### 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.

### Two Cubes

Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]

### Square Mean

Is the mean of the squares of two numbers greater than, or less than, the square of their means?

# A Third of the Area

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

Using the relationship between the areas
The area of the small square is $\frac13$ of the area of the large square, so $$\frac {x^2}{(x+y)^2}=\frac13\Rightarrow\left(\frac x{x+y}\right)^2=\frac13\Rightarrow\frac x{x+y}=\dfrac1{\sqrt3}$$
We can make $\dfrac x y$ the subject of this equation:
\begin{align}\frac{x}{x+y}&=\frac1{\sqrt3}\\ \Rightarrow x\times\sqrt3&=1\times(x+y)\\ \Rightarrow\sqrt3x&=x+y\\ \Rightarrow\sqrt3x-x&=y\\ \Rightarrow(\sqrt3-1)x&=y\\ \Rightarrow x&=\frac y{\sqrt3-1}\\ \Rightarrow\frac x y &=\frac 1 {\sqrt3-1}\end{align}
To see how to express this as a fraction with a whole number denominator, see the bottom of the solution.

Using the relationship for lengths
If the area scale factor between the small square and the large square is $3$, then the length scale factor between the small square and the large square is $\sqrt3$.

So $\dfrac x{x+y}=\dfrac1{\sqrt3}.$

We can make $\dfrac xy$ the subject of this equation as shown above.

Splitting the larger square into smaller squares and rectangles
In the diagram the square has been split into a red square with area $y^2,$ a green square with area $x^2,$ and two blue rectangles with sides of length $x$ and $y,$ so area $xy.$

The green square occupies a third of the total area, so the red and blue areas are two thirds of the total area - or twice the area of the green square.

So $y^2+2xy=2x^2$.

Dividing both sides of this equation by $y^2$ gives $$1+2\frac xy=2\frac{x^2}{y^2}$$
This is a quadratic equation in $\dfrac xy$. This is easier to see when we replace $\dfrac xy$ with $w$ and remember that $w^2=\left(\dfrac xy\right)^2=\dfrac{x^2}{y^2}$: $$1+2w=2w^2\Rightarrow 2w^2-2w-1=0.$$
This can be solved using the quadratic formula or by completing the square. The method shown below is completing the square.
\begin{align}&2w^2-2w-1=0&\\ \Rightarrow&2\left(w^2-w-\tfrac12\right)=0&\\ \Rightarrow&w^2-w-\tfrac12=0\div2=0&\\ \Rightarrow&\left(w-\tfrac12\right)^2-\tfrac14-\tfrac12=0\hspace{10mm}&\text{since }\left(w-\tfrac12\right)^2=w^2-w+\tfrac14\\ \Rightarrow&\left(w-\tfrac12\right)^2=\tfrac14+\tfrac12=\tfrac34&\\ \Rightarrow &w-\tfrac12=\pm\tfrac{\sqrt3}{2}&\\ \Rightarrow &w=\tfrac12\pm\tfrac{\sqrt3}{2}&\text{take the positive root since }w>0\\ \Rightarrow&w=\tfrac{1+\sqrt3}2\end{align}

To see that this is the same as the solution above, $\frac1{\sqrt3-1}$, see the bottom of the page.

Splitting the larger square into trapeziums
In the diagram the square has been split into a green square with area $x^2,$ and two trapeziums. Both trapeziums have parallel sides along the side of the large square and the side of the green square. So the parallel sides have length $(x+y)$ and $x,$ and the trapeziums have 'height' $y.$

So the area of each trapezium is $\frac12\left(\left(x+y\right)+x\right)y$

Since the trapeziums are congruent and the green square occupies a third of the area, the three shapes have equal area. In particular, $\frac12\left(\left(x+y\right)+x\right)y=x^2$

Multiplying both sides of this equation by $2$ and expanding gives $2xy+y^2=2x^2$.

$\dfrac xy$ can be found from this equation as shown above.

Expressing $\dfrac1{\sqrt3-1}$ as a fraction with a whole number as the denominator
Notice that $\left(\sqrt3-1\right)\left(\sqrt3+1\right)=\sqrt3\times\sqrt3+1\times\sqrt3-1\times\sqrt3-1\times1=3-1=2.$

So $\dfrac1{\sqrt3-1}=\dfrac{1\times\left(\sqrt3+1\right)}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}=\dfrac{\left(\sqrt3+1\right)}2$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.