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Problems
Contests
International Contests
Iranian Geometry Olympiad
2023 Iranian Geometry Olympiad
2023 Iranian Geometry Olympiad
Part of
Iranian Geometry Olympiad
Subcontests
(5)
4
3
Hide problems
AD + DC>= AB if CD = BC = BE and ABCD is convex with E =AC\cap BD
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral. Let
E
E
E
be the intersection of its diagonals. Suppose that
C
D
=
B
C
=
B
E
CD = BC = BE
C
D
=
BC
=
BE
. Prove that
A
D
+
D
C
≥
A
B
AD + DC\ge AB
A
D
+
D
C
≥
A
B
.Proposed by Dominik Burek - Poland
AB, SH , TR concurrent wanted , HAPQ and SACQ are #
Let
A
B
C
ABC
A
BC
be a triangle and
P
P
P
be the midpoint of arc
B
A
C
BAC
B
A
C
of circumcircle of triangle
A
B
C
ABC
A
BC
with orthocenter
H
H
H
. Let
Q
,
S
Q, S
Q
,
S
be points such that
H
A
P
Q
HAPQ
H
A
PQ
and
S
A
C
Q
SACQ
S
A
CQ
are parallelograms. Let
T
T
T
be the midpoint of
A
Q
AQ
A
Q
, and
R
R
R
be the intersection point of the lines
S
Q
SQ
SQ
and
P
B
PB
PB
. Prove that
A
B
AB
A
B
,
S
H
SH
S
H
and
T
R
TR
TR
are concurrent.Proposed by Dominik Burek - Poland
line AI bisects segment XY , XY _|_ IP
Let
A
B
C
ABC
A
BC
be a triangle with bisectors
B
E
BE
BE
and
C
F
CF
CF
meet at
I
I
I
. Let
D
D
D
be the projection of
I
I
I
on the
B
C
BC
BC
. Let M and
N
N
N
be the orthocenters of triangles
A
I
F
AIF
A
I
F
and
A
I
E
AIE
A
I
E
, respectively. Lines
E
M
EM
EM
and
F
N
FN
FN
meet at
P
.
P.
P
.
Let
X
X
X
be the midpoint of
B
C
BC
BC
. Let
Y
Y
Y
be the point lying on the line
A
D
AD
A
D
such that
X
Y
⊥
I
P
XY \perp IP
X
Y
⊥
I
P
. Prove that line
A
I
AI
A
I
bisects the segment
X
Y
XY
X
Y
.Proposed by Tran Quang Hung - Vietnam
3
3
Hide problems
tangent (ACE) and (BDM), <B = 3<C
Let
ω
\omega
ω
be the circumcircle of the triangle
A
B
C
ABC
A
BC
with
∠
B
=
3
∠
C
\angle B = 3\angle C
∠
B
=
3∠
C
. The internal angle bisector of
∠
A
\angle A
∠
A
, intersects
ω
\omega
ω
and
B
C
BC
BC
at
M
M
M
and
D
D
D
, respectively. Point
E
E
E
lies on the extension of the line
M
C
MC
MC
from
M
M
M
such that
M
E
ME
ME
is equal to the radius of
ω
\omega
ω
. Prove that circumcircles of triangles
A
C
E
ACE
A
CE
and
B
D
M
BDM
B
D
M
are tangent.Proposed by Mehran Talaei - Iran
square can be cut into 10 triangles of equal area , with vertex P inside ABCD
Let
A
B
C
D
ABCD
A
BC
D
be a square with side length
1
1
1
. How many points
P
P
P
inside the square (not on its sides) have the property that the square can be cut into
10
10
10
triangles of equal area such that all of them have
P
P
P
as a vertex?Proposed by Josef Tkadlec - Czech Republic
Easy combo geo at IGO
There are several discs whose radii are no more that
1
1
1
, and whose centers all lie on a segment with length
l
{l}
l
. Prove that the union of all the discs has a perimeter not exceeding
4
l
+
8
4l+8
4
l
+
8
.Proposed by Morteza Saghafian - Iran
5
3
Hide problems
circumcircle of every triangle used in triangulation contains entire polygo
A polygon is decomposed into triangles by drawing some non-intersecting interior diagonals in such a way that for every pair of triangles of the triangulation sharing a common side, the sum of the angles opposite to this common side is greater than
18
0
o
180^o
18
0
o
. a) Prove that this polygon is convex. b) Prove that the circumcircle of every triangle used in the decomposition contains the entire polygon.Proposed by Morteza Saghafian - Iran
IGO 2023 Intermidiate P5
There are
n
n
n
points in the plane such that at least
99
%
99\%
99%
