International Mathematics Olympiad Problems - 2017 | International Mathematics Olympiad (IMO) for Class 10 PDF Download

 Page 1


Tuesday, July 18, 2017
Problem 1. For each integer a
0
> 1, de?ne the sequence a
0
, a
1
, a
2
, ... by:
a
n+1
=

p
a
n
if
p
a
n
is an integer;
a
n
+3 otherwise,
for each n> 0.
Determine all values of a
0
for which there is a number A such that a
n
=A for in?nitely many values
of n.
Problem 2. Let R be the set of real numbers. Determine all functions f:R!R such that, for
all real numbers x and y,
f (f(x)f(y))+f(x+y) =f(xy):
Problem 3. A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit’s
starting point, A
0
, and the hunter’s starting point, B
0
, are the same. Aftern1 rounds of the game,
the rabbit is at point A
n1
and the hunter is at point B
n1
. In the n
th
round of the game, three
things occur in order.
(i) The rabbit moves invisibly to a point A
n
such that the distance between A
n1
and A
n
is
exactly 1.
(ii) AtrackingdevicereportsapointP
n
tothehunter. Theonlyguaranteeprovidedbythetracking
device to the hunter is that the distance between P
n
and A
n
is at most 1.
(iii) The hunter moves visibly to a point B
n
such that the distance between B
n1
and B
n
is
exactly 1.
Is it always possible, no matter how the rabbit moves, and no matter what points are reported
by the tracking device, for the hunter to choose her moves so that after 10
9
rounds she can ensure
that the distance between her and the rabbit is at most 100?
Page 2


Tuesday, July 18, 2017
Problem 1. For each integer a
0
> 1, de?ne the sequence a
0
, a
1
, a
2
, ... by:
a
n+1
=

p
a
n
if
p
a
n
is an integer;
a
n
+3 otherwise,
for each n> 0.
Determine all values of a
0
for which there is a number A such that a
n
=A for in?nitely many values
of n.
Problem 2. Let R be the set of real numbers. Determine all functions f:R!R such that, for
all real numbers x and y,
f (f(x)f(y))+f(x+y) =f(xy):
Problem 3. A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit’s
starting point, A
0
, and the hunter’s starting point, B
0
, are the same. Aftern1 rounds of the game,
the rabbit is at point A
n1
and the hunter is at point B
n1
. In the n
th
round of the game, three
things occur in order.
(i) The rabbit moves invisibly to a point A
n
such that the distance between A
n1
and A
n
is
exactly 1.
(ii) AtrackingdevicereportsapointP
n
tothehunter. Theonlyguaranteeprovidedbythetracking
device to the hunter is that the distance between P
n
and A
n
is at most 1.
(iii) The hunter moves visibly to a point B
n
such that the distance between B
n1
and B
n
is
exactly 1.
Is it always possible, no matter how the rabbit moves, and no matter what points are reported
by the tracking device, for the hunter to choose her moves so that after 10
9
rounds she can ensure
that the distance between her and the rabbit is at most 100?
Wednesday, July 19, 2017
Problem 4. Let R and S be di?erent points on a circle 
 such that RS is not a diameter. Let `
be the tangent line to 
 at R. Point T is such that S is the midpoint of the line segment RT. Point
J is chosen on the shorter arc RS of 
 so that the circumcircle  of triangle JST intersects ` at two
distinct points. Let A be the common point of  and ` that is closer to R. Line AJ meets 
 again
at K. Prove that the line KT is tangent to .
Problem 5. An integer N> 2 is given. A collection of N(N + 1) soccer players, no two of whom
are of the same height, stand in a row. Sir Alex wants to remove N(N  1) players from this row
leaving a new row of 2N players in which the following N conditions hold:
(1) no one stands between the two tallest players,
(2) no one stands between the third and fourth tallest players,
.
.
.
(N) no one stands between the two shortest players.
Show that this is always possible.
Problem 6. An ordered pair (x;y) of integers is a primitive point if the greatest common divisor
of x and y is 1. Given a ?nite set S of primitive points, prove that there exist a positive integer n
and integers a
0
, a
1
, ..., a
n
such that, for each (x;y) in S, we have:
a
0
x
n
+a
1
x
n1
y +a
2
x
n2
y
2
+


 +a
n1
xy
n1
+a
n
y
n
= 1: