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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: .$\displaystyle \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: .$\displaystyle 2a + 2b + 2c + 2d + 2e \:=\:334$

. . and we have: .$\displaystyle a + b + c + d + e \:=\:167\;\;{\color{blue}[6]}$

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

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

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

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

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