, 2023

From the given information, if $a$ and $b$ are in two consecutive squares, then $a+b$ goes in the circle between them. Since all of the numbers that we can use are positive, then $a+b$ is larger than both $a$ and $b$. This means that the largest integer in the list, which is 13, cannot be either $x$ or $y$ (and in fact cannot be placed in any square). This is because the number in the circle next to it must be smaller than 13 (because 13 is the largest number in the list) and so cannot be the sum of 13 and another positive number from the list. Thus, for $x+y$ to be as large as possible, we would have $x$ and $y$ equal to 10 and 11 in some order. But here we have the same problem: there is only one larger number from the list (namely 13) that can go in the circles next to 10 and 11, and so we could not fill in the circle next to both 10 and 11. Therefore, the next largest possible value for $x+y$ is when $x=9$ and $y=11$. (We could also swap $x$ and $y$.) Here, we could have $13 = 11+2$ and $10 = 9 + 1$, giving the following partial list: [[IMAGE0]] The remaining integers (4, 5 and 6) can be put in the shapes in the following way that satisfies the requirements. [[IMAGE1]] This tells us that the largest possible value of $x+y$ is 20.

From the given information, if $a$ and $b$ are in two consecutive squares, then $a+b$ goes in the circle between them. Since all of the numbers that we can use are positive, then $a+b$ is larger than both $a$ and $b$. This means that the largest integer in the list, which is 13, cannot be either $x$ or $y$ (and in fact cannot be placed in any square). This is because the number in the circle next to it must be smaller than 13 (because 13 is the largest number in the list) and so cannot be the sum of 13 and another positive number from the list. Thus, for $x+y$ to be as large as possible, we would have $x$ and $y$ equal to 10 and 11 in some order. But here we have the same problem: there is only one larger number from the list (namely 13) that can go in the circles next to 10 and 11, and so we could not fill in the circle next to both 10 and 11. Therefore, the next largest possible value for $x+y$ is when $x=9$ and $y=11$. (We could also swap $x$ and $y$.) Here, we could have $13 = 11+2$ and $10 = 9 + 1$, giving the following partial list: [[IMAGE0]] The remaining integers (4, 5 and 6) can be put in the shapes in the following way that satisfies the requirements. [[IMAGE1]] This tells us that the largest possible value of $x+y$ is 20.

From the given information, if $a$ and $b$ are in two consecutive squares, then $a+b$ goes in the circle between them. Since all of the numbers that we can use are positive, then $a+b$ is larger than both $a$ and $b$. This means that the largest integer in the list, which is 13, cannot be either $x$ or $y$ (and in fact cannot be placed in any square). This is because the number in the circle next to it must be smaller than 13 (which is the largest number in the list) and so cannot be the sum of 13 and another positive number from the list. Thus, for $x+y$ to be as large as possible, we would have $x$ and $y$ equal to 10 and 11 in some order. But here we have the same problem: there is only one larger number from the list (namely 13) that can go in the circles next to 10 and 11, and so we could not fill in the circle next to both 10 and 11. Therefore, the next largest possible value for $x+y$ is when $x=9$ and $y=11$. (We could also swap $x$ and $y$.) Here, we could have $13 = 11+2$ and $10 = 9 + 1$, giving the following partial list: [[IMAGE0]] The remaining integers (4, 5 and 6) can be put in the shapes in the following way that satisfies the requirements. [[IMAGE1]] This tells us that the largest possible value of $x+y$ is 20.