MathDB
Problems
Contests
National and Regional Contests
India Contests
India IMO Training Camp
2024 India IMOTC
2024 India IMOTC
Part of
India IMO Training Camp
Subcontests
(24)
24
1
Hide problems
The coin covering problem..., but with a twist
There are
n
>
1
n > 1
n
>
1
distinct points marked in the plane. Prove that there exists a set of circles
C
\mathcal C
C
such that[color=#FFFFFF]___
∙
\bullet
∙
Each circle in
C
\mathcal C
C
has unit radius. [color=#FFFFFF]___
∙
\bullet
∙
Every marked point lies in the (strict) interior of some circle in
C
\mathcal C
C
. [color=#FFFFFF]___
∙
\bullet
∙
There are less than
0.3
n
0.3n
0.3
n
pairs of circles in
C
\mathcal C
C
that intersect in exactly
2
2
2
points.Note: Weaker results with
0.3
n
\it{0.3n}
0.3n
replaced by
c
n
\it{cn}
cn
may be awarded points depending on the value of the constant
c
>
0.3
\it{c > 0.3}
c
>
0.3
.Proposed by Siddharth Choppara, Archit Manas, Ananda Bhaduri, Manu Param
23
1
Hide problems
What function is this, part 2???
Prove that there exists a function
f
:
N
↦
N
f : \mathbb{N} \mapsto \mathbb{N}
f
:
N
↦
N
that satisfies the following: [color=#FFFFFF]___1. For all positive integers
m
,
n
m, n
m
,
n
we have
gcd
(
∣
f
(
m
)
−
f
(
n
)
∣
,
f
(
m
n
)
)
=
f
(
gcd
(
m
,
n
)
)
\gcd(|f(m)-f(n)|, f(mn)) = f(\gcd(m, n))
g
cd
(
∣
f
(
m
)
−
f
(
n
)
∣
,
f
(
mn
))
=
f
(
g
cd
(
m
,
n
))
[color=#FFFFFF]___2. For all positive integers
m
m
m
, we have
f
(
f
(
m
)
)
=
f
(
m
)
f(f(m)) = f(m)
f
(
f
(
m
))
=
f
(
m
)
. [color=#FFFFFF]___3. For all positive integers
k
k
k
, there exists a positive integer
n
n
n
with
202
4
k
∣
f
(
n
)
2024^{k} \mid f(n)
202
4
k
∣
f
(
n
)
.Proposed by MV Adhitya, Archit Manas
22
1
Hide problems
60 deg strikes again
Let
A
B
C
ABC
A
BC
be a triangle with circumcenter
O
O
O
and
∠
B
A
C
=
6
0
∘
\angle BAC = 60^{\circ}
∠
B
A
C
=
6
0
∘
. The internal angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
meets line
B
C
BC
BC
and the circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
in points
M
,
L
M,L
M
,
L
respectively. Let
K
K
K
denote the reflection of
B
L
∩
A
C
BL\cap AC
B
L
∩
A
C
over the line
B
C
BC
BC
. Let
D
D
D
be on the line
C
O
CO
CO
with
D
M
DM
D
M
perpendicular to
K
L
KL
K
L
. Prove that points
K
,
A
,
D
K,A,D
K
,
A
,
D
are collinear.Proposed by Sanjana Philo Chacko
21
1
Hide problems
0 points on 0 point geo
Let
Δ
0
\Delta_0
Δ
0
be an equilateral triangle with incircle
ω
\omega
ω
. A point on
ω
\omega
ω
is reflected in the sides of
Δ
0
\Delta_0
Δ
0
to obtain a new triangle
Δ
1
\Delta_1
Δ
1
. The same point is then reflected over the sides of
Δ
1
\Delta_1
Δ
1
to obtain another triangle
Δ
2
\Delta_2
Δ
2
. Prove that the circumcircle of
Δ
2
\Delta_2
Δ
2
is tangent to
ω
\omega
ω
.Proposed by Siddharth Choppara
20
1
Hide problems
The circusmaster and his LMAOnkey
A circus act consists of
2024
2024
2024
bamboo sticks of pairwise different heights placed in some order, with a monkey standing atop one of them. The circus master can then give commands to the monkey as follows:[color=#FFFFFF]___
∙
\bullet
∙
Left! : When given this command, the monkey locates the closest bamboo stick to the left taller than the one it is currently atop, and jumps to it. If there is no such stick, the monkey stays put. [color=#FFFFFF]___
∙
\bullet
∙
Right! : When given this command, the monkey locates the closest bamboo stick to the right taller than the one it is currently atop, and jumps to it. If there is no such stick, the monkey stays put.The circus master claims that given any two bamboo sticks, if the monkey is originally atop the shorter stick, then after giving at most
c
c
c
commands he can reposition the monkey atop the taller stick. What is the smallest possible value of
c
c
c
?Proposed by Archit Manas
19
1
Hide problems
What function is this??
