### Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

### From All Corners

Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.

### Star Gazing

Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.

# Mixing More Paints

##### Age 14 to 16 Challenge Level:

We received several incorrect solutions like the ones below:

Combining paints A ($1:4$) and B ($1:5$):

 Required Ratio Amount of paint A Amount of paint B $2:9$ $1$ $1$ $3:14$ $1$ $2$ $10:43$ $7$ $3$

Combining paints C ($1:3$) and D ($1:7$):

 Required Ratio Amount of paint C Amount of paint D $2:9$ $5$ $3$ $3:14$ $7$ $5$ $10:43$ $27$ $13$
They are based on the misconception that you can add the ratios to work out the necessary combinations. The solutions given have assumed that the 'parts' in the ratios are of equal size so that a can in the ratio $1:3$ contains half the amount of the one in the ratio $1:7$.

However, one can of paint C and one can of paint D does not produce paint in the ratio $2:10$ (or $1:5$), since that would assume that the one part blue in can C has the same volume as the one part blue in can D.

This can't be the case since there are $4$ parts in can C and $8$ parts in can D,
so $1/4$ of can C is blue and $1/8$ of can D is blue.

Ian and Amos from South Island School in Hong Kong sent an picutre of their work for making paint in the ratio $2 : 9.$ There is also an explanation about what they have done.

$\frac{11}{30}$ refers to the case where $a$ and $b$ are both $1,$ so $1$ can of each paint. This is the amount of blue paint in the mixture.

The desired proportion of blue paint is $\frac 2 {11}.$ This is equivalent to $\frac{60}{330}.$ It is multiplied by $\frac{30}{30}$ because the amount of blue paint will be $\frac{6a+5b}{30}.$

The proportion of blue paint will be $\dfrac{\frac{6a+5b}{30}}{a+b}$ which is equal to $\dfrac{6a+5b}{(a+b)\times30},$ so they have equated this to $\frac{60}{330}.$ This gives $a+b=11$ and $6a+5b=60.$

Navjot from Sherborne Qatar used algebra to find a general solution. Navjot's solution begins by using the simpler case in Mixing Paints.
It is possible to find a ratio of paint $1:z$ with 2 colours of paint with ratios $1:x$ and $1:y$ where $x\lt z \lt y$.

To start off, we first need to know the fact that the mixing of $2$ paints essentially gives us the average of the two ratios. Eg; mixing $1:2$  and $1:5$ gives us $1:3,$ which is $\dfrac{\frac13+\frac16}2.$

So, let:
The number of pots of paint A $= a,$
The number of pots of paint B $= b,$
The ratio of paint A $= 1:x,$
The ratio of paint B $= 1:y,$
And, the target ratio $= 1:z.$

Since we know that the mixing of paints A and B will give us the mean of the two ratios:
$\dfrac{a\frac1{x+1}+b\frac1{y+1}}{a+b} = \frac1{z+1}$

(I am using $x+1, y+1,$ and $z+1$ rather than $x, y,$ and $z$ because both terms of a ratio add up to make a whole. Eg; $1:3$ means that something is divided up into $\frac14$ and $\frac34,$ not $\frac13$ and $\frac33$)

Now, we split the equation above into 2 parts, one being the numerator of both sides of the equation and the other being the denominator:
\begin{align} & a\tfrac1{x+1}+b\tfrac1{y+1}=1\\ \Rightarrow &\tfrac{a(y+1)+b(x+1)}{(x+1)(y+1)}=1\\ \Rightarrow &a(y+1)+b(x+1)=(x+1)(y+1)\end{align} and $$a+b = z+1\Rightarrow a=(z+1)-b$$

Substituting 2) into 1),
\begin{align}& ((z+1) - b)(y+1) + b(x+1) = (x+1)(y+1)\\ \Rightarrow &(z+1)(y+1) - b(y+1) + b(x+1) = (x+1)(y+1)\\ \Rightarrow &b((x+1)-(y+1)) = (x+1)(y+1)-(z+1)(y+1)\\ \Rightarrow &b(x-y) = (y+1)((x+1)-(z+1))\\ \Rightarrow& b = \frac{(y+1)(x-z)}{x-y}\end{align}

Therefore, $a = z+1 - \dfrac{(y+1)(x-z)}{x-y}$

However, this general solution does give answers in fractions or in unsimplified form.

