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Problems
Contests
National and Regional Contests
Iran Contests
Iran MO (2nd Round)
1992 Iran MO (2nd round)
1992 Iran MO (2nd round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
3
2
Hide problems
Cities and the river [Iran Second Round 1992]
There are some cities in both sides of a river and there are some sailing channels between the cities. Each sailing channel connects exactly one city from a side of the river to a city on the other side. Each city has exactly
k
k
k
sailing channels. For every two cities, there's a way which connects them together. Prove that if we remove any (just one) sailing channel, then again for every two cities, there's a way that connect them together.
(
k
≥
2
)
( k \geq 2)
(
k
≥
2
)
Number of the members of the set is divisible by p [Iran 92]
Let
X
≠
∅
X \neq \varnothing
X
=
∅
be a finite set and let
f
:
X
→
X
f: X \to X
f
:
X
→
X
be a function such that for every
x
∈
X
x \in X
x
∈
X
and a fixed prime
p
p
p
we have
f
p
(
x
)
=
x
.
f^p(x)=x.
f
p
(
x
)
=
x
.
Let
Y
=
{
x
∈
X
∣
f
(
x
)
≠
x
}
.
Y=\{x \in X | f(x) \neq x\}.
Y
=
{
x
∈
X
∣
f
(
x
)
=
x
}
.
Prove that the number of the members of the set
Y
Y
Y
is divisible by
p
.
p.
p
.
Note.
f
p
(
x
)
=
x
=
f
(
f
(
f
(
⋯
(
(
f
⏟
p
times
(
x
)
)
⋯
)
)
)
.
{f^p(x)=x = \underbrace{f(f(f(\cdots ((f}_{ p \text{ times}}(x) ) \cdots )))} .
f
p
(
x
)
=
x
=
p
times
f
(
f
(
f
(
⋯
((
f
(
x
))
⋯
)))
.
1
2
Hide problems
Prove that S_{BCDE}=2S_{BIC} [Iran Second Round 1992]
Let
A
B
C
ABC
A
BC
be a right triangle with
∠
A
=
9
0
∘
.
\angle A=90^\circ.
∠
A
=
9
0
∘
.
The bisectors of the angles
B
B
B
and
C
C
C
meet each other in
I
I
I
and meet the sides
A
C
AC
A
C
and
A
B
AB
A
B
in
D
D
D
and
E
E
E
, respectively. Prove that
S
B
C
D
E
=
2
S
B
I
C
,
S_{BCDE}=2S_{BIC},
S
BC
D
E
=
2
S
B
I
C
,
where
S
S
S
is the area function. [asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttqqcc = rgb(0.2,0,0.8); pen qqwuqq = rgb(0,0.39,0); pen xdxdff = rgb(0.49,0.49,1); pen fftttt = rgb(1,0.2,0.2); pen ccccff = rgb(0.8,0.8,1); draw((1.89,4.08)--(1.89,4.55)--(1.42,4.55)--(1.42,4.08)--cycle,qqwuqq); draw((1.42,4.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(1.42,4.08),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(4,4.09),fftttt+linewidth(1.2pt)); draw((7.42,4.1)--(1.41,6.24),fftttt+linewidth(1.2pt)); draw((1.41,6.24)--(4,4.09),ccccff+linetype("5pt 5pt")); dot((1.42,4.08),ds); label("
A
A
A
", (1.1,3.66),NE*lsf); dot((7.42,4.1),ds); label("
B
B
B
", (7.15,3.75),NE*lsf); dot((1.4,10.08),ds); label("
C
C
C
", (1.49,10.22),NE*lsf); dot((4,4.09),ds); label("
E
E
E
", (3.96,3.46),NE*lsf); dot((1.41,6.24),ds); label("
D
D
D
", (0.9,6.17),NE*lsf); dot((3.37,5.54),ds); label("
I
I
I
", (3.45,5.69),NE*lsf); clip((-6.47,-7.49)--(-6.47,11.47)--(16.06,11.47)--(16.06,-7.49)--cycle); [/asy]
Prove that the number is divisible by 18 [Iran 1992]
Prove that for any positive integer
t
,
t,
t
,
1
+
2
t
+
3
t
+
⋯
+
9
t
−
3
(
1
+
6
t
+
8
t
)
1+2^t+3^t+\cdots+9^t - 3(1 + 6^t +8^t )
1
+
2
t
+
3
t
+
⋯
+
9
t
−
3
(
1
+
6
t
+
8
t
)
is divisible by
18.
18.
18.
2
2
Hide problems
Sequence Inequality
In the sequence
{
a
n
}
n
=
0
∞
\{a_n\}_{n=0}^{\infty}
{
a
n
}
n
=
0
∞
we have
a
0
=
1
a_0=1
a
0
=
1
,
a
1
=
2
a_1=2
a
1
=
2
and
a
n
+
1
=
a
n
+
a
n
−
1
1
+
a
n
−
1
2
∀
n
≥
1
a_{n+1}=a_n+\dfrac{a_{n-1}}{1+a_{n-1}^2} \qquad \forall n \geq 1
a
n
+
1
=
a
n
+
1
+
a
n
−
1
2
a
n
−
1
∀
n
≥
1
Prove that
52
<
a
1371
<
65
52 < a_{1371} < 65
52
<
a
1371
<
65
Prove that MDC is not greater than 45 degrees [Iran 1992]
In triangle
A
B
C
,
ABC,
A
BC
,
we have
∠
A
≤
9
0
∘
\angle A \leq 90^\circ
∠
A
≤
9
0
∘
and
∠
B
=
2
∠
C
.
\angle B = 2 \angle C.
∠
B
=
2∠
C
.
The interior bisector of the angle
C
C
C
meets the median
A
M
AM
A
M
in
D
.
D.
D
.
Prove that
∠
M
D
C
≤
4
5
∘
.
\angle MDC \leq 45^\circ.
∠
M
D
C
≤
4
5
∘
.
When does equality hold? [asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqqqzz = rgb(0,0,0.6); pen ffqqtt = rgb(1,0,0.2); pen qqwuqq = rgb(0,0.39,0); draw((3.14,5.81)--(3.23,0.74),ffqqtt+linewidth(2pt)); draw((3.23,0.74)--(9.73,1.32),ffqqtt+linewidth(2pt)); draw((9.73,1.32)--(3.14,5.81),ffqqtt+linewidth(2pt)); draw((9.73,1.32)--(3.19,2.88)); draw((3.14,5.81)--(6.71,1.05)); pair parametricplot0_cus(real t){ return (0.7*cos(t)+5.8,0.7*sin(t)+2.26); } draw(graph(parametricplot0_cus,-0.9270442638657642,-0.23350086562377703)--(5.8,2.26)--cycle,qqwuqq); dot((3.14,5.81),ds); label("
A
A
A
", (2.93,6.09),NE*lsf); dot((3.23,0.74),ds); label("
B
B
B
", (2.88,0.34),NE*lsf); dot((9.73,1.32),ds); label("
C
C
C
", (10.11,1.04),NE*lsf); dot((3.19,2.88),ds); label("
X
X
X
", (2.77,2.94),NE*lsf); dot((6.71,1.05),ds); label("
M
M
M
", (6.75,0.5),NE*lsf); dot((5.8,2.26),ds); label("
D
D
D
", (5.89,2.4),NE*lsf); label("
α
\alpha
α
", (6.52,1.65),NE*lsf); clip((-3.26,-11.86)--(-3.26,8.36)--(20.76,8.36)--(20.76,-11.86)--cycle); [/asy]