IMO 2014/1

Let a0<a1<a2 a_0 < a_1 < a_2 \ldots be an infinite sequence of positive integers. Prove that there exists a unique integer n1 n \geq 1 such that

an<a0+a1++annan+1. a_n < \frac{a_0 + a_1 + \ldots + a_n } { n} \leq a_{n+1} .

Note by Calvin Lin
5 years, 1 month ago

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Let's call Sk=a0+a1+...+ak S_k = a_0 + a_1 + ... + a_k and suppose Skk>ak+1  mk \frac { { S }_{ k } }{ k } >{ a }_{ k+1 } \; \forall m \leq k Then we have kSk+Sk>kak+1+kSkSkk>Sk+1k+1Sk+1k+1<Skk<Sk1k1<...<S1=a0+a1 k\cdot { S }_{ k }+{ S }_{ k }>k\cdot { a }_{ k+1 }+k\cdot{ S }_{ k }\quad \Leftrightarrow \quad \frac { { S }_{ k } }{ k } >\frac { { S }_{ k+1 } }{ k+1 }\\ \quad \Leftrightarrow \quad \frac { { S }_{ k+1 } }{ k+1 } <\frac { { S }_{ k } }{ k } <\frac { { S }_{ k-1 } }{ k-1 } <...<{ S }_{ 1 }={ a }_{ 0 }+{ a }_{ 1 } Hence, we prove by induction 2(a0+a1)>a0+a1+a2a0+a1>a2kak+1<Sk<k(a0+a1)ak+1<(a0+a1) 2(a_{ 0 }+{ a }_{ 1 })>{ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }\quad \Leftrightarrow \quad { a }_{ 0 }+{ a }_{ 1 }>{ a }_{ 2 }\\ k{ a }_{ k+1 }<{ S }_{ k }<k(a_{ 0 }+{ a }_{ 1 })\quad \Leftrightarrow \quad { a }_{ k+1 }<(a_{ 0 }+{ a }_{ 1 }) Because a0+a1 a_{ 0 }+{ a }_{ 1 } is a sum of integer number, there must be a unique number in the sequence for which aq+1a0+a1 a_{q+1} \geq a_{ 0 }+{ a }_{ 1 } and aq<a0+a1 a_{q} < a_{ 0 }+{ a }_{ 1 } . We now show that q q is the required number. We have alredy proved that Sqqaq+1 \frac { { S }_{ q } }{q } \leq { a }_{ q+1 } by contraddiction! Thus we have only to show that aq<Sqqqaqaq<Sqaqaq<Sq1q1{ a }_{ q }<\frac { S_{ q } }{ q } \quad \Leftrightarrow \quad q{ \cdot a }_{ q }-a_{ q }<S_{ q }-a_{ q }\quad \Leftrightarrow \quad { a }_{ q }<\frac { S_{ q-1 } }{ q-1 } And because Sq1(q1)+aq(q1)<Sq1+Sq1(q1)Sqq<Sq1q1Sqq<Sq1q1<Sq2q2<...<S1=a0+a1 S_{ q-1 }(q-1)+{ a }_{ q }(q-1)<S_{ q-1 }+S_{ q-1 }(q-1)\quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 }\\ \quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 } <\frac { S_{ q-2 } }{ q-2 } <...<S_{ 1 }={ a }_{ 0 }+{ a }_{ 1 } Excactly as done beofre we must have aq<a0+a1 a_q < a_0 + a_1 . That is true. And we are done.

Andrea Gallese - 5 years, 1 month ago

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Beautiful solution

Scrub Lord - 2 years, 2 months ago

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Question 1 was way easier than I expected it to be. Bummed that there were no proper number theory problems.

Mursalin Habib - 5 years, 1 month ago

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In recent years, they have made question 1 (and sometimes 4) pretty easy and direct to approach, so that all of the participants (who come from a wide range of ability level) have something to do, and have a chance of geting a HM.

Calvin Lin Staff - 5 years, 1 month ago

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This was indeed very nice, but also pretty easy, even for an IMO P1. I have posted my solution in AoPS.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3542324&#p3542324

Sreejato Bhattacharya - 5 years, 1 month ago

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Here's a conjecture:

The last possible nn that satisfies an<a0+a1++anna_n < \dfrac{a_0+a_1+\cdots +a_n}{n} also satisfies a0+a1++annan+1\dfrac{a_0+a_1+\cdots +a_n}{n}\le a_{n+1}. This is just a guess though.

These later posts will be what I have came up with.

Daniel Liu - 5 years, 1 month ago

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Update: letting a0=xa_0=x and a1=ya_1=y, we find that in order for ak<a0+a1++akka_k < \dfrac{a_0+a_1+\cdots +a_k}{k}, we must have ak<x+ya_k < x+y. However, since x+yx+y is finite, and the sequence is strictly increasing, we must have that there is a maximum kk that satisfies this condition.

So let kk be the maximum such that it still satisfies the condition. Thus, ak+1x+ya_{k+1} \ge x+y. However, a0+a1++akk<x+y+(x+y)++(x+y)k=x+y\dfrac{a_0+a_1+\cdots +a_k}{k} < \dfrac{x+y+(x+y)+\cdots +(x+y)}{k}=x+y. Thus, a0+a1++akk<ak+1\dfrac{a_0+a_1+\cdots +a_k}{k} < a_{k+1}.

This seems really sketchy though. Something seems wrong... one of my inequalities was strict when forming the right inequality, but the inequality in the problem is not strict...

EDIT: found out why mine was strict and the problem's wasn't. As long as the maximum kk satisfies k2k\ge 2, then the upper bound inequality is strict, just like my result. However, the special case of the maximum kk is k=1k=1 yields a non-strict inequality.

TO DO: prove uniqueness of satisfying both inequalities.

Daniel Liu - 5 years, 1 month ago

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I realize that I have to tighten the bound for each ak<x+ya_k < x+y to something more strict, or something like that in order to prove the uniqueness of that one kk that satisfies both inequalities. I feel like I'm almost there, but just can't get over the last little bump. I need to somehow use the ak<ak+1a_k < a_{k+1} to my advantage...

Daniel Liu - 5 years, 1 month ago

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