A
(
1
,
2
)
A (1,2)
A
(
1
,
2
)
and B
(
3
,
1
)
(3, 1)
(
3
,
1
)
were marked, and then the coordinate system was erased. Restore the coordinate system at the two marked points.
2014.15
1
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right triangle construction (2014 From Tasks to Tasks, p15 Ukr)
Construct a right triangle given the hypotenuse and the median drawn to the leg.
2014.9
1
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tangent bisects AD, where AD is diameter (2014 From Tasks to Tasks, p9 Ukr)
On a circle with diameter
A
B
AB
A
B
we marked an arbitrary point
C
C
C
, which does not coincide with
A
A
A
and
B
B
B
. The tangent to the circle at point
A
A
A
intersects the line
B
C
BC
BC
at point
D
D
D
. Prove that the tangent to the circle at point
C
C
C
bisects the segment
A
D
AD
A
D
.
2014.4
1
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computational, square on hypotenuse (2014 From Tasks to Tasks, p4 Ukr)
In the triangle
A
B
C
ABC
A
BC
it is known that
A
C
=
21
AC = 21
A
C
=
21
cm,
B
C
=
28
BC = 28
BC
=
28
cm and
∠
C
=
9
0
o
\angle C = 90^o
∠
C
=
9
0
o
. On the hypotenuse
A
B
AB
A
B
, we construct a square
A
B
M
N
ABMN
A
BMN
with center
O
O
O
such that the segment
C
O
CO
CO
intersects the hypotenuse
A
B
AB
A
B
at the point
K
K
K
. Find the lengths of the segments
A
K
AK
A
K
and
K
B
KB
K
B
.
2013.13
1
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ABCD, <ABC+<DBC=<ADC+<BDC=180^o (2013 From Tasks to Tasks, p13 Ukr)
In the quadrilateral
A
B
C
D
ABCD
A
BC
D
it is known that
A
B
C
+
D
B
C
=
18
0
o
ABC + DBC = 180^o
A
BC
+
D
BC
=
18
0
o
and
A
D
C
+
B
D
C
=
18
0
o
ADC + BDC = 180^o
A
D
C
+
B
D
C
=
18
0
o
. Prove that the center of the circle circumscribed around the triangle
B
C
D
BCD
BC
D
lies on the diagonal
A
C
AC
A
C
.
2013.9
1
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trisect diagonal in rhombus (2013 From Tasks to Tasks, p9 Ukr)
The perpendicular bisectors of the sides
A
B
AB
A
B
and
C
D
CD
C
D
of the rhombus
A
B
C
D
ABCD
A
BC
D
are drawn. It turned out that they divided the diagonal
A
C
AC
A
C
into three equal parts. Find the altitude of the rhombus if
A
B
=
1
AB = 1
A
B
=
1
.
2013.4
1
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cut in4 trapezoid, 3 conguent right isosceles(2013 From Tasks to Tasks, p4 Ukr)
The trapezoid is composed of three conguent right isosceles triangles as shown in the figure. It is necessary to cut it into
4
4
4
equal parts. How to do it? https://cdn.artofproblemsolving.com/attachments/f/e/87b07ae823190f26b70bfa22824679a829e649.png
2012.13
1
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circumradii =? consecutive sides, inradii (2012 From Tasks to Tasks, p13 Ukr)
The sides of a triangle are consecutive natural numbers, and the radius of the inscribed circle is
4
4
4
. Find the radius of the circumscribed circle.
2012.9
1
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equilateral if median=altitude, <A=60^o (2012 From Tasks to Tasks, p9 Ukr)
In the triangle
A
B
C
ABC
A
BC
, the angle
A
A
A
is equal to
6
0
o
60^o
6
0
o
, and the median
B
D
BD
B
D
is equal to the altitude
C
H
CH
C
H
. Prove that this triangle is equilateral.
2012.4
1
Hide problems
angle chasing in isosceles trapezoid (2012 From Tasks to Tasks, p4 Ukr)
Let
A
B
C
D
ABCD
A
BC
D
be an isosceles trapezoid (
A
D
∥
B
C
AD\parallel BC
A
D
∥
BC
),
∠
B
A
D
=
8
0
o
\angle BAD = 80^o
∠
B
A
D
=
8
0
o
,
∠
B
D
A
=
6
0
o
\angle BDA = 60^o
∠
B
D
A
=
6
0
o
. Point
P
P
P
lies on
C
D
CD
C
D
and
∠
P
A
D
=
5
0
o
\angle PAD = 50^o
∠
P
A
D
=
5
0
o
. Find
∠
P
B
C
\angle PBC
∠
PBC
2012.2
1
Hide problems
locus, 2 isosceles, 1 equilateral (2012 From Tasks to Tasks, p2 Ukr)
The triangle
A
B
C
ABC
A
BC
is equilateral. Find the locus of the points
M
M
M
such that the triangles
A
B
M
ABM
A
BM
and
A
C
M
ACM
A
CM
are both isosceles.
