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National and Regional Contests
Moldova Contests
Moldova Team Selection Test
1998 Moldova Team Selection Test
5
$$\max_{1\leq i\leq n} |a_i-f(a_i)| \geq \max_{1\leq i\leq n} |a_i-b_i|$$
$$\max_{1\leq i\leq n} |a_i-f(a_i)| \geq \max_{1\leq i\leq n} |a_i-b_i|$$
Source: Moldova TST 1998
August 8, 2023
function
Problem Statement
Let
A
=
{
a
1
,
a
2
,
…
,
a
n
}
A=\{a_1,a_2,\ldots,a_n\}
A
=
{
a
1
,
a
2
,
…
,
a
n
}
be a set with
a
1
<
a
2
…
<
a
n
a_1<a_2\ldots<a_n
a
1
<
a
2
…
<
a
n
and
B
=
{
b
1
,
b
2
,
…
,
b
n
}
B=\{b_1,b_2,\ldots,b_n\}
B
=
{
b
1
,
b
2
,
…
,
b
n
}
be a set with
b
1
<
b
2
…
<
b
n
b_1<b_2\ldots<b_n
b
1
<
b
2
…
<
b
n
. Show that for every bijective function
f
:
A
→
B
f:A\rightarrow B
f
:
A
→
B
the following relation takes place
max
1
≤
i
≤
n
∣
a
i
−
f
(
a
i
)
∣
≥
max
1
≤
i
≤
n
∣
a
i
−
b
i
∣
.
\max_{1\leq i\leq n} |a_i-f(a_i)| \geq \max_{1\leq i\leq n} |a_i-b_i|.
1
≤
i
≤
n
max
∣
a
i
−
f
(
a
i
)
∣
≥
1
≤
i
≤
n
max
∣
a
i
−
b
i
∣.
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