Denote by
S
\mathbb{S}
S
the set of all proper subsets of
Z
>
0
\mathbb{Z}_{>0}
Z
>
0
. Find all functions
f
:
S
↦
Z
>
0
f : \mathbb{S} \mapsto \mathbb{Z}_{>0}
f
:
S
↦
Z
>
0
that satisfy the following:\\[color=#FFFFFF]___1. For all sets
A
,
B
∈
S
A, B \in \mathbb{S}
A
,
B
∈
S
we have
f
(
A
∩
B
)
=
min
(
f
(
A
)
,
f
(
B
)
)
.
f(A \cap B) = \text{min}(f(A), f(B)).
f
(
A
∩
B
)
=
min
(
f
(
A
)
,
f
(
B
))
.
[color=#FFFFFF]___2. For all positive integers
n
n
n
we have
∑
X
⊆
[
1
,
n
]
f
(
X
)
=
2
n
+
1
−
1.
\sum \limits_{X \subseteq [1, n]} f(X) = 2^{n+1}-1.
X
⊆
[
1
,
n
]
∑
f
(
X
)
=
2
n
+
1
−
1.
(Here, by a proper subset
X
X
X
of
Z
>
0
\mathbb{Z}_{>0}
Z
>
0
we mean
X
⊂
Z
>
0
X \subset \mathbb{Z}_{>0}
X
⊂
Z
>
0
with
X
≠
Z
>
0
X \ne \mathbb{Z}_{>0}
X
=
Z
>
0
. It is allowed for
X
X
X
to have infinite size.) \\Proposed by MV Adhitya, Kanav Talwar, Siddharth Choppara, Archit Manas
18
1
Hide problems
Quadrilateral admits an excircle
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral which admits an incircle. Let
A
B
AB
A
B
produced beyond
B
B
B
meet
D
C
DC
D
C
produced towards
C
C
C
, at
E
E
E
. Let
B
C
BC
BC
produced beyond
C
C
C
meet
A
D
AD
A
D
produced towards
D
D
D
, at
F
F
F
. Let
G
G
G
be the point on line
A
B
AB
A
B
so that
F
G
∥
C
D
FG \parallel CD
FG
∥
C
D
, and let
H
H
H
be the point on line
B
C
BC
BC
so that
E
H
∥
A
D
EH \parallel AD
E
H
∥
A
D
. Prove that the (concave) quadrilateral
E
G
F
H
EGFH
EGF
H
admits an excircle tangent to
E
G
‾
,
E
H
‾
,
F
G
→
,
F
H
→
\overline{EG}, \overline{EH}, \overrightarrow{FG}, \overrightarrow{FH}
EG
,
E
H
,
FG
,
F
H
.Proposed by Rijul Saini
17
1
Hide problems
Triples of polynomials
Fix a positive integer
a
>
1
a > 1
a
>
1
. Consider triples
(
f
(
x
)
,
g
(
x
)
,
h
(
x
)
)
(f(x), g(x), h(x))
(
f
(
x
)
,
g
(
x
)
,
h
(
x
))
of polynomials with integer coefficients, such that 1.