of quadrilaterals with vertices from these points are convex. Can we find a convex polygon in the plane having at least
90
%
90\%
90%
of the points as vertices?Proposed by Morteza Saghafian - Iran
tangent (ART) , (PEF) wanted
In triangle
A
B
C
ABC
A
BC
points
M
M
M
and
N
N
N
are the midpoints of sides
A
C
AC
A
C
and
A
B
AB
A
B
, respectively and
D
D
D
is the projection of
A
A
A
into
B
C
BC
BC
. Point
O
O
O
is the circumcenter of
A
B
C
ABC
A
BC
and circumcircles of
B
O
C
BOC
BOC
,
D
M
N
DMN
D
MN
intersect at points
R
,
T
R, T
R
,
T
. Lines
D
T
DT
D
T
,
D
R
DR
D
R
intersect line
M
N
MN
MN
at
E
E
E
and
F
F
F
, respectively. Lines
C
T
CT
CT
,
B
R
BR
BR
intersect at
K
K
K
. A point
P
P
P
lies on
K
D
KD
KD
such that
P
K
PK
P
K
is the angle bisector of
∠
B
P
C
\angle BPC
∠
BPC
. Prove that the circumcircles of
A
R
T
ART
A
RT
and
P
E
F
PEF
PEF
are tangent.Proposed by Mehran Talaei - Iran
2
3
Hide problems
KM +ML = BC if AB = AC, <A = 30^o, AL = CM, <AMK = 45^o, <LMC = 75^&omicron;
In an isosceles triangle
A
B
C
ABC
A
BC
with
A
B
=
A
C
AB = AC
A
B
=
A
C
and
∠
A
=
3
0
o
\angle A = 30^o
∠
A
=
3
0
o
, points
L
L
L
and
M
M
M
lie on the sides
A
B
AB
A
B
and
A
C
AC
A
C
, respectively such that
A
L
=
C
M
AL = CM
A
L
=
CM
. Point
K
K
K
lies on
A
B
AB
A
B
such that
∠
A
M
K
=
4
5
o
\angle AMK = 45^o
∠
A
M
K
=
4
5
o
. If
∠
L
M
C
=
7
5
o
\angle LMC = 75^o
∠
L
MC
=
7
5
o
, prove that
K
M
+
M
L
=
B
C
KM +ML = BC
K
M
+
M
L
=
BC
.Proposed by Mahdi Etesamifard - Iran
IGO 2023 Intermidiate P2
A convex hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
with an interior point
P
P
P
is given. Assume that
B
C
E
F
BCEF
BCEF
is a square and both
A
B
P
ABP
A
BP
and
P
C
D
PCD
PC
D
are right isosceles triangles with right angles at
B
B
B
and
C
C
C
, respectively. Lines
A
F
AF
A
F
and
D
E
DE
D
E
intersect at
G
G
G
. Prove that
G
P
GP
GP
is perpendicular to
B
C
BC
BC
.Proposed by Patrik Bak - Slovakia
Why are equal areas so difficult?
Let
I
{I}
I
be the incenter of
△
A
B
C
\triangle {ABC}
△
A
BC
and
B
X
{BX}
BX
,
C
Y
{CY}
C
Y
are its two angle bisectors.
M
{M}
M
is the midpoint of arc
B
A
C
⌢
\overset{\frown}{BAC}
B
A
C
⌢
. It is known that
M
X
I
Y
MXIY
MX
I
Y
are concyclic. Prove that the area of quadrilateral
M
B
I
C
MBIC
MB
I
C
is equal to that of pentagon
B
X
I
Y
C
BXIYC
BX
I
Y
C
.Proposed by Dominik Burek - Poland
1
3
Hide problems
isosceles trapezoid wanted when all polygons are regular
All of the polygons in the figure below are regular. Prove that
A
B
C
D
ABCD
A
BC
D
is an isosceles trapezoid. https://cdn.artofproblemsolving.com/attachments/e/a/3f4de32becf4a90bf0f0b002fb4d8e724e8844.pngProposed by Mahdi Etesamifard - Iran
IGO 2023 Intermidiate P1
Points
M
M
M
and
N
N
N
are the midpoints of sides
A
B
AB
A
B
and
B
C
BC
BC
of the square
A
B
C
D
ABCD
A
BC
D
. According to the fgure, we have drawn a regular hexagon and a regular
12
12
12
-gon. The points
P
,
Q
P, Q
P
,
Q
and
R
R
R
are the centers of these three polygons. Prove that
P
Q
R
S
PQRS
PQRS
is a cyclic quadrilateral.Proposed by Mahdi Etesamifard - Iran
4 concyclic wanted, 5 given
We are given an acute triangle
A
B
C
ABC
A
BC
. The angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
cuts
B
C
BC
BC
at
P
P
P
. Points
D
D
D
and
E
E
E
lie on segments
A
B
AB
A
B
and
A
C
AC
A
C
, respectively, so that
B
C
∥
D
E
BC \parallel DE
BC
∥
D
E
. Points
K
K
K
and
L
L
L
lie on segments
P
D
PD
P
D
and
P
E
PE
PE
, respectively, so that points
A
A
A
,
D
D
D
,
E
E
E
,
K
K
K
,
L
L
L
are concyclic. Prove that points
B
B
B
,
C
C
C
,
K
K
K
,
L
L
L
are also concyclic.Proposed by Patrik Bak, Slovakia