MathDB
Problems
Contests
National and Regional Contests
Ireland Contests
Ireland National Math Olympiad
1988 Irish Math Olympiad
1988 Irish Math Olympiad
Part of
Ireland National Math Olympiad
Subcontests
(12)
11
1
Hide problems
Determining Decimal Expansion without Division
If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of
1
/
a
1/a
1/
a
,
a
>
0
a>0
a
>
0
, to use the iteration x_{k+1}=x_k(2-ax_k), k=0,1,2,\dots , where
x
0
x_0
x
0
is a selected “starting” value. Find the limitations, if any, on the starting values
x
0
x_0
x
0
, in order that the above iteration converges to the desired value
1
/
a
1/a
1/
a
.
10
1
Hide problems
Interesting Inequality
Let
0
≤
x
≤
1
0\le x\le 1
0
≤
x
≤
1
. Show that if
n
n
n
is any positive integer, then
(
1
+
x
)
n
≥
(
1
−
x
)
n
+
2
n
x
(
1
−
x
2
)
n
−
1
2
(1+x)^n\ge (1-x)^n+2nx(1-x^2)^{\frac{n-1}{2}}
(
1
+
x
)
n
≥
(
1
−
x
)
n
+
2
n
x
(
1
−
x
2
)
2
n
−
1
.
9
1
Hide problems
Peculiar Years
The year
1978
1978
1978
was “peculiar” in that the sum of the numbers formed with the first two digits and the last two digits is equal to the number formed with the middle two digits, i.e.,
19
+
78
=
97
19+78=97
19
+
78
=
97
. What was the last previous peculiar year, and when will the next one occur?
8
1
Hide problems
Sequence with all Integers
Let
x
1
,
x
2
,
x
3
,
…
x_1,x_2,x_3,\dots
x
1
,
x
2
,
x
3
,
…
be sequence of nonzero real numbers satisfying x_n=\frac{x_{n-2}x_{n-1}}{2x_{n-2}-x_{n-1}}, n=3,4,5,\dots Establish necessary and sufficient conditions on
x
1
,
x
2
x_1,x_2
x
1
,
x
2
for
x
n
x_n
x
n
to be an integer for infinitely many values of
n
n
n
.
7
1
Hide problems
Cubic with Horizontal Chord
A function
f
f
f
, defined on the set of real numbers
R
\mathbb{R}
R
is said to have a horizontal chord of length
a
>
0
a>0
a
>
0
if there is a real number
x
x
x
such that
f
(
a
+
x
)
=
f
(
x
)
f(a+x)=f(x)
f
(
a
+
x
)
=
f
(
x
)
. Show that the cubic f(x)=x^3-x (x\in \mathbb{R}) has a horizontal chord of length
a
a
a
if, and only if,
0
<
a
≤
2
0<a\le 2
0
<
a
≤
2
.
6
1
Hide problems
Group Among n Blocks with Weights
Suppose you are given
n
n
n
blocks, each of which weighs an integral number of pounds, but less than
n
n
n
pounds. Suppose also that the total weight of the
n
n
n
blocks is less than
2
n
2n
2
n
pounds. Prove that the blocks can be divided into two groups, one of which weighs exactly
n
n
n
pounds.
3
2
Hide problems
Triangle Inscribed in Circle
A
B
C
ABC
A
BC
is a triangle inscribed in a circle, and
E
E
E
is the mid-point of the arc subtended by
B
C
BC
BC
on the side remote from
A
A
A
. If through
E
E
E
a diameter
E
D
ED
E
D
is drawn, show that the measure of the angle
D
E
A
DEA
D
E
A
is half the magnitude of the difference of the measures of the angles at
B
B
B
and
C
C
C
.
Finding Number of Bus Routes in a City
A city has a system of bus routes laid out in such a way that (a) there are exactly
11
11
11
bus stops on each route; (b) it is possible to travel between any two bus stops without changing routes; (c) any two bus routes have exactly one bus stop in common. What is the number of bus routes in the city?
1
2
Hide problems
Pyramid Glued to Tetrahedron
A pyramid with a square base, and all its edges of length
2
2
2
, is joined to a regular tetrahedron, whose edges are also of length
2
2
2
, by gluing together two of the triangular faces. Find the sum of the lengths of the edges of the resulting solid.
Conditions on two Triangles
The triangles
A
B
G
ABG
A
BG
and
A
E
F
AEF
A
EF
are in the same plane. Between them the following conditions hold: (a)
E
E
E
is the mid-point of
A
B
AB
A
B
; (b) points
A
,
G
A,G
A
,
G
and
F
F
F
are on the same line; (c) there is a point
C
C
C
at which
B
G
BG
BG
and
E
F
EF
EF
intersect; (d)
∣
C
E
∣
=
1
|CE|=1
∣
CE
∣
=
1
and
∣
A
C
∣
=
∣
A
E
∣
=
∣
F
G
∣
|AC|=|AE|=|FG|
∣
A
C
∣
=
∣
A
E
∣
=
∣
FG
∣
. Show that if
∣
A
G
∣
=
x
|AG|=x
∣
A
G
∣
=
x
, then
∣
A
B
∣
=
x
3
|AB|=x^3
∣
A
B
∣
=
x
3
.