f
f
f
is a monic polynomial with
deg
f
≥
1
\deg f \ge 1
de
g
f
≥
1
. 2. There exists a positive integer
N
N
N
such that
g
(
x
)
>
0
g(x)>0
g
(
x
)
>
0
for
x
≥
N
x \ge N
x
≥
N
and for all positive integers
n
≥
N
n \ge N
n
≥
N
, we have
f
(
n
)
∣
a
g
(
n
)
+
h
(
n
)
f(n) \mid a^{g(n)} + h(n)
f
(
n
)
∣
a
g
(
n
)
+
h
(
n
)
.Find all such possible triples.Proposed by Mainak Ghosh and Rijul Saini
16
1
Hide problems
Cities and flights
There are
n
n
n
cities in a country, one of which is the capital. An airline operates bi-directional flights between some pairs of cities such that one can reach any city from any other city. The airline wants to close down some (possibly zero) number of flights, such that the number of flights needed to reach any particular city from the capital does not increase. Suppose that there are an odd number of ways that the airline can do this. Prove that the set of cities can be divided into two groups, such that there is no flight between two cities of the same group.Proposed by Pranjal Srivastava
15
1
Hide problems
Conference of mathematicians
In a conference, mathematicians from
11
11
11
different countries participate and they have integer-valued ages between
27
27
27
and
33
33
33
years (including
27
27
27
and
33
33
33
). There is at least one mathematician from each country, and there is at least one mathematician of each possible age between
27
27
27
and
33
33
33
. Show that we can find at least five mathematicians
m
1
,
…
,
m
5
m_1, \ldots, m_5
m
1
,
…
,
m
5
such that for any
i
∈
{
1
,
…
,
5
}
i \in \{1, \ldots, 5 \}
i
∈
{
1
,
…
,
5
}
there are more mathematicians in the conference having the same age as
m
i
m_i
m
i
than those having the same nationality as
m
i
m_i
m
i
.Proposed by S. Muralidharan
14
1
Hide problems
Familiar cyclic quad config
Let
A
B
C
D
ABCD
A
BC
D
be a convex cyclic quadrilateral with circumcircle
ω
\omega
ω
. Let
B
A
BA
B
A
produced beyond
A
A
A
meet
C
D
CD
C
D
produced beyond
D
D
D
, at
L
L
L
. Let
ℓ
\ell
ℓ
be a line through
L
L
L
meeting
A
D
AD
A
D
and
B
C
BC
BC
at
M
M
M
and
N
N
N
respectively, so that
M
,
D
M,D
M
,
D
(respectively
N
,
C
N,C
N
,
C
) are on opposite sides of
A
A
A
(resp.
B
B
B
). Suppose
K
K
K
and
J
J
J
are points on the arc
A
B
AB
A
B
of
ω
\omega
ω
not containing
C
,
D
C,D
C
,
D
so that
M
K
,
N
J
MK, NJ
M
K
,
N
J
are tangent to
ω
\omega
ω
. Prove that
K
,
J
,
L
K,J,L
K
,
J
,
L
are collinear.Proposed by Rijul Saini
13
1
Hide problems
Beware the trap!
Find all functions
f
:
R
→
R
f:\mathbb R \to \mathbb R
f
:
R
→
R
such that
x
f
(
x
f
(
y
)
+
y
f
(
x
)
)
=
x
2
f
(
y
)
+
y
f
(
x
)
2
,
xf(xf(y)+yf(x))= x^2f(y)+yf(x)^2,
x
f
(
x
f
(
y
)
+
y
f
(
x
))
=
x
2
f
(
y
)
+
y
f
(
x
)
2
,
for all real numbers
x
,
y
x,y
x
,
y
.Proposed by B.J. Venkatachala
12
1
Hide problems
Nice orthocenter config
Let
A
B
C
ABC
A
BC
be an acute-angled triangle with
A
B
<
A
C
AB<AC
A
B
<
A
C
, and let
O
,
H
O,H
O
,
H
be its circumcentre and orthocentre respectively. Points
Z
,
Y
Z,Y
Z
,
Y
lie on segments
A
B
,
A
C
AB,AC
A
B
,
A
C
respectively, such that
∠
Z
O
B
=
∠
Y
O
C
=
9
0
∘
.
\angle ZOB=\angle YOC = 90^{\circ}.
∠
ZOB
=
∠
Y
OC
=
9
0
∘
.
The perpendicular line from
H
H
H
to line
Y
Z
YZ
Y
Z
meets lines
B
O
BO
BO
and
C
O
CO
CO
at
Q
,
R
Q,R
Q
,
R
respectively. Let the tangents to the circumcircle of
△
A
Y
Z
\triangle AYZ
△
A
Y
Z
at points
Y
Y
Y
and
Z
Z
Z
meet at point
T
T
T
. Prove that
Q
,
R
,
O
,
T
Q, R, O, T
Q
,
R
,
O
,
T
are concyclic. Proposed by Kazi Aryan Amin and K.V. Sudharshan
10
1
Hide problems
Primordial polynomials
Let
r
>
0
r>0
r
>
0
be a real number. We call a monic polynomial with complex coefficients
r
r
r
-good if all of its roots have absolute value at most
r
r
r
. We call a monic polynomial with complex coefficients primordial if all of its coefficients have absolute value at most
1
1
1
. a) Prove that any
1
1
1
-good polynomial has a primordial multiple. b) If
r
>
1
r>1
r
>
1
, prove that there exists an
r
r
r
-good polynomial that does not have a primordial multiple.Proposed by Pranjal Srivastava
9
1
Hide problems
Cute FE, but you all get 0
Find all functions
f
:
R
→
R
f : \mathbb{R} \to \mathbb{R}
f
:
R
→
R
such that for all real numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
, we have
f
(
a
+
b
+
c
)
f
(
a
b
+
b
c
+
c
a
)
−
f
(
a
)
f
(
b
)
f
(
c
)
=
f
(
a
+
b
)
f
(
b
+
c
)
f
(
c
+
a
)
.