2011.14
1
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area of cyclic octagon, 4 sides->4,others->6 (2011 From Tasks to Tasks, p14 Ukr)
The lengths of the four sides of an cyclic octagon are
4
4
4
cm, the lengths of the other four sides are
6
6
6
cm. Find the area of the octagon.
2011.8
1
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// angle bisectors wanted, isosceles (2011 From Tasks to Tasks, p8 Ukr)
On the median
A
D
AD
A
D
of the isosceles triangle
A
B
C
ABC
A
BC
, point
E
E
E
is marked. Point
F
F
F
is the projection of point
E
E
E
on the line
B
C
BC
BC
, point
M
M
M
lies on the segment
E
F
EF
EF
, points
N
N
N
and
P
P
P
are projections of point
M
M
M
on the lines
A
C
AC
A
C
and
A
B
AB
A
B
, respectively. Prove that the bisectors of the angles
P
M
N
PMN
PMN
and
P
E
N
PEN
PEN
are parallel.
2011.3
1
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AP=AB wanted, circumcenter, bisector (2011 From Tasks to Tasks, p3 Ukr)
Let
O
O
O
be the center of the circumcircle, and
A
D
AD
A
D
be the angle bisector of the acute triangle
A
B
C
ABC
A
BC
. The perpendicular drawn from point
D
D
D
on the line
A
O
AO
A
O
intersects the line
A
C
AC
A
C
at the point
P
P
P
. Prove that
A
P
=
A
B
AP = AB
A
P
=
A
B
.
2010.13
1
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3 right angles, tangential pentagon (2010 From Tasks to Tasks, p13 Ukr )
You can inscribe a circle in the pentagon
A
B
C
D
E
ABCDE
A
BC
D
E
. It is also known that
∠
A
B
C
=
∠
B
A
E
=
∠
C
D
E
=
9
0
o
\angle ABC = \angle BAE = \angle CDE = 90^o
∠
A
BC
=
∠
B
A
E
=
∠
C
D
E
=
9
0
o
. Find the measure of the angle
A
D
B
ADB
A
D
B
.
2010.9
1
Hide problems
#in #, is area half?, use only compass (2010 From Tasks to Tasks, p9 Ukr )
On the sides
A
B
,
B
C
,
C
D
AB, BC, CD
A
B
,
BC
,
C
D
and
D
A
DA
D
A
of the parallelogram
A
B
C
D
ABCD
A
BC
D
marked the points
M
,
N
,
K
M, N, K
M
,
N
,
K
and
F
F
F
. respectively. Is it possible to determine, using only compass, whether the area of the quadrilateral
M
N
K
F
MNKF
MN
K
F
is equal to half the area of the parallelogram
A
B
C
D
ABCD
A
BC
D
?
2010.5
1
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computational, in a right triangle (2010 From Tasks to Tasks, p5 Ukr )
In a right triangle
A
B
C
ABC
A
BC
(
∠
C
=
9
0
o
\angle C = 90^o
∠
C
=
9
0
o
) it is known that
A
C
=
4
AC = 4
A
C
=
4
cm,
B
C
=
3
BC = 3
BC
=
3
cm. The points
A
1
,
B
1
A_1, B_1
A
1
,
B
1
and
C
1
C_1
C
1
are such that
A
A
1
∥
B
C
AA_1 \parallel BC
A
A
1
∥
BC
,
B
B
1
∥
A
1
C
BB_1\parallel A_1C
B
B
1
∥
A
1
C
,
C
C
1
∥
A
1
B
1
CC_1\parallel A_1B_1
C
C
1
∥
A
1
B
1
,
A
1
B
1
C
1
=
9
0
o
A_1B_1C_1= 90^o
A
1
B
1
C
1
=
9
0
o
,
A
1
B
1
=
1
A_1B_1= 1
A
1
B
1
=
1
cm. Find
B
1
C
1
B_1C_1
B
1
C
1
.