IMO 2011 #1

by Wolstenholme, Oct 29, 2014, 2:53 PM

Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$ . Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$ . Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$ .

Solution:

Assume WLOG that $a_1 < a_2 < a_3 < a_4$ . Then since $s_A > a_3 + a_4 > a_2 + a_4 > \frac{s_A}{2}$ we have that $a_3 + a_4$ and $a_2 + a_4$ cannot divide $s_A$ . Therefore the largest possible value of $n_A$ is $4$ . From now on, assume $n_A = 4$ .

Then since $a_1 + a_4 \vert s_A$ we have that $a_1 + a_4 \vert a_2 + a_3$ and vice-versa, so $a_4 = a_2 + a_3 - a_1$ .

So after some manipulation we get the identities $a_1 + a_2 \vert 2(a_2 + a_3)$ and $a_1 + a_3 \vert 2(a_2 + a_3).$ But since $a_1 + a_3 > \frac{a_2 + a_3}{2}$ we have that $2(a_2 + a_3) = k(a_1 + a_3)$ where $k = 3$ or $k = 4$ . We now proceed with casework.

Case 1: $k = 3$

Then $2(a_2 + a_3) = 3(a_1 + a_3)$ so $a_3 = 2a_2 - 3a_1$ . Plugging this into our other identity and manipulating we obtain $a_1 + a_2 \vert 12a_1.$ But since $a_1 + a_2 > 2a_1$ we have that $12a_1 = j(a_1 + a_2)$ where $j = 1$ or $2$ or $3$ or $4$ or $5.$

Case 1.1: $j = 1$

We get the solution set $(n, 11n, 19n, 29n)$ for all $n \in \mathbb{N}$

Case 1.2 $j = 2$

We get the solution set $(n, 5n, 7n, 11n)$ for all $n \in \mathbb{N}$

Case 1.3 $j = 3$

We get $a_3 = a_2,$ contradiction!

Case 1.4: $j = 4$

We get $a_3 = a_1,$ contradiction!

Case 1.5: $j = 5$

We get that $a_3$ is negative, contradiction!

Case 2: $k = 4$

Then $2(a_2 + a_3) = 4(a_1 + a_3)$ so $a_3 = a_2 - 2a_1$ . Plugging this into our other identity and manipulating we obtain $a_1 + a_2 \vert 8a_1.$ But since $a_1 + a_2 > 2a_1$ we have that $8a_1 = j(a_1 + a_2)$ where $j = 1$ or $2$ or $3.$

Case 2.1: $j = 1$

We get that $a_3 < a_2,$ contradiction!

Case 2.2: $j = 2$

We get that $a_3 = a_1,$ contradiction!

Case 2.3: $j = 3$

We get that $a_3$ is negative, contradiction!

Therefore the only solutions are $\boxed{(a_1, a_2, a_3, a_4) = (n, 5n, 7n, 11n), (n, 11n, 19n, 29n), n \in \mathbb{N}}$ and their permutations.

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
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by fukano_2, Jun 14, 2020, 6:24 AM
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by mathleticguyyy, Jun 5, 2019, 8:26 PM
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by songssari, Jun 12, 2016, 8:21 AM
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by briantix, Mar 18, 2016, 9:57 PM
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by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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IMO 2011 #1

by Wolstenholme, Oct 29, 2014, 2:53 PM

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