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Contests
National and Regional Contests
Indonesia Contests
Indonesia MO
2019 Indonesia MO
4
4
Part of
2019 Indonesia MO
Problems
(1)
INAMO 2019 P4 - Triangle equivalence
Source: INAMO 2019 P4
7/3/2019
Let us define a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
t
r
i
a
n
g
l
e
e
q
u
i
v
a
l
e
n
c
e
<
/
s
p
a
n
>
<span class='latex-italic'>triangle equivalence</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
t
r
ian
g
l
ee
q
u
i
v
a
l
e
n
ce
<
/
s
p
an
>
a group of numbers that can be arranged as shown
a
+
b
=
c
a+b=c
a
+
b
=
c
d
+
e
+
f
=
g
+
h
d+e+f=g+h
d
+
e
+
f
=
g
+
h
i
+
j
+
k
+
l
=
m
+
n
+
o
i+j+k+l=m+n+o
i
+
j
+
k
+
l
=
m
+
n
+
o
and so on...Where at the
j
j
j
-th row, the left hand side has
j
+
1
j+1
j
+
1
terms and the right hand side has
j
j
j
terms.Now, we are given the first
N
2
N^2
N
2
positive integers, where
N
N
N
is a positive integer. Suppose we eliminate any one number that has the same parity with
N
N
N
.Prove that the remaining
N
2
−
1
N^2-1
N
2
−
1
numbers can be formed into a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
t
r
i
a
n
g
l
e
e
q
u
i
v
a
l
e
n
c
e
<
/
s
p
a
n
>
<span class='latex-italic'>triangle equivalence</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
t
r
ian
g
l
ee
q
u
i
v
a
l
e
n
ce
<
/
s
p
an
>
.For example, if
10
10
10
is eliminated from the first
16
16
16
numbers, the remaining numbers can be arranged into a
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
t
r
i
a
n
g
l
e
e
q
u
i
v
a
l
e
n
c
e
<
/
s
p
a
n
>
<span class='latex-italic'>triangle equivalence</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
i
t
a
l
i
c
′
>
t
r
ian
g
l
ee
q
u
i
v
a
l
e
n
ce
<
/
s
p
an
>
as shown.
1
+
3
=
4
1+3=4
1
+
3
=
4
2
+
5
+
8
=
6
+
9
2+5+8=6+9
2
+
5
+
8
=
6
+
9
7
+
11
+
12
+
14
=
13
+
15
+
16
7+11+12+14=13+15+16
7
+
11
+
12
+
14
=
13
+
15
+
16
algebra