12
1
Hide problems
Irish 1988 paper 1 #12
Prove that if
n
n
n
is a positive integer ,then
c
o
s
4
π
2
n
+
1
+
c
o
s
4
2
π
2
n
+
1
+
⋯
+
c
o
s
4
n
π
2
n
+
1
=
6
n
−
5
16
.
cos^4\frac{\pi}{2n+1}+cos^4\frac{2\pi}{2n+1}+\cdots+cos^4\frac{n\pi}{2n+1}=\frac{6n-5}{16}.
co
s
4
2
n
+
1
π
+
co
s
4
2
n
+
1
2
π
+
⋯
+
co
s
4
2
n
+
1
nπ
=
16
6
n
−
5
.
5
1
Hide problems
Choosing Problem
Problem: A person has seven friends and invites a different subset of three friends to dinner every night for one week (seven days). In how many ways can this be done so that all friends are invited at least once?
4
1
Hide problems
Triangle Log Problem
Problem: A mathematical moron is given the values b; c; A for a triangle ABC and is required to find a. He does this by using the cosine rule
a
2
=
b
2
+
c
2
−
2
b
c
c
o
s
A
a^2 = b^2 + c^2 - 2bccosA
a
2
=
b
2
+
c
2
−
2
b
ccos
A
and misapplying the low of the logarithm to this to get
l
o
g
a
2
=
l
o
g
b
2
+
l
o
g
c
2
−
l
o
g
(
2
b
c
c
o
s
A
)
log a^2 = log b^2 + log c^2 - log(2bc cos A)
l
o
g
a
2
=
l
o
g
b
2
+
l
o
g
c
2
−
l
o
g
(
2
b
ccos
A
)
He proceeds to evaluate the right-hand side correctly, takes the anti-logarithms and gets the correct answer. What can be said about the triangle ABC?
2
2
Hide problems
Square and Circumcircle Problem
A; B; C; D are the vertices of a square, and P is a point on the arc CD of its circumcircle. Prove that
∣
P
A
∣
2
−
∣
P
B
∣
2
=
∣
P
B
∣
.
∣
P
D
∣
−
∣
P
A
∣
.
∣
P
C
∣
|PA|^2 - |PB|^2 = |PB|.|PD| -|PA|.|PC|
∣
P
A
∣
2
−
∣
PB
∣
2
=
∣
PB
∣.∣
P
D
∣
−
∣
P
A
∣.∣
PC
∣
Can anyone here find the solution? I'm not great with geometry, so i tried turning it into co-ordinate geometry equations, but sadly to no avail. Thanks in advance.
this question be real
2. Let
x
1
,
.
.
.
,
x
n
x_1, . . . , x_n
x
1
,
...
,
x
n
be
n
n
n
integers, and let
p
p
p
be a positive integer, with
p
<
n
p < n
p
<
n
. Put
S
1
=
x
1
+
x
2
+
.
.
.
+
x
p
S_1 = x_1 + x_2 + . . . + x_p
S
1
=
x
1
+
x
2
+
...
+
x
p
T
1
=
x
p
+
1
+
x
p
+
2
+
.
.
.
+
x
n
T_1 = x_{p+1} + x_{p+2} + . . . + x_n
T
1
=
x
p
+
1
+
x
p
+
2
+
...
+
x
n
S
2
=
x
2
+
x
3
+
.
.
.
+
x
p
+
1
S_2 = x_2 + x_3 + . . . + x_{p+1}
S
2
=
x
2
+
x
3
+
...
+
x
p
+
1
T
2
=
x
p
+
2
+
x
p
+
3
+
.
.
.
+
x
n
+
x
1
T_2 = x_{p+2} + x_{p+3} + . . . + x_n + x_1
T
2
=
x
p
+
2
+
x
p
+
3
+
...
+
x
n
+
x
1
.
.
.
...
...
S
n
=
x
n
+
x
1
+
.
.
.
+
x
p
−
1
S_n=x_n+x_1+...+x_{p-1}
S
n
=
x
n
+
x
1
+
...
+
x
p
−
1
T
n
=
x
p
+
x
p
+
1
+
.
.
.
+
x
n
−
1
T_n=x_p+x_{p+1}+...+x_{n-1}
T
n
=
x
p
+
x
p
+
1
+
...
+
x
n
−
1
For
a
=
0
,
1
,
2
,
3
a = 0, 1, 2, 3
a
=
0
,
1
,
2
,
3
, and
b
=
0
,
1
,
2
,
3
b = 0, 1, 2, 3
b
=
0
,
1
,
2
,
3
, let
m
(
a
,
b
)
m(a, b)
m
(
a
,
b
)
be the number of numbers
i
i
i
,
1
≤
i
≤
n
1 \leq i \leq n
1
≤
i
≤
n
, such that
S
i
S_i
S
i
leaves remainder
a
a
a
on division by
4
4
4
and
T
i
T_i
T
i
leaves remainder
b
b
b
on division by
4
4
4
. Show that
m
(
1
,
3
)
m(1, 3)
m
(
1
,
3
)
and
m
(
3
,
1
)
m(3, 1)
m
(
3
,
1
)
leave the same remainder when divided by
4
4
4
if, and only if,
m
(
2
,
2
)
m(2, 2)
m
(
2
,
2
)
is even.