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Problems
Contests
National and Regional Contests
Russia Contests
Oral Moscow Geometry Olympiad
2009 Oral Moscow Geometry Olympiad
2009 Oral Moscow Geometry Olympiad
Part of
Oral Moscow Geometry Olympiad
Subcontests
(6)
6
2
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locus of incenter is a line, starting with non intersecting circles, ext. each
Fixed two circles
w
1
w_1
w
1
and
w
2
w_2
w
2
,
ℓ
\ell
ℓ
one of their external tangent and
m
m
m
one of their internal tangent . On the line
m
m
m
, a point
X
X
X
is chosen, and on the line
ℓ
\ell
ℓ
, points
Y
Y
Y
and
Z
Z
Z
are constructed so that
X
Y
XY
X
Y
and
X
Z
XZ
XZ
touch
w
1
w_1
w
1
and
w
2
w_2
w
2
, respectively, and the triangle
X
Y
Z
XYZ
X
Y
Z
contains circles
w
1
w_1
w
1
and
w
2
w_2
w
2
. Prove that the centers of the circles inscribed in triangles
X
Y
Z
XYZ
X
Y
Z
lie on one line.(P. Kozhevnikov)
symmedian wanted, starting with intersecting circles
To two circles
r
1
r_1
r
1
and
r
2
r_2
r
2
, intersecting at points
A
A
A
and
B
B
B
, their common tangent
C
D
CD
C
D
is drawn (
C
C
C
and
D
D
D
are tangency points, respectively, point
B
B
B
is closer to line
C
B
CB
CB
than
A
A
A
). Line passing through
A
A
A
, intersects
r
1
r_1
r
1
and
r
2
r_2
r
2
for second time at points
K
K
K
and
L
L
L
, respectively (
A
A
A
lies between
K
K
K
and
L
L
L
). Lines
K
C
KC
K
C
and
L
D
LD
L
D
intersect at point
P
P
P
. Prove that
P
B
PB
PB
is the symmedian of triangle
K
P
L
KPL
K
P
L
.(Yu. Blinkov)
5
2
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treasure hunting by a mathematician , inside a circular island of radius 1
A treasure is buried at some point on a round island with a radius of
1
1
1
km. On the coast of the island there is a mathematician with a device that indicates the direction to the treasure when the distance to the treasure does not exceed
500
500
500
m. In addition, the mathematician has a map of the island, on which he can record all his movements, perform measurements and geometric constructions. The mathematician claims that he has an algorithm for how to get to the treasure after walking less than
4
4
4
km. Could this be true?(B. Frenkin)
any convex polyhedron has three edges that can be used to form a triangle.
Prove that any convex polyhedron has three edges that can be used to form a triangle.(Barbu Bercanu, Romania)
4
2
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triangle construction, given 1 side, inradii, respective exradii
Construct a triangle given a side, the radius of the inscribed circle, and the radius of the exscribed circle tangent to that side. (Research is not required.)
concurrent wanted, circles with medians as diamaters intersect in pairs
Three circles are constructed on the medians of a triangle as on diameters. It is known that they intersect in pairs. Let
C
1
C_1
C
1
be the intersection point of the circles built on the medians
A
M
1
AM_1
A
M
1
and
B
M
2
BM_2
B
M
2
, which is more distant from the vertex
C
C
C
. Points
A
1
A_1
A
1
and
B
1
B_1
B
1
are defined similarly. Prove that the lines
A
A
1
,
B
B
1
AA_1, BB_1
A
A
1
,
B
B
1
and
C
C
1
CC_1
C
C
1
intersect at one point.(D. Tereshin)
3
2
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<AA_1K=<BB_1M wanted, B_1K// BC, A_1M// AC, altitudes AA_1,BB_1
In the triangle
A
B
C
ABC
A
BC
,
A
A
1
AA_1
A
A
1
and
B
B
1
BB_1
B
B
1
are altitudes. On the side
A
B
AB
A
B
, points
M
M
M
and
K
K
K
are selected so that
B
1
K
∥
B
C
B_1K \parallel BC
B
1
K
∥
BC
and
A
1
M
∥
A
C
A_1M \parallel AC
A
1
M
∥
A
C
. Prove that the angle
A
A
1
K
AA_1K
A
A
1
K
is equal to the angle
B
B
1
M
BB_1M
B
B
1
M
.(D. Prokopenko)
incenter lies on perpendicular, other incircle related
Altitudes
A
A
1
AA_1
A
A
1
and
B
B
1
BB_1
B
B
1
are drawn in the acute-angled triangle
A
B
C
ABC
A
BC
. Prove that the perpendicular drawn from the touchpoint of the inscribed circle with the side
B
C
BC
BC
, on the line
A
C
AC
A
C
passes through the center of the inscribed circle of the triangle
A
1
C
B
1
A_1CB_1
A
1
C
B
1
.(V. Protasov)
2
2
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intersection point of diagonals of rectangle lies on diagonal of square
A square and a rectangle of the same perimeter have a common corner. Prove that the intersection point of the diagonals of the rectangle lies on the diagonal of the square.(Yu. Blinkov)
gray area is equal to sum of black areas, trapezoid, parallelogram
Trapezium
A
B
C
D
ABCD
A
BC
D
and parallelogram
M
B
D
K
MBDK
MB
DK
are located so that the sides of the parallelogram are parallel to the diagonals of the trapezoid (see fig.). Prove that the area of the gray part is equal to the sum of the areas of the black part.(Yu. Blinkov) https://cdn.artofproblemsolving.com/attachments/b/9/cfff83b1b85aea16b603995d4f3d327256b60b.png
1
2
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conctruct line to a parallelogram such that folded you can create a rhombus
The figure shows a parallelogram and the point
P
P
P
of intersection of its diagonals is marked. Draw a straight line through
P
P
P
so that it breaks the parallelogram into two parts, from which you can fold a rhombus. https://1.bp.blogspot.com/-Df2tIBthcmI/X2ZwIx3R4vI/AAAAAAAAMhQ/8Zkxfq30H8MSCdc66tm33n6jt-QKfGMowCLcBGAsYHQ/s0/2009%2Boral%2Bmoscow%2Bj1.png
smaller sides but larger diagonals of 2 quadrilaterals
Are there two such quadrangles that the sides of the first are less than the corresponding sides of the second, and the corresponding diagonals are larger?(Arseniy Akopyan)