Find $y+z$ in the magic square, understanding the solution

Question

I am having trouble understanding the alternate solution in my algebra book on system of equations for the problem below.

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.

$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]">

I solved this using a few system of equations which matches the answer, but my book proposes an alternate quick solution as well below:

A nice thing to know is that any $3$ numbers that goes through the middle forms an arithmetic sequence.

Using this, we know that $x=24+z/2$, or $2x=24+z$ because $x$ would be the average.

We also know that because $x$ is the average the magic sum would be $3x$, so we can also write the equation $3x-46=z$ using the bottom row.

Solving for x in this system we get $x=22$, so now using the arithmetic sequence knowledge we find that $y=26$ and $z=20$.

Adding these we get $46$

My question

Im a bit confused, on which numbers are creating an arithmetic sequence. Is it saying that $24,x,z$ for example are in an arithmetic sequence?

Answer 1

Denote the magic sum by $m$. Adding up all rows (or all columns) shows that the sum $s$ of all numbers is $3m$. Adding up the four lines through the centre yields $s+3x=4m$, and thus $x=\frac m3$.

Answer 2

This is a nice Observation about any magic square.

Suppose you had the square:

$\begin{array} \text{A} &B&C\\D&E&F\\G&H&I\end{array}$

And you are told it is a magic square.

If all the rows columns and diagonals equal $M$ then

$(A+E+I) + (B+E +H) + (C+E+G) = 3M$

$(A+B+C) +(G+H+I) + 3E = 3M$

$2M + 3E = 3M$

So $M = 3E$.

....

And that's the magic observation. If every row, column, and diagonal add to $3E$ then the average square value in every row, column, and diagonal average to $E$ and the center square IS $E$ so.....

f $E-A = k$ then $A+E+I = (E-k) + E + I = 3E$ and $I = E+k$.

If $E-B = j$ then $B+E+H =(E-j)+E + H = 3E$ and $H= E+j$.

And so on.

All rows, columns, and diagonals that pass through the center are in arithmethic sequence as the middle square is the average of the $3$.

.....

(Maybe be worth convincing yourself that if $\frac {a_1 + a_2 + a_3}3 = a_2$ then $a_1, a_2, a_3$ are in arithmetic sequence with $d = a_2-a_1 = a_3 - a_2$ and $a_2 = a_1 + d$ and $a_3 = a_2 + d$.)

Answer 3

It is true that in a 3 by 3 magic square the entries in any line through the middle make an AP (and in order), but you can solve this problem without having to know that.

\begin{align*} w &= 18+25-24=19\tag{top row, left column}\\ \implies x &= 24+19-21=22\tag{top row, diagonal}\\ \implies y &= 22+25-21=26;\tag{diagonal, right column}\\ z &= 22+19-21=20\tag{diagonal, bottom row}\\ \implies y+z &= 26+20=46. \end{align*}

Answer 4

The following may help you to visualize the arithmetic sequences.

First, consider a generic $3\times 3$ magic square. Let $x$ be the middle entry—as in your problem. Mow from the previous response, we have that the magic sum is three times the middle entry, that is $3x$. But since the sum is $3x$ and one of the terms is $x$, then the other two terms have to be equally distant from $x$ in opposite directions. Let these distances on the two diagonals be $a$ and $b$. But these are just the common differences of arithmetic sequences.

We have:

$ \begin{bmatrix} x+a & . & x+b \\ . & x & . \\ x-b & . & x-a \\ \end{bmatrix} $

Next, given that all column and row sums have to be $3x$, we solve for the missing entry in the first and third columns, giving $x-a+b$ and $x+a-b$, and then the missing rowsyielding $x-a-b$ and $x+a+b$.

Notice that the middle row and column again form arithmetic progressions with common differences of $a+b$ and $a-b$ respectively.

$ \begin{bmatrix} x+a & x-(a+b) & x+b \\ x-(a-b) & x & x+(a-b) \\ x-b & x+(a+b) & x-a \\ \end{bmatrix} $

I'm not sure that leads to the easiest solution, and it wouldn't have been how I would have solved the problem, but hopefully it sheds some light on the arithmetic sequences the book points to.