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National and Regional Contests
Iran Contests
Iran MO (3rd Round)
2023 Iran MO (3rd Round)
2023 Iran MO (3rd Round)
Part of
Iran MO (3rd Round)
Subcontests
(6)
6
1
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Another triangle geo
In the acute triangle
△
A
B
C
\triangle ABC
△
A
BC
,
H
H
H
is the orthocenter.
S
S
S
is a point on
(
A
H
C
)
(AHC)
(
A
H
C
)
st
∠
A
S
B
=
90
\angle ASB = 90
∠
A
SB
=
90
.
P
P
P
is on
A
C
AC
A
C
and not on the extention of
A
C
AC
A
C
from
A
A
A
, st
∠
A
P
S
=
∠
B
A
S
\angle APS=\angle BAS
∠
A
PS
=
∠
B
A
S
.Prove that
C
S
CS
CS
, the circle
(
B
P
C
)
(BPC)
(
BPC
)
and the circle with diameter
A
C
AC
A
C
are concurrent.
5
1
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Life-changing troll
There is
n
n
n
black points in the plane.We do the following algorithm: Start from any point from those
n
n
n
points and colour it red. Then connect this point to the nearest black point available and colour this new point red. Then do the same with this point but at any step , but you are never allowed to draw a line which intersects on of the current drawn segments. If you reach an intersection , the algorithm is over. Is it true that for any
n
n
n
and at any initial position , we can start from a point st in the algorithm , we reach all the points?
4
1
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Another easy FE
For any function
f
:
N
→
N
f:\mathbb{N}\to\mathbb{N}
f
:
N
→
N
we define
P
(
n
)
=
f
(
1
)
f
(
2
)
.
.
.
f
(
n
)
P(n)=f(1)f(2)...f(n)
P
(
n
)
=
f
(
1
)
f
(
2
)
...
f
(
n
)
. Find all functions
f
:
N
→
N
f:\mathbb{N}\to\mathbb{N}
f
:
N
→
N
st for each
a
,
b
a,b
a
,
b
:
P
(
a
)
+
P
(
b
)
∣
a
!
+
b
!
P(a)+P(b) | a! + b!
P
(
a
)
+
P
(
b
)
∣
a
!
+
b
!
3
5
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2
5
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1
5
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