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Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

Overturning Fracsum

Can you solve the system of equations to find the values of x, y and z?

Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

Always Two

Age 14 to 18
Challenge Level

 

Well done to Nishad from Thomas Estley Community College in England, who sent in this full solution to the main problem:

Firstly we get that the equations are:

$ab+c=2$
$ac+b=2$
$bc+a=2$


Now notice that $$(ba+c)-(bc+a)=2-2 \Rightarrow b(a-c)-(a-c)=0 \Rightarrow (b-1)(a-c)=0$$
This gives us 2 cases namely $b=1$ or $a=c$

Case 1: $b=1$

We can substitute this into the original equations that we had 
$a+c=2 \Rightarrow c=2-a$
$ac=1$
Hence
$a(2-a)=1 \Rightarrow 0=a^2 -2a +1 = (a-1)^2$

This gives us the solutions $a=1$,$c=1$

Hence 1 set of solutions $(a,b,c)$ is 
$(1,1,1)$

Case 2: $a=c$

We can substitute this into the original equations that we had 
$ab+a=2 \Rightarrow b=\frac{2}{a}-1$
$a^2+b=2$
Hence
$a^2 + \frac{2}{a}-1 = 2 \Rightarrow a^{3} - 3a +2 = 0$

Since $a=1$ is a root of the equation $(a-1)$ is a factor

$a^3 -3a +2 = (a-1)(a^2+a-2) = (a-1)(a-1)(a+2) = 0$

Using $a=1$ gives us the same set of solutions as before but $a=-2$ gives us a new set, namely 
$(-2,-2,-2)$

Nayanika from The Tiffin Girls' School in the UK began the first part of the extension:

Here, it should say $ac+1=6,$ so $c(6-c) +1 = 6c-c^2+1=6,$ therefore $c^2-6c+5=0.$ From here, Clare continued:

$c^2-6c+5=0\Rightarrow (c-1)(c-5)=0$ so $c=1$ or $c=5.$
$a+c=6,$ so $a$ is the other out of $1$ and $5.$
This means $a,b,c$ are $1,1,5$ (in any order).

Case 2 - $a=c$
Now $ab+a=6$ and $a^2+b=6,$ so $b=6-a^2$
Therefore $a(6-a^2)+a=6\Rightarrow 6a-a^3+a=6 \Rightarrow a^3-7a+6=0$
We can see that $a=1$ is a solution (and this is not a surprise because we already know $b$ could be $5$ and $a$ and $c$ could both be $1$), so $a-1$ will be a factor:
$a^3-7a+6=(a-1)(a^2+a-6)=(a-1)(a+3)(a-2)=0$
Therefore $a=c$ could also be $-3$ or $2.$

If $a=c=-3,$ then $ac+b=6$ becomes $9+b=6$ so $b=-3$ as well. This will clearly work for all of the equations.
If $a=c=2,$ then $ac+b=6$ becomes $2+b=6$ so $b=2$ as well.

Therefore $a,b,c$ could be $1,1,5$ or $2,2,2$ or $-3,-3,-3.$

For the second part of the extension, Mattia wrote:

Nishad completed the extension by beginning with the further extension:

$ab+c=k$
$ac+b=k$
$bc+a=k$


Now notice that 
$(ba+c)-(bc+a)=k-k \Rightarrow b(a-c)-(a-c)=0 \Rightarrow (b-1)(a-c)=0$ so we get the same equation in the general case

Case 1: $b=1$

We can substitute this into the original equations that we had 
$a+c=k \Rightarrow c=k-a$
$ac=k-1$
Hence
$a(k-a)=k-1 \Rightarrow 0=a^2 -ka +k-1 = (a-1)(a-(k-1))$

This gives us the solutions $a=1$, $c=k-1$ (and $a=k-1$,$c=1$)

Hence 1 set of solutions $(a,b,c)$ is 
$(1,1,(k-1))$
but due to the symmetry of the problem all 3 permutations of this set are solutions, namely $((k-1),1,1)$, $(1,(k-1),1)$ as well.

