MathDB
Problems
Contests
National and Regional Contests
India Contests
India National Olympiad
2021 India National Olympiad
2021 India National Olympiad
Part of
India National Olympiad
Subcontests
(6)
6
1
Hide problems
Roots, bounding and other delusions
Let
R
[
x
]
\mathbb{R}[x]
R
[
x
]
be the set of all polynomials with real coefficients. Find all functions
f
:
R
[
x
]
→
R
[
x
]
f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]
f
:
R
[
x
]
→
R
[
x
]
satisfying the following conditions: [*]
f
f
f
maps the zero polynomial to itself, [*] for any non-zero polynomial
P
∈
R
[
x
]
P \in \mathbb{R}[x]
P
∈
R
[
x
]
,
deg
f
(
P
)
≤
1
+
deg
P
\text{deg} \, f(P) \le 1+ \text{deg} \, P
deg
f
(
P
)
≤
1
+
deg
P
, and [*] for any two polynomials
P
,
Q
∈
R
[
x
]
P, Q \in \mathbb{R}[x]
P
,
Q
∈
R
[
x
]
, the polynomials
P
−
f
(
Q
)
P-f(Q)
P
−
f
(
Q
)
and
Q
−
f
(
P
)
Q-f(P)
Q
−
f
(
P
)
have the same set of real roots. Proposed by Anant Mudgal, Sutanay Bhattacharya, Pulkit Sinha
5
1
Hide problems
Incredible vanilla geometry again
In a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
,
∠
A
B
D
=
3
0
∘
\angle ABD=30^\circ
∠
A
B
D
=
3
0
∘
,
∠
B
C
A
=
7
5
∘
\angle BCA=75^\circ
∠
BC
A
=
7
5
∘
,
∠
A
C
D
=
2
5
∘
\angle ACD=25^\circ
∠
A
C
D
=
2
5
∘
and
C
D
=
C
B
CD=CB
C
D
=
CB
. Extend
C
B
CB
CB
to meet the circumcircle of triangle
D
A
C
DAC
D
A
C
at
E
E
E
. Prove that
C
E
=
B
D
CE=BD
CE
=
B
D
.Proposed by BJ Venkatachala
4
1
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Deduction card battle
A Magician and a Detective play a game. The Magician lays down cards numbered from
1
1
1
to
52
52
52
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise. Prove that the Detective can guarantee a win if and only if she is allowed to ask at least
50
50
50
questions. Proposed by Anant Mudgal
3
1
Hide problems
Straightedge to draw midpoints
Betal marks
2021
2021
2021
points on the plane such that no three are collinear, and draws all possible segments joining these. He then chooses any
1011
1011
1011
of these segments, and marks their midpoints. Finally, he chooses a segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint using only a straightedge. Can Vikram always complete this challenge?Note. A straightedge is an infinitely long ruler without markings, which can only be used to draw the line joining any two given distinct points. Proposed by Prithwijit De and Sutanay Bhattacharya
2
1
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Dual cubics with integer roots
Find all pairs of integers
(
a
,
b
)
(a,b)
(
a
,
b
)
so that each of the two cubic polynomials
x
3
+
a
x
+
b
and
x
3
+
b
x
+
a
x^3+ax+b \, \, \text{and} \, \, x^3+bx+a
x
3
+
a
x
+
b
and
x
3
+
b
x
+
a
has all the roots to be integers.Proposed by Prithwijit De and Sutanay Bhattacharya
1
1
Hide problems
Integers with determinant \pm 1
Suppose
r
≥
2
r \ge 2
r
≥
2
is an integer, and let
m
1
,
n
1
,
m
2
,
n
2
,
…
,
m
r
,
n
r
m_1, n_1, m_2, n_2, \dots, m_r, n_r
m
1
,
n
1
,
m
2
,
n
2
,
…
,
m
r
,
n
r
be
2
r
2r
2
r
integers such that
∣
m
i
n
j
−
m
j
n
i
∣
=
1
\left|m_in_j-m_jn_i\right|=1
∣
m
i
n
j
−
m
j
n
i
∣
=
1
for any two integers
i
i
i
and
j
j
j
satisfying
1
≤
i
<
j
≤
r
1 \le i<j \le r
1
≤
i
<
j
≤
r
. Determine the maximum possible value of
r
r
r
. Proposed by B Sury