f(a+b+c)f(ab+bc+ca) - f(a)f(b)f(c) = f(a+b)f(b+c)f(c+a).
f
(
a
+
b
+
c
)
f
(
ab
+
b
c
+
c
a
)
−
f
(
a
)
f
(
b
)
f
(
c
)
=
f
(
a
+
b
)
f
(
b
+
c
)
f
(
c
+
a
)
.
Proposed by Mainak Ghosh and Rijul Saini
8
1
Hide problems
Power tower sum
Let
a
a
a
and
n
n
n
be positive integers such that: 1.
a
2
n
−
a
a^{2^n}-a
a
2
n
−
a
is divisible by
n
n
n
, 2.
∑
k
=
1
n
k
2024
a
2
k
\sum\limits_{k=1}^{n} k^{2024}a^{2^k}
k
=
1
∑
n
k
2024
a
2
k
is not divisible by
n
n
n
.Prove that
n
n
n
has a prime factor smaller than
2024
2024
2024
.Proposed by Shantanu Nene
7
1
Hide problems
Yet another incenter problem
Let
A
B
C
ABC
A
BC
be an acute-angled triangle with
A
B
<
A
C
AB<AC
A
B
<
A
C
, incentre
I
I
I
, and let
M
M
M
be the midpoint of major arc
B
A
C
BAC
B
A
C
. Suppose the perpendicular line from
A
A
A
to segment
B
C
BC
BC
meets lines
B
I
BI
B
I
,
C
I
CI
C
I
, and
M
I
MI
M
I
at points
P
P
P
,
Q
Q
Q
, and
K
K
K
respectively. Prove that the
A
−
A-
A
−
median line in
△
A
I
K
\triangle AIK
△
A
I
K
passes through the circumcentre of
△
P
I
Q
\triangle PIQ
△
P
I
Q
.Proposed by Pranjal Srivastava and Rohan Goyal
6
1
Hide problems
Time for an IMOTC party!
At an IMOTC party, all people have pairwise distinct ages. Some pairs of people are friends and friendship is mutual. Call a person junior if they are younger than all their friends, and senior if they are older than all their friends. A person with no friends is both junior and senior. A sequence of pairwise distinct people
A
1
,
…
,
A
m
A_1, \dots, A_m
A
1
,
…
,
A
m
is called photogenic if: 1.
A
1
A_1
A
1
is junior, 2.
A
m
A_m
A
m
is senior, and 3.
A
i
A_i
A
i
and
A
i
+
1
A_{i+1}
A
i
+
1
are friends, and
A
i
+
1
A_{i+1}
A
i
+
1
is older than
A
i
A_i
A
i
for all
1
≤
i
≤
m
−
1
1 \leq i \leq m-1
1
≤
i
≤
m
−
1
.Let
k
k
k
be a positive integer such that for every photogenic sequence
A
1
,
…
,
A
m
A_1, \dots, A_m
A
1
,
…
,
A
m
,
m
m
m
is not divisible by
k
k
k
. Prove that the people at the party can be partitioned into
k
k
k
groups so that no two people in the same group are friends.Proposed by Shantanu Nene
11
1
Hide problems
Physics disguised as math
There are
n
≥
3
n\ge 3
n
≥
3
particles on a circle situated at the vertices of a regular
n
n
n
-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before.Let
s
s
s
be the smallest number of collisions after which all particles return to their original positions. Find
s
s
s
.Proposed by N.V. Tejaswi
5
1
Hide problems
45 degrees everywhere
Let
A
B
C
ABC
A
BC
be an acute angled triangle with
A
C
>
A
B
AC>AB
A
C
>
A
B
and incircle
ω
\omega
ω
. Let
ω
\omega
ω
touch the sides
B
C
,
C
A
,
BC, CA,
BC
,
C
A
,
and
A
B
AB
A
B
at
D
,
E
,
D, E,
D
,
E
,
and
F
F
F
respectively. Let
X
X
X
and
Y
Y
Y
be points outside
△
A
B
C
\triangle ABC
△
A
BC
satisfying
∠
B
D
X
=
∠
X
E
A
=
∠
Y
D
C
=
∠
A
F
Y
=
4
5
∘
.
