# Difference between revisions of "2001 AMC 10 Problems/Problem 22"

## Problem

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 $y + z$.

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("25",(0.5,0.5)); label("z",(1.5,0.5)); label("21",(2.5,0.5)); label("18",(0.5,1.5)); label("x",(1.5,1.5)); label("y",(2.5,1.5)); label("v",(0.5,2.5)); label("24",(1.5,2.5)); label("w",(2.5,2.5));[/asy]$

$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$

## Solutions

-DaBob

### Solution 1

We know that $y+z=2v$, so we could find one variable rather than two.

$v+24+w=43+v$

$24+w=43$

$w=19$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("25",(0.5,0.5)); label("z",(1.5,0.5)); label("21",(2.5,0.5)); label("18",(0.5,1.5)); label("x",(1.5,1.5)); label("y",(2.5,1.5)); label("v",(0.5,2.5)); label("24",(1.5,2.5)); label("19",(2.5,2.5));[/asy]$

$44+x=24+x+z \implies z=20$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("25",(0.5,0.5)); label("20",(1.5,0.5)); label("21",(2.5,0.5)); label("18",(0.5,1.5)); label("x",(1.5,1.5)); label("y",(2.5,1.5)); label("v",(0.5,2.5)); label("24",(1.5,2.5)); label("19",(2.5,2.5));[/asy]$

The sum per row is $25+21+20=66$.

Thus $66-18-25=66-43=v=23$.

Since we needed $2v$ and we know $v=23$, $23 \times 2 = \boxed{\textbf{(D)}\ 46}$.

### Solution 2

$v+24+w=43+v$

$24+w=43$

$w=19$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("25",(0.5,0.5)); label("z",(1.5,0.5)); label("21",(2.5,0.5)); label("18",(0.5,1.5)); label("x",(1.5,1.5)); label("y",(2.5,1.5)); label("v",(0.5,2.5)); label("24",(1.5,2.5)); label("19",(2.5,2.5));[/asy]$

$44+x=24+x+z \implies z=20$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("25",(0.5,0.5)); label("20",(1.5,0.5)); label("21",(2.5,0.5)); label("18",(0.5,1.5)); label("x",(1.5,1.5)); label("y",(2.5,1.5)); label("v",(0.5,2.5)); label("24",(1.5,2.5)); label("19",(2.5,2.5));[/asy]$

The magic sum is determined by the bottom row. $25+20+21=66$.

Solving for $y$:

$y=66-19-21=66-40=26$.

To find our answer, we need to find $y+z$. $y+z=20+26 = \boxed{\textbf{(D)}\ 46}$.

### Really Easy Solution

A nice thing to know is that any $3$ numbers that go through the middle form 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 $\boxed{\textbf{(D)}\ 46}$.

-harsha12345

## Systems of Equations

Create an equation for every row, column, and diagonal. Let $e$ be the sum of the rows, columns, and diagonals. $$w+v+24=e$$ $$x+y+18=e$$ $$z+46=e$$ $$v+43=e$$ $$x+z+24=e$$ $$w+y+21=e$$ $$x+w+25=e$$ $$x+v+21=e$$.

Notice that $z+46=e$ and $x+z+24=e$ both have $z$. Equate them and you get that $x=22$. Using that same strategy, we use $v+43=e$ instead. $w+v+24=e$ is good for our purposes. It turns out that $w=19$. Since we already know those numbers, and $x+w+25=e$, We can say that $e$ will be $66$. We are now able to solve: $x+z+24=e$, $w+y+21=e$, $x+v+21=e$, and $x+y+18=e$. Respectively, $v=23$, $w=19$, $x=22$, $y=26$, and $z=20$. We only require The sum of $y+z$, which is $26+20=46$. We get that the sum of $y$ and $z$ respectively is $\boxed{\textbf{(D)}\ 46}$

-OofPirate

~savannahsolver