MathDB
Problems
Contests
National and Regional Contests
India Contests
ISI Entrance Examination
2023 ISI Entrance UGB
2023 ISI Entrance UGB
Part of
ISI Entrance Examination
Subcontests
(8)
3
1
Hide problems
ISI summons a ladder to help students
In
△
A
B
C
\triangle ABC
△
A
BC
, consider points
D
D
D
and
E
E
E
on
A
C
AC
A
C
and
A
B
AB
A
B
, respectively, and assume that they do not coincide with any of the vertices
A
A
A
,
B
B
B
,
C
C
C
. If the segments
B
D
BD
B
D
and
C
E
CE
CE
intersect at
F
F
F
, consider areas
w
w
w
,
x
x
x
,
y
y
y
,
z
z
z
of the quadrilateral
A
E
F
D
AEFD
A
EF
D
and the triangles
B
E
F
BEF
BEF
,
B
F
C
BFC
BFC
,
C
D
F
CDF
C
D
F
, respectively. [*] Prove that
y
2
>
x
z
y^2 > xz
y
2
>
x
z
. [*] Determine
w
w
w
in terms of
x
x
x
,
y
y
y
,
z
z
z
. [asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.8465032978885407, xmax = 9.445649196374966, ymin = -1.7618066305534972, ymax = 4.389732795464592; /* image dimensions */ draw((3.8295013012181283,2.816337276198864)--(-0.7368327629589799,-0.5920813291311117)--(5.672613975760373,-0.636902634996282)--cycle, linewidth(0.5)); /* draw figures */ draw((3.8295013012181283,2.816337276198864)--(-0.7368327629589799,-0.5920813291311117), linewidth(0.5)); draw((-0.7368327629589799,-0.5920813291311117)--(5.672613975760373,-0.636902634996282), linewidth(0.5)); draw((5.672613975760373,-0.636902634996282)--(3.8295013012181283,2.816337276198864), linewidth(0.5)); draw((-0.7368327629589799,-0.5920813291311117)--(4.569287648059735,1.430279997142299), linewidth(0.5)); draw((5.672613975760373,-0.636902634996282)--(1.8844000180622977,1.3644681598392678), linewidth(0.5)); label("
y
y
y
",(2.74779188172294,0.23771684184669772),SE*labelscalefactor); label("
w
w
w
",(3.2941097703568736,1.8657441499758196),SE*labelscalefactor); label("
x
x
x
",(1.6660824622277512,1.0025618859342047),SE*labelscalefactor); label("
z
z
z
",(4.288408327670633,0.8168138037986672),SE*labelscalefactor); /* dots and labels */ dot((3.8295013012181283,2.816337276198864),dotstyle); label("
A
A
A
", (3.8732067323088435,2.925600853925651), NE * labelscalefactor); dot((-0.7368327629589799,-0.5920813291311117),dotstyle); label("
B
B
B
", (-1.1,-0.7565817154670613), NE * labelscalefactor); dot((5.672613975760373,-0.636902634996282),dotstyle); label("
C
C
C
", (5.763466626982254,-0.7784344310124186), NE * labelscalefactor); dot((4.569287648059735,1.430279997142299),dotstyle); label("
D
D
D
", (4.692683565259744,1.5051743434774234), NE * labelscalefactor); dot((1.8844000180622977,1.3644681598392678),dotstyle); label("
E
E
E
", (1.775346039954538,1.4942479857047448), NE * labelscalefactor); dot((2.937230516274804,0.8082418657164665),linewidth(4.pt) + dotstyle); label("
F
F
F
", (2.889834532767763,0.954), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
5
1
Hide problems
LOTS of recurrence!
There is a rectangular plot of size
1
×
n
1 \times n
1
×
n
. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size
1
×
1
1 \times 1
1
×
1
, the blue tiles are of size
1
×
1
1 \times 1
1
×
1
and the black tiles are of size
1
×
2
1 \times 2
1
×
2
. Let
t
n
t_n
t
n
denote the number of ways this can be done. For example, clearly
t
1
=
2
t_1 = 2
t
1
=
2
because we can have either a red or a blue tile. Also
t
2
=
5
t_2 = 5
t
2
=
5
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile. [*]Prove that
t
2
n
+
1
=
t
n
(
t
n
−
1
+
t
n
+
1
)
t_{2n+1} = t_n(t_{n-1} + t_{n+1})
t
2
n
+
1
=
t
n
(
t
n
−
1
+
t
n
+
1
)
for all
n
>
1
n > 1
n
>
1
.[*]Prove that
t
n
=
∑
d
≥
0
(
n
−
d
d
)
2
n
−
2
d
t_n = \sum_{d \ge 0} \binom{n-d}{d}2^{n-2d}
t
n
=
∑
d
≥
0
(
d
n
−
d
)
2
n
−
2
d
for all
n
>
0
n >0
n
>
0
.Here,
(
m
r
)
=
{
m
!
r
!
(
m
−
r
)
!
,
if
0
≤
r
≤
m
,
0
,
otherwise
\binom{m}{r} = \begin{cases} \dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\ 0, &\text{ otherwise} \end{cases}
(
r
m
)
=
⎩
⎨
⎧
r
!
(
m
−
r
)!
m
!
,
0
,
if 0
≤
r
≤
m
,
otherwise
for integers
m
,
r
m,r
m
,
r
.