Case 2: $a=c$

We can substitute this into the original equations that we had 
$ab+a=k \Rightarrow b=\frac{k}{a}-1$
$a^2+b=k$
Hence
$a^2 + \frac{k}{a}-1 = k \Rightarrow a^3 - (k+1)a +k = 0$

Since $a=1$ is a root of the equation $(a-1)$ is a factor
$a^3 -(k+1)a +k = (a-1)(a^2+a-k) = (a-1)(a-\phi)(a+(\phi+1)) = 0$ for some $\phi,$ where
$\phi (\phi +1) =k \Rightarrow \phi^2+\phi-k$
$\phi = \frac{-1+\sqrt{4k+1}}{2}$
$-(\phi +1) = \frac{-1-\sqrt{4k+1}}{2}$

Now we have new 2 new sets of solutions

When $a=\frac{-1+\sqrt{4k+1}}{2}$ gives us the set of solutions $$\left(
\frac{-1+\sqrt{4k+1}}{2},\frac{-1+\sqrt{4k+1}}{2},\frac{-1+\sqrt{4k+1}}{2}
\right)$$

When $a=\frac{-1 -\sqrt{4k+1}}{2}$ gives us the set of solutions 
$$\left(
\frac{-1-\sqrt{4k+1}}{2},\frac{-1-\sqrt{4k+1}}{2},\frac{-1-\sqrt{4k+1}}{2}
\right)$$

Remark: To find $b$ we could substitute the value of $a$ into 
$b=\frac{k}{a} - 1$
For the case where $a=\frac{-1+\sqrt{4k+1}}{2}$:
$$\begin{split}b&= \frac{2k}{-1+\sqrt{4k+1}} -1\\
&=\frac{2k+1-\sqrt{4k+1}}{-1+\sqrt{4k+1}}\\&=\frac{2k+1-\sqrt{4k+1}}{-1+\sqrt{4k+1}} \times
\frac{1+\sqrt{4k+1}}{1+\sqrt{4k+1}}\\ &=
\frac{2k+1-\sqrt{4k+1}+(2k+1)\sqrt{4k+1} -4k-1}{4k+1-1}\\&=\frac{-2k-\sqrt{4k+1}+(2k+1)\sqrt{4k+1}}{4k}\\& =\frac{-2k+2k\sqrt{4k+1}}{4k}\\& = \frac{-1+\sqrt{4k+1}}{2} \\&= a\end{split}$$
Which is why in fact $a=b=c$ (A very similar rationalizing of the denominator works for the case where $a=\frac{-1-\sqrt{4k+1}}{2}$) 

Extension 1:

After having done the further extension for the general case we can substitute in $k=6$ to get the sets $(5,1,1)$ (and its permutations), $(2,2,2)$, $(-3,-3,-3)$

Extension 2:

$ab-c=2$
$ac-b=2$
$bc-a=2$

Now notice that 
$(ba-c)-(bc-a)=2-2 \Rightarrow b(a-c)+(a-c)=0 \Rightarrow (b+1)(a-c)=0$

This gives us 2 cases namely $b=-1$ or $a=c$

Case 1: $b=-1$

We can substitute this into the original equations that we had 
$-a-c=2 \Rightarrow c=-2-a$
$ac=1$
Hence
$a(-2-a)=1 \Rightarrow 0=a^2 +2a +1 = (a+1)^2$

This gives us the solutions $a=-1$, $c=-1$

Hence 1 set of solutions $(a,b,c)$ is 
$(-1,-1,-1)$

Case 2: $a=c$

We can substitute this into the original equations that we had 
$ab-a=2 \Rightarrow b=\frac{2}{a}+1$
$a^2-b=2$
Hence
$a^2 - \frac{2}{a}-1 = 2 \Rightarrow a^{3} - 3a -2 = 0$

Since $a=-1$ is a root of the equation $(a+1)$ is a factor

$a^3 -3a -2 = (a+1)(a^2-a-2) = (a+1)(a+1)(a+2) = 0$

Using $a=-1$ gives us the same set of solutions as before but $a=2$ gives
us a new set, namely $(2,2,2)$