\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.
∠
B
D
X
=
∠
XE
A
=
∠
Y
D
C
=
∠
A
F
Y
=
4
5
∘
.
Prove that the circumcircles of
△
A
X
Y
,
△
A
E
F
\triangle AXY, \triangle AEF
△
A
X
Y
,
△
A
EF
and
△
A
B
C
\triangle ABC
△
A
BC
meet at a point
Z
≠
A
Z\ne A
Z
=
A
.Proposed by Atul Shatavart Nadig and Shantanu Nene
4
1
Hide problems
All residues modulo (n+1)^2
Let
n
n
n
be a positive integer. Let
s
:
N
→
{
1
,
…
,
n
}
s: \mathbb N \to \{1, \ldots, n\}
s
:
N
→
{
1
,
…
,
n
}
be a function such that
n
n
n
divides
m
−
s
(
m
)
m-s(m)
m
−
s
(
m
)
for all positive integers
m
m
m
. Let
a
0
,
a
1
,
a
2
,
…
a_0, a_1, a_2, \ldots
a
0
,
a
1
,
a
2
,
…
be a sequence such that
a
0
=
0
a_0=0
a
0
=
0
and
a
k
=
a
k
−
1
+
s
(
k
)
for all
k
≥
1.
a_{k}=a_{k-1}+s(k) \text{ for all }k\ge 1.
a
k
=
a
k
−
1
+
s
(
k
)
for all
k
≥
1.
Find all
n
n
n
for which this sequence contains all the residues modulo
(
n
+
1
)
2
(n+1)^2
(
n
+
1
)
2
.Proposed by N.V. Tejaswi
3
1
Hide problems
Impossible Infinite Sequence
Let
P
(
x
)
∈
Q
[
x
]
P(x) \in \mathbb{Q}[x]
P
(
x
)
∈
Q
[
x
]
be a polynomial with rational coefficients and degree
d
≥
2
d\ge 2
d
≥
2
. Prove there is no infinite sequence
a
0
,
a
1
,
…
a_0, a_1, \ldots
a
0
,
a
1
,
…
of rational numbers such that
P
(
a
i
)
=
a
i
−
1
+
i
P(a_i)=a_{i-1}+i
P
(
a
i
)
=
a
i
−
1
+
i
for all
i
≥
1
i\ge 1
i
≥
1
.Proposed by Pranjal Srivastava and Rohan Goyal
2
1
Hide problems
Asymmetric inequality
Let
x
1
,
x
2
…
,
x
2024
x_1, x_2 \dots, x_{2024}
x
1
,
x
2
…
,
x
2024
be non-negative real numbers such that
x
1
≤
x
2
⋯
≤
x
2024
x_1 \le x_2\cdots \le x_{2024}
x
1
≤
x
2
⋯
≤
x
2024
, and
x
1
3
+
x
2
3
+
⋯
+
x
2024
3
=
2024
x_1^3 + x_2^3 + \dots + x_{2024}^3 = 2024
x
1
3
+
x
2
3
+
⋯
+
x
2024
3
=
2024
. Prove that
∑
1
≤
i
<
j
≤
2024
(
−
1
)
i
+
j
x
i
2
x
j
≥
−
1012.
\sum_{1 \le i < j \le 2024} (-1)^{i+j} x_i^2 x_j \ge -1012.
1
≤
i
<
j
≤
2024
∑
(
−
1
)
i
+
j
x
i
2
x
j
≥
−
1012.
Proposed by Shantanu Nene
1
1
Hide problems
Hunter rabbit makes a comeback
A sleeping rabbit lies in the interior of a convex
2024
2024
2024
-gon. A hunter picks three vertices of the polygon and he lays a trap which covers the interior and the boundary of the triangular region determined by them. Determine the minimum number of times he needs to do this to guarantee that the rabbit will be trapped.Proposed by Anant Mudgal and Rohan Goyal