7
1
Hide problems
Integer Symmetric Polynomials
(a) Let
n
≥
1
n \geq 1
n
≥
1
be an integer. Prove that
X
n
+
Y
n
+
Z
n
X^n+Y^n+Z^n
X
n
+
Y
n
+
Z
n
can be written as a polynomial with integer coefficients in the variables
α
=
X
+
Y
+
Z
\alpha=X+Y+Z
α
=
X
+
Y
+
Z
,
β
=
X
Y
+
Y
Z
+
Z
X
\beta= XY+YZ+ZX
β
=
X
Y
+
Y
Z
+
ZX
and
γ
=
X
Y
Z
\gamma = XYZ
γ
=
X
Y
Z
. (b) Let
G
n
=
x
n
sin
(
n
A
)
+
y
n
sin
(
n
B
)
+
z
n
sin
(
n
C
)
G_n=x^n \sin(nA)+y^n \sin(nB)+z^n \sin(nC)
G
n
=
x
n
sin
(
n
A
)
+
y
n
sin
(
n
B
)
+
z
n
sin
(
n
C
)
, where
x
,
y
,
z
,
A
,
B
,
C
x,y,z, A,B,C
x
,
y
,
z
,
A
,
B
,
C
are real numbers such that
A
+
B
+
C
A+B+C
A
+
B
+
C
is an integral multiple of
π
\pi
π
. Using (a) or otherwise show that if
G
1
=
G
2
=
0
G_1=G_2=0
G
1
=
G
2
=
0
, then
G
n
=
0
G_n=0
G
n
=
0
for all positive integers
n
n
n
.
2
1
Hide problems
Half angle cosine sequence its limit at infinity!
Let
a
0
=
1
2
a_0 = \frac{1}{2}
a
0
=
2
1
and
a
n
a_n
a
n
be defined inductively by
a
n
=
1
+
a
n
−
1
2
,
n
≥
1
.
a_n = \sqrt{\frac{1+a_{n-1}}{2}} \text{, $n \ge 1$.}
a
n
=
2
1
+
a
n
−
1
,
n
≥
1.
[*] Show that for
n
=
0
,
1
,
2
,
…
,
n = 0,1,2, \ldots,
n
=
0
,
1
,
2
,
…
,
a
n
=
cos
(
θ
n
)
for some
0
<
θ
n
<
π
2
,
a_n = \cos(\theta_n) \text{ for some $0 < \theta_n < \frac{\pi}{2}$, }
a
n
=
cos
(
θ
n
)
for some 0
<
θ
n
<
2
π
,
and determine
θ
n
\theta_n
θ
n
. [*] Using (a) or otherwise, calculate
lim
n
→
∞
4
n
(
1
−
a
n
)
.
\lim_{n \to \infty} 4^n (1 - a_n).
n
→
∞
lim
4
n
(
1
−
a
n
)
.
4
1
Hide problems
2023 factors and perfect cube
Let
n
1
,
n
2
,
⋯
,
n
51
n_1, n_2, \cdots , n_{51}
n
1
,
n
2
,
⋯
,
n
51
be distinct natural numbers each of which has exactly
2023
2023
2023
positive integer factors. For instance,
2
2022
2^{2022}
2
2022
has exactly
2023
2023
2023
positive integer factors
1
,
2
,
2
2
,
2
3
,
⋯
2
2021
,
2
2022
1,2, 2^{2}, 2^{3}, \cdots 2^{2021}, 2^{2022}
1
,
2
,
2
2
,
2
3
,
⋯
2
2021
,
2
2022
. Assume that no prime larger than
11
11
11
divides any of the
n
i
n_{i}
n
i
's. Show that there must be some perfect cube among the
n
i
n_{i}
n
i
's.
6
1
Hide problems
Reciprocal Sums and Boring Bounds
Let
{
u
n
}
n
≥
1
\{u_n\}_{n \ge 1}
{
u
n
}
n
≥
1
be a sequence of real numbers defined as
u
1
=
1
u_1 = 1
u
1
=
1
and
u
n
+
1
=
u
n
+
1
u
n
for all
n
≥
1
.
u_{n+1} = u_n + \frac{1}{u_n} \text{ for all $n \ge 1$.}
u
n
+
1
=
u
n
+
u
n
1
for all
n
≥
1.
Prove that
u
n
≤
3
n
2
u_n \le \frac{3\sqrt{n}}{2}
u
n
≤
2
3
n
for all
n
n
n
.
1
1
Hide problems
The odd digit powers
Determine all integers
n
>
1
n>1
n
>
1
such that every power of
n
n
n
has an odd number of digits.
8
1
Hide problems
Inequality in MVT?
Let
f
:
[
0
,
1
]
→
R
f \colon [0,1] \to \mathbb{R}
f
:
[
0
,
1
]
→
R
be a continuous function which is differentiable on
(
0
,
1
)
(0,1)
(
0
,
1
)
. Prove that either
f
(
x
)
=
a
x
+
b
f(x) = ax + b
f
(
x
)
=
a
x
+
b
for all
x
∈
[
0
,
1
]
x \in [0,1]
x
∈
[
0
,
1
]
for some constants
a
,
b
∈
R
a,b \in \mathbb{R}
a
,
b
∈
R
or there exists
t
∈
(
0
,
1
)
t \in (0,1)
t
∈
(
0
,
1
)
such that
∣
f
(
1
)
−
f
(
0
)
∣
<
∣
f
′
(
t
)
∣
|f(1) - f(0)| < |f'(t)|
∣
f
(
1
)
−
f
(
0
)
∣
<
∣
f
′
(
t
)
∣
.