Key Concepts: Combinatorial Construction and Counting

Hello young mathematicians! Today, we’re going to dive into a fun and interesting math problem. This problem comes from the world of Math Olympiads, but don’t worry, it’s not as scary as it sounds. We’ll break it down step by step, and I promise you’ll have fun along the way!

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Imagine you have a row of eight boxes, and you need to fill them with the numbers 1 to 8. Sounds easy, right? But here’s the twist: between each pair of boxes, there’s a circle. In each circle, you need to write a number that is equal to de diffrence of twice the number in the box on the left and the number in the box on the right(the bigger minus the smaller).

So, if we have the numbers $a_1, a_2, …, a_8$ in the boxes, the numbers in the circles will be $\left|2a_1 - a_2\right|, \left|2a_2 - a_3\right|, …, \left|2a_7 - a_8\right|$.

We’re looking for the maximum possible sum of the numbers in the circles, which we’ll call $S$. So, $S = \left|2a_1 - a_2\right| + \left|2a_2 - a_3\right| + … + \left|2a_7 - a_8\right|$.

Now, let’s think about this. Each term in the sum, $\left|2a_i - a_{i+1}\right|$, can be written as $b_i - c_i$, where $b_i$ is either $2a_i$ or $a_{i+1}$, and $c_i$ is the other one.

So, we can rewrite $S$ as $b_1 + b_2 + … + b_7 - (c_1 + c_2 + … + c_7)$.

The possible values for $b_i$ and $c_i$ are the numbers 1 to 8 and their doubles. If we want to maximize $S$, we should make the $b_i$’s as large as possible and the $c_i$’s as small as possible.

So, we sort all possible values in ascending order: $(1,2,2 * 1,3,4,2 * 2,5,6,2 * 3,7,8,2 * 4,2 * 5,2 * 6,2 * 7 ,2 * 8)$.

We assign the smallest seven numbers to the $c_i$’s and the largest seven numbers to the $b_i$’s.

This gives us $b_i$ in $(2 * 8,2 * 7,2 * 6,2 * 5,2 * 4,8,7)$ and $c_i$ in $(1,2,2 * 1,3,4,2 * 2,5)$.

So, the maximum value of $S$ is $ \displaylines{ (2 * 8+2 * 7+2 * 6+2 * 5+2 * 4+8+7) - (1+2+2 * 1+3+4+2 * 2+5) \newline = 75 - 21 = 54 } $.

Finally, let’s see how to fill the boxes to achieve this maximum sum. We notice that the number $3$ appears only once (in $c_i$), and the number $2 * 6$ appears only once (in $b_i$). So, we put $6$ in the leftmost box and $3$ in the rightmost box. The remaining numbers can be placed in any order.

For example, if we fill the boxes with the numbers 6, 8, 7, 5, 4, 2, 1, 3, we get the maximum sum $S = 54$.

And there you have it! A fun and challenging Math Olympiad problem, solved step by step. Remember, math is all about problem-solving and logical thinking, so keep practicing and you’ll become a math whiz in no time!