MathDB
Problems
Contests
National and Regional Contests
Iran Contests
Iran MO (2nd Round)
2024 Iran MO (2nd Round)
2024 Iran MO (2nd Round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
3
2
Hide problems
Concurrent circles in incircle structure
In a triangle
A
B
C
ABC
A
BC
the incenter, the
B
B
B
-excenter and the
C
C
C
-excenter are
I
,
K
I, K
I
,
K
and
L
L
L
, respectively. The perpendiculars at
B
B
B
and
C
C
C
to
B
C
BC
BC
intersect the lines
A
C
AC
A
C
and
A
B
AB
A
B
at
E
E
E
and
F
F
F
, respectively. Prove that the circumcircles of
A
E
F
,
F
I
L
,
E
I
K
AEF, FIL, EIK
A
EF
,
F
I
L
,
E
I
K
concur.
Another Number Theory!
Find all natural numbers
x
,
y
>
1
x,y>1
x
,
y
>
1
and primes
p
p
p
that satisfy
x
2
−
1
y
2
−
1
=
(
p
+
1
)
2
.
\frac{x^2-1}{y^2-1}=(p+1)^2.
y
2
−
1
x
2
−
1
=
(
p
+
1
)
2
.
1
2
Hide problems
Clock jumping 34 or 47 seconds will become a perfect square
Kimia has a weird clock; the clock's second hand moves 34 or 47 seconds forward instead of each regular second, at random. As an example, if the clock displays the time as
12:23:05
\text{12:23:05}
12:23:05
, the following times could be displayed in this order:
12:23:39, 12:24:13, 12:25:00, 12:25:34, 12:26:21,
…
\text{12:23:39, 12:24:13, 12:25:00, 12:25:34, 12:26:21,\dots}
12:23:39, 12:24:13, 12:25:00, 12:25:34, 12:26:21,
…
Prove that the clock's second hand would eventually land on a perfect square.
Altitude and midpoint circle
In the triangle
A
B
C
ABC
A
BC
,
M
M
M
is the midpoint of
A
B
AB
A
B
and
B
′
B'
B
′
is the foot of
B
B
B
-altitude.
C
B
′
M
CB'M
C
B
′
M
intersects the line
B
C
BC
BC
for the second time at
D
D
D
. Circumcircles of
C
B
′
M
CB'M
C
B
′
M
and
A
B
D
ABD
A
B
D
intersect each other again at
K
K
K
. The parallel to
A
B
AB
A
B
through
C
C
C
intersects the
C
B
′
M
CB'M
C
B
′
M
circle again at
L
L
L
. Prove that
K
L
KL
K
L
cuts
C
M
CM
CM
in half.
2
2
Hide problems
Iranian sequence
Find all sequences
(
a
n
)
n
≥
1
(a_n)_{n\geq 1}
(
a
n
)
n
≥
1
of positive integers such that for all integers
n
≥
3
n\geq 3
n
≥
3
we have
1
a
1
a
3
+
1
a
2
a
4
+
⋯
+
1
a
n
−
2
a
n
=
1
−
1
a
1
2
+
a
2
2
+
⋯
+
a
n
−
1
2
.
\dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n}= 1 - \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}.
a
1
a
3
1
+
a
2
a
4
1
+
⋯
+
a
n
−
2
a
n
1
=
1
−
a
1
2
+
a
2
2
+
⋯
+
a
n
−
1
2
1
.
Recoloring Rows Alternatively
Sahand and Gholam play on a
1403
×
1403
1403\times 1403
1403
×
1403
table. Initially all the unit square cells are white. For each row and column there is a key for it (totally 2806 keys). Starting with Sahand players take turn alternatively pushing a button that has not been pushed yet, until all the buttons are pushed. When Sahand pushes a button all the cells of that row or column become black, regardless of the previous colors. When Gholam pushes a button all the cells of that row or column become red, regardless of the previous colors. Finally, Gholam's score equals the number of red squares minus the number of black squares and Sahand's score equals the number of black squares minus the number of red squares. Determine the minimum number of scores Gholam can guarantee without if both players play their best moves.