Difference between revisions of "2007 IMO Problems/Problem 1"
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<cmath>x_p-x_q>a_p-a_q-d=a_p-a_q-a_p+a_q=0</cmath> | <cmath>x_p-x_q>a_p-a_q-d=a_p-a_q-a_p+a_q=0</cmath> | ||
− | <math>x_p>x_q</math> --- contradiction. | + | <math>x_p>x_q</math> --- contradiction (<math>p\le q \rightarrow x_p\le x_q</math>). |
<br/> | <br/> | ||
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<math>-\dfrac{d}{2} \le (a_m-a_i)\dfrac{d}{2} \le \dfrac{d}{2}</math> | <math>-\dfrac{d}{2} \le (a_m-a_i)\dfrac{d}{2} \le \dfrac{d}{2}</math> | ||
− | <math>\left|(a_m-a_i)\dfrac{d}{2}\right|=|x_i-a_i|\le \frac{d}{2}</math> | + | <math>\left|(a_m-a_i)-\dfrac{d}{2}\right|=|x_i-a_i|\le \frac{d}{2}</math> |
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Latest revision as of 23:15, 31 March 2010
Problem
Real numbers are given. For each
(
) define
and let
.
(a) Prove that, for any real numbers ,
(b) Show that there are real numbers such that equality holds in (*)
Solution
Since , all
can be expressed as
, where
.
Thus, can be expressed as
for some
and
,
Lemma)
Assume for contradiction that , then for all
,
Then, is a non-decreasing function, which means,
, and
, which means,
.
Then, and contradiction.
a)
Case 1)
If ,
is the maximum of a set of non-negative number, which must be at least
.
Case 2) (We can ignore
because of lemma)
Using the fact that can be expressed as
for some
and
,
.
Assume for contradiction that .
Then, ,
.
, and
Thus, and
.
Subtracting the two inequality, we will obtain:
--- contradiction (
).
Thus,
(b)
A set of where the equality in (*) holds is:
Since is a non-decreasing function,
is non-decreasing.
:
Let ,
.
Thus, (
because
is the max of a set including
)
Since and
,
This is written by Mo Lam--- who is a horrible proof writer, so please fix the proof for me. Thank you. O, also the formatting.
--> Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2007 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |
- <url>viewtopic.php?p=894656#p894656 Discussion on AoPS/MathLinks</url>