# Magic Pentagon - all sides add up to 100

• May 9th 2008, 12:20 AM
maltaman
Magic Pentagon - all sides add up to 100
Magic Pentagon - All sides add up to 100
Michael and his family are utterly stumped at the question below. We have looked at similar problems on your site and been able to get as far as the below. Can you indicate to me where to go next?

Here is the problem and where I have gone

Complete the magic pentagon by putting numbers in the empty circles so that the sum of the three numbers along each of the sides of the pentagon equals 100. Show that there is only one solution to this problem.

The pentagon is
- 21 -
10 31
- -
56 47
-
I found this confusing because instead of the usual one digit numbers in magic pentagon there are two digit numbers as well.

When I add the pentagon’s sides as all the numbers on 1 side add up to 100 and there are 5 sides the numbers must all add up to 500 but that will be the same as adding all ten numbers and then adding the five corner numbers again. This says that 136 (the sum of the known numbers) plus double the sum of the corner numbers must equal to 500, so double the sum of the corner numbers is 364 and the sum of the corner numbers is 182. So, ...............

Michael
• May 9th 2008, 01:58 AM
angel.white
Quote:

Originally Posted by maltaman
Magic Pentagon - All sides add up to 100
Michael and his family are utterly stumped at the question below. We have looked at similar problems on your site and been able to get as far as the below. Can you indicate to me where to go next?

Here is the problem and where I have gone

Complete the magic pentagon by putting numbers in the empty circles so that the sum of the three numbers along each of the sides of the pentagon equals 100. Show that there is only one solution to this problem.

The pentagon is
- 21 -
10 31
- -
56 47
-
I found this confusing because instead of the usual one digit numbers in magic pentagon there are two digit numbers as well.

When I add the pentagon’s sides as all the numbers on 1 side add up to 100 and there are 5 sides the numbers must all add up to 500 but that will be the same as adding all ten numbers and then adding the five corner numbers again. This says that 136 (the sum of the known numbers) plus double the sum of the corner numbers must equal to 500, so double the sum of the corner numbers is 364 and the sum of the corner numbers is 182. So, ...............

Michael

Assuming I correctly understand the problem.

Each side takes the form of #1 + #1 + unknown = 100. You simply fill in the numbers, and solve for unknown.

ie first side is
31 + 10 + unknown = 100

41 + unknown = 100

unknown = 100 - 41

unknown = 59

The 59 plugs in as one of the two numbers on the next side, allowing you to follow the same process to solve for that side, which allows you to solve the next side and so on. There is only one solution because 59 is the only answer that will make the side with 10 on it be equal to 100. And since that gives two values for the next side, there is only one solution for that side (20) and this gives you two numbers on the next side, meaning only a single solution for that one (33), etc.
• May 9th 2008, 02:04 PM
maltaman
I must have explained this incorrectly. this is the actual pentagon.

? 21 ?

10 32

? ?

56 47
?
• May 9th 2008, 03:59 PM
Soroban
Hello, maltaman!

Quote:

Magic Pentagon - All sides add up to 100

Complete the magic pentagon by putting numbers in the empty circles
so that the sum of the three numbers on each side of the pentagon equals 100.
Show that there is only one solution to this problem.

Code:

            (a) 21  (b)             *        *              10          32           *            *         (e)            (c)             *          *               6      44                 *  *                 (d)

We have: . $\begin{array}{ccccccc}a + 21 + b \:=\:100 & \Rightarrow & a + b \:=\:79 & {\color{blue}[1]}\\
b + 32 + c \:=\:100 & \Rightarrow & b + c \:=\:68 & {\color{blue}[2]}\\
c + 47 + d \:=\:100 & \Rightarrow & c + d \:=\:53 & {\color{blue}[3]}\\
d + 56 + e \:=\:100 & \Rightarrow & d + e \:=\:44 & {\color{blue}[4]}\\
e + 10 + a \:=\:100 & \Rightarrow & a + e \:=\:90 & {\color{blue}[5]}\end{array}$

Add the five equations: . $2a + 2b + 2c + 2d + 2e \:=\:334$
. . and we have: . $a + b + c + d + e \:=\:167\;\;{\color{blue}[6]}$

Add [2] and [4]:. . $\begin{array}{ccc}b + c &=&68 \\ d+e &=&44 \end{array}$

. . and we have: . $b + c + d + e \:=\:112\;\;{\color{blue}[7]}$

Subtract [7] from [6]: . $\begin{array}{ccc} a +b+c+d+e &=& 167 \\ \quad\;\; b+c+d+e &=& 112 \end{array}$

. . And we have: . $\boxed{a \:=\:55}$

Substitute into [1]: . $55 + b \:=\:79\quad\Rightarrow\quad \boxed{b \,=\,24}\quad\hdots$ and so on.

• May 9th 2008, 08:24 PM
angel.white
Quote:

Originally Posted by Soroban
Hello, maltaman!

Code:

            (a) 21  (b)             *        *              10          32           *            *         (e)            (c)             *          *               6      44                 *  *                 (d)

We have: . $\begin{array}{ccccccc}a + 21 + b \:=\:100 & \Rightarrow & a + b \:=\:79 & {\color{blue}[1]}\\
b + 32 + c \:=\:100 & \Rightarrow & b + c \:=\:68 & {\color{blue}[2]}\\
c + 47 + d \:=\:100 & \Rightarrow & c + d \:=\:53 & {\color{blue}[3]}\\
d + 56 + e \:=\:100 & \Rightarrow & d + e \:=\:44 & {\color{blue}[4]}\\
e + 10 + a \:=\:100 & \Rightarrow & a + e \:=\:90 & {\color{blue}[5]}\end{array}$

Add the five equations: . $2a + 2b + 2c + 2d + 2e \:=\:334$
. . and we have: . $a + b + c + d + e \:=\:167\;\;{\color{blue}[6]}$

Add [2] and [4]:. . $\begin{array}{ccc}b + c &=&68 \\ d+e &=&44 \end{array}$

. . and we have: . $b + c + d + e \:=\:112\;\;{\color{blue}[7]}$

Subtract [7] from [6]: . $\begin{array}{ccc} a +b+c+d+e &=& 167 \\ \quad\;\; b+c+d+e &=& 112 \end{array}$

. . And we have: . $\boxed{a \:=\:55}$

Substitute into [1]: . $55 + b \:=\:79\quad\Rightarrow\quad \boxed{b \,=\,24}\quad\hdots$ and so on.

Oh, I see, I thought they were given in linear order around the pentagon.