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Source: Bosnia and Herzegovina 2011

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118 posts

RaleD

#1 h May 16, 2011, 10:35 AM • 2 Y

Y by Adventure10, Mango247

Find maximum value of number $a$ such that for any arrangement of numbers $1,2,\ldots ,10$ on a circle, we can find three consecutive numbers such their sum bigger or equal than $a$ .

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#2 h May 17, 2011, 12:02 PM • 2 Y

Y by Adventure10, Mango247

We will prove that the maximum value of $a$ is larger than or equal to 18.

First, define $x_1, x_2, \cdots, x_{10}$ to be the numbers on the circle consecutively, $x_{11} = x_1$ , $x_{12} = x_2$ , and $y_k = x_k + x_{k+1} + x_{k+2}$ for $1 \le k \le 10$ .

Note that since $\sum x_i = 55$ , $\sum y_i = 165$ , so that there exists some $y_i$ such that $y_i \ge 17$ . Then $a \ge 17$ .

Let us assume the contrary; that is, $a \le 17$ . So since $a \ge 17$ , $a$ must be equal to 17, so all $y_i$ must be less than or equal to 17.

Note that $y_i$ and $y_{i+1}$ can't be equal for any $1 \le i \le 10$ (assume $y_{11} = y_1$ ), because if they are equal, then $x_i$ and $x_{i+3}$ are equal (assume $x_{13} = x_3$ ). So there are at most five $y_i$ which are equal to 17; WLOG assume $y_1 = y_3 = y_5 = y_7 = y_9 = 17$ . This causes $y_2, y_4, y_6, y_8, y_{10} \le 16$ . However, since $\sum y_i = 165$ , $y_2 + y_4 + y_6 + y_8 + y_{10} = 80$ , so $y_2 = y_4 = y_6 = y_8 = y_{10} = 16$ .

Now, note that $x_1 = x_4 + 1$ , and $x_4 = x_7 - 1$ by comparing $y_1$ against $y_2$ and $y_4$ against $y_5$ . This means $x_1 = x_7$ , contradiction. So $a$ cannot be equal to 17, contradiction.

Now, this leaves us with the necessity of the proof that $a \le 18$ . Note the following arrangement:
10 1 7 9 2 6 4 8 3 5
No three consecutive numbers has a sum of larger than 18, so $a \le 18$ .

So the maximum value of $a$ is $\boxed{18}$ .

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#3 h Jun 30, 2011, 11:41 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

There is a much simpler argument for $a\geq 18$ . The other nine numbers than $1$ form three groups of three adjacent numbers each, of total sum $54=3\cdot 18$ , so (at least) one of the grup sums to (at least) $18$ .

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RaleD

#4 h Jul 11, 2011, 2:40 PM • 1 Y

Y by Adventure10

Yes, mavropnevma, that was official solution. My was same as the one of chaotic_iak.

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#5 h Dec 15, 2013, 2:50 AM • 2 Y

Y by Adventure10, Mango247

RaleD wrote:

Find maximum value of number $a$ such that for any arrangement of numbers $1,2,\ldots ,10$ on a circle, we can find three consecutive numbers such their sum bigger or equal than $a$ .

Find maximum value of number $a$ such that for any arrangement of numbers $1,2,\ldots ,10$ on a line, we can find three consecutive numbers such their sum bigger or equal than $a$ ?.

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