This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

Let's call $S_k = a_0 + a_1 + ... + a_k$ and suppose
$\frac { { S }_{ k } }{ k } >{ a }_{ k+1 } \; \forall m \leq k$
Then we have
$k\cdot { S }_{ k }+{ S }_{ k }>k\cdot { a }_{ k+1 }+k\cdot{ S }_{ k }\quad \Leftrightarrow \quad \frac { { S }_{ k } }{ k } >\frac { { S }_{ k+1 } }{ k+1 }\\ \quad \Leftrightarrow \quad \frac { { S }_{ k+1 } }{ k+1 } <\frac { { S }_{ k } }{ k } <\frac { { S }_{ k-1 } }{ k-1 } <...<{ S }_{ 1 }={ a }_{ 0 }+{ a }_{ 1 }$
Hence, we prove by induction
$2(a_{ 0 }+{ a }_{ 1 })>{ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }\quad \Leftrightarrow \quad { a }_{ 0 }+{ a }_{ 1 }>{ a }_{ 2 }\\ k{ a }_{ k+1 }<{ S }_{ k }<k(a_{ 0 }+{ a }_{ 1 })\quad \Leftrightarrow \quad { a }_{ k+1 }<(a_{ 0 }+{ a }_{ 1 })$
Because $a_{ 0 }+{ a }_{ 1 }$ is a sum of integer number, there must be a unique number in the sequence for which $a_{q+1} \geq a_{ 0 }+{ a }_{ 1 }$ and $a_{q} < a_{ 0 }+{ a }_{ 1 }$. We now show that $q$ is the required number.
We have alredy proved that
$\frac { { S }_{ q } }{q } \leq { a }_{ q+1 }$
by contraddiction! Thus we have only to show that
${ a }_{ q }<\frac { S_{ q } }{ q } \quad \Leftrightarrow \quad q{ \cdot a }_{ q }-a_{ q }<S_{ q }-a_{ q }\quad \Leftrightarrow \quad { a }_{ q }<\frac { S_{ q-1 } }{ q-1 }$
And because
$S_{ q-1 }(q-1)+{ a }_{ q }(q-1)<S_{ q-1 }+S_{ q-1 }(q-1)\quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 }\\ \quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 } <\frac { S_{ q-2 } }{ q-2 } <...<S_{ 1 }={ a }_{ 0 }+{ a }_{ 1 }$
Excactly as done beofre we must have $a_q < a_0 + a_1$. That is true.
And we are done.

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.

The last possible $n$ that satisfies $a_n < \dfrac{a_0+a_1+\cdots +a_n}{n}$ also satisfies $\dfrac{a_0+a_1+\cdots +a_n}{n}\le a_{n+1}$. This is just a guess though.

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

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

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

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 $k$ satisfies $k\ge 2$, then the upper bound inequality is strict, just like my result. However, the special case of the maximum $k$ is $k=1$ yields a non-strict inequality.

TO DO: prove uniqueness of satisfying both inequalities.

I realize that I have to tighten the bound for each $a_k < x+y$ to something more strict, or something like that in order to prove the uniqueness of that one $k$ 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 $a_k < a_{k+1}$ to my advantage...

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet's call $S_k = a_0 + a_1 + ... + a_k$ and suppose $\frac { { S }_{ k } }{ k } >{ a }_{ k+1 } \; \forall m \leq k$ Then we have $k\cdot { S }_{ k }+{ S }_{ k }>k\cdot { a }_{ k+1 }+k\cdot{ S }_{ k }\quad \Leftrightarrow \quad \frac { { S }_{ k } }{ k } >\frac { { S }_{ k+1 } }{ k+1 }\\ \quad \Leftrightarrow \quad \frac { { S }_{ k+1 } }{ k+1 } <\frac { { S }_{ k } }{ k } <\frac { { S }_{ k-1 } }{ k-1 } <...<{ S }_{ 1 }={ a }_{ 0 }+{ a }_{ 1 }$ Hence, we prove by induction $2(a_{ 0 }+{ a }_{ 1 })>{ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }\quad \Leftrightarrow \quad { a }_{ 0 }+{ a }_{ 1 }>{ a }_{ 2 }\\ k{ a }_{ k+1 }<{ S }_{ k }<k(a_{ 0 }+{ a }_{ 1 })\quad \Leftrightarrow \quad { a }_{ k+1 }<(a_{ 0 }+{ a }_{ 1 })$ Because $a_{ 0 }+{ a }_{ 1 }$ is a sum of integer number, there must be a unique number in the sequence for which $a_{q+1} \geq a_{ 0 }+{ a }_{ 1 }$ and $a_{q} < a_{ 0 }+{ a }_{ 1 }$. We now show that $q$ is the required number. We have alredy proved that $\frac { { S }_{ q } }{q } \leq { a }_{ q+1 }$ by contraddiction! Thus we have only to show that ${ a }_{ q }<\frac { S_{ q } }{ q } \quad \Leftrightarrow \quad q{ \cdot a }_{ q }-a_{ q }<S_{ q }-a_{ q }\quad \Leftrightarrow \quad { a }_{ q }<\frac { S_{ q-1 } }{ q-1 }$ And because $S_{ q-1 }(q-1)+{ a }_{ q }(q-1)<S_{ q-1 }+S_{ q-1 }(q-1)\quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 }\\ \quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 } <\frac { S_{ q-2 } }{ q-2 } <...<S_{ 1 }={ a }_{ 0 }+{ a }_{ 1 }$ Excactly as done beofre we must have $a_q < a_0 + a_1$. That is true. And we are done.

Log in to reply

Beautiful solution

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

Here's a conjecture:

The last possible $n$ that satisfies $a_n < \dfrac{a_0+a_1+\cdots +a_n}{n}$ also satisfies $\dfrac{a_0+a_1+\cdots +a_n}{n}\le a_{n+1}$. This is just a guess though.

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

Log in to reply

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

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

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 $k$ satisfies $k\ge 2$, then the upper bound inequality is strict, just like my result. However, the special case of the maximum $k$ is $k=1$ yields a non-strict inequality.

TO DO: prove uniqueness of satisfying both inequalities.

Log in to reply

I realize that I have to tighten the bound for each $a_k < x+y$ to something more strict, or something like that in order to prove the uniqueness of that one $k$ 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 $a_k < a_{k+1}$ to my advantage...

Log in to reply