The only difference here is that the target ratios are in the form $a:b$ rather than $1:x.$ However, this $m:n,$ a ratio simplified to its simplest terms and where $a$ and $b$ are natural
numbers, can be converted into a unitary ratio (a ratio in the form $1:x$), dividing both terms by $m$ gives us the ratio $1:\frac nm,$ which is in the form of $1:x.$

Using paints of ratios $1:4$ and $1:5$
Target ratio:
$2:9 = 1:\frac92 = 1:4.5$
$x =4, y = 5, z = 4.5$
$a = 5.5 - \frac{6\times-0.5}{-1}=\frac52$
$b = \frac{6(-0.5)}{-1} = 3$

So the decorator needs $\frac52$ pots of paint A and $3$ pots of paint B.
However, the answer gives us a fraction rather than a natural number, and we can’t buy parts of pots of paint, so we find a common denominator for both the numbers of pots and then remove the common factor, which would be the common denominator.

$\frac52, \frac31$ are equivalent to $\frac52$ and $\frac 62$
Answer = $5$ pots of A and $6$ pots of B.

Keeping in mind how to handle the fractions we get in this solution, let’s move on to the next answers.

$3:14 = 1:\frac{14}3$
$x = 4, y = 5, z = \frac{14}3$
$a = \frac{14}3+1 - \dfrac{6\times\frac23}{-1}=\frac53$
$b = \dfrac{6\times\frac23}{-1}=4$

$\frac53$ of A and $4$ of B
$= 5$ of A and $12$ of B

$10:43 = 1: \frac{43}{10} = 1:4.3$
$x = 4, y = 5, z = 4.3$
$a = 5.3 - \frac{6(-0.3)}{-1} = \frac72$
$b = \frac{6(-0.3)}{-1} = \frac95$
$\frac72$ pots of A and $\frac95$ pots of B
Or, $\frac{35}{10}$ pots of A and $\frac{18}{10}$ pots of B
Or, $35$ pots of A and $18$ pots of B

Using paints with ratios $1:3$ and $1:7$
Target ratio:
$2:9 = 1:4.5$
$x =3, y = 7, z = 4.5$
$a = 5.5 - \frac{8(-1.5)}{-4} = \frac52$
$b = 8\frac{-1.5}{-4} = 3$

$\frac52$ of A and $3$ of B
Or, $5$ of A and $6$ of B

$3:14 = 1:\frac{14}3$
$x = 3, y = 7, z = \frac{14}3$
$a = \frac{14}3+1 - \dfrac{8(\frac{-5}3}{-4} = \frac73$
$b = \dfrac{8(\frac{-5}3}{-4} = \frac{10}3$

$\frac73$ of A and $\frac{10}3$ of B
Or, $7$ of A and $10$ of B

$10:43 = 1:4.3$
$x = 3, y = 7, z = 4.3$
$a = 5.3 - \frac{8(-1.3)}{-4} = \frac{27}{10}$
$b = \frac{8(-1.3)}{-4} = \frac{13}5$
$\frac{27}{10}$ pots of A and $\frac{13}5$ pots of B
Or, $27$ pots of A and $26$ pots of B

It is possible to achieve a ratio of $a:b$ by using paints of ratios $1:x$ and $1:y$ only if $x\lt\frac ba\lt y.$
In an algebraic sense too, let’s consider the general solution for $b,$ which is $\frac{(y+1)(x-z)}{x-y}.$ Since $x \lt y,$ the denominator is negative, but if $z\lt x,$ the numerator remains positive, thus giving a negative answer, but we can’t have negative number of pots.

If $z\gt y,$ the value for $b$ will be positive since the numerator will be negative like the denominator, but, the value for $b$ is greater than $z+1,$ and the general solution for $a = z+1 - \frac{(y+1)(x-z)}{x-y} = (z+1) - b,$ since $b \gt z+1,$ the value for $a$ is negative, so it is not a valid solution.

Therefore, to get a ratio $a:b$ using paints with ratio $1:x$ and $1: y$ where $x\lt y,$ $x\lt \frac ab \lt y.$