Question

Find the number of triplets (a, b, c) such that a, b, c are three distinct positive numbers and a, b, c, b + c − a, c + a − b, a + b − c and a + b + c form a seven term arithmetic progression in some order.

Answer 1

0

Answer 2

Let the set of seven distinct positive numbers be $S = \{a, b, c, b+c-a, c+a-b, a+b-c, a+b+c\}$. These numbers form a seven-term arithmetic progression (AP) in some order. Let the AP be $x_1, x_2, \dots, x_7$ with common difference $d$. Since the terms are distinct, $d \ne 0$.

Since $a, b, c$ are positive, $a+b+c$ is strictly greater than $a, b,$ and $c$. Also, $a+b+c = a+(b+c) > a$, etc. For the terms $b+c-a, c+a-b, a+b-c$ to be positive, $a, b, c$ must satisfy the triangle inequality ($a+b>c, a+c>b, b+c>a$). If $a, b, c$ satisfy the triangle inequality, then $a+b+c > a+b-c, a+b+c > a+c-b, a+b+c > b+c-a$. Thus, $a+b+c$ is the largest term in the set $S$.

In an AP, the largest term is $x_7$ if $d>0$ or $x_1$ if $d<0$. Let's assume $d>0$. So, $x_7 = a+b+c$. The terms of the AP are $x_1, x_1+d, \dots, x_1+6d$. The set of differences between the largest term $x_7$ and the other six terms $\{x_1, \dots, x_6\}$ is $\{x_7-x_1, x_7-x_2, \dots, x_7-x_6\} = \{6d, 5d, 4d, 3d, 2d, d\}$. The set of differences between $a+b+c$ and the other six terms in $S$ is: $\{ (a+b+c)-a, (a+b+c)-b, (a+b+c)-c, (a+b+c)-(b+c-a), (a+b+c)-(c+a-b), (a+b+c)-(a+b-c) \}$ $= \{b+c, a+c, a+b, 2a, 2b, 2c\}$. So, the set $\{2a, 2b, 2c, a+b, a+c, b+c\}$ must be equal to the set $\{d, 2d, 3d, 4d, 5d, 6d\}$.

The sum of the elements in the first set is $2a+2b+2c+a+b+a+c+b+c = 4(a+b+c)$. The sum of the elements in the second set is $d+2d+3d+4d+5d+6d = 21d$. Thus, $4(a+b+c) = 21d$.

The sum of the terms in the AP is $7x_1 + 21d$. The sum of the terms in $S$ is $a+b+c + (b+c-a) + (c+a-b) + (a+b-c) + (a+b+c) = 3(a+b+c)$. So, $7x_1 + 21d = 3(a+b+c)$. Substitute $a+b+c = 21d/4$ into this equation: $7x_1 + 21d = 3(21d/4) = 63d/4$. $7x_1 = 63d/4 - 21d = (63d - 84d)/4 = -21d/4$. $x_1 = -3d/4$.

Since the terms of the AP must be positive, the smallest term $x_1$ must be positive. If $d>0$, then $x_1 = -3d/4$ is negative. This contradicts $x_1>0$. If $d<0$, let $d' = -d > 0$. The AP is $x_1, x_1-d', \dots, x_1-6d'$. The largest term is $x_1=a+b+c$. The set of differences from $x_1$ is $\{d', 2d', \dots, 6d'\}$. $\{2a, 2b, 2c, a+b, a+c, b+c\} = \{d', 2d', \dots, 6d'\}$. Summing gives $4(a+b+c) = 21d'$. $4x_1 = 21d'$. $x_1 = 21d'/4$. The terms of the AP are $x_1, x_1-d', \dots, x_1-6d'$. The smallest term is $x_7 = x_1 - 6d' = 21d'/4 - 6d' = (21-24)d'/4 = -3d'/4$. Since $d'>0$, the smallest term $-3d'/4$ is negative. This contradicts the requirement that all terms must be positive.

In both cases ($d>0$ and $d<0$), we find that the AP must contain negative terms, which is not allowed as $a, b, c$ are positive and $b+c-a, c+a-b, a+b-c, a+b+c$ must also be positive. Therefore, there are no such triplets $(a, b, c)$. The number of such triplets (a, b, c) is 0.