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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.

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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.]

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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:

Answer: $\dfrac{1}{\sqrt3-1} = \dfrac{1+\sqrt3}2 =1.366$ to 3 d.p.

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:
\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.

Splitting the larger square into smaller squares and rectangles
Green = $\frac13$ of total
$\therefore$ red + blue = green $\times$ 2

$\Rightarrow y^2+2xy=2x^2$
$\Rightarrow 1+2\dfrac xy=2\dfrac{x^2}{y^2}$

Let $\dfrac xy=w$:
$$\begin{align}1+2w=2w^2&\Rightarrow 2w^2-2w-1=0\\
&\Rightarrow w^2-w-\tfrac12=0\\
&\Rightarrow w-\tfrac12=\pm\tfrac{\sqrt3}{2}\\
&\Rightarrow w=\tfrac12\pm\tfrac{\sqrt3}{2}\text{  take the positive root since }w>0\\
&\Rightarrow w=\tfrac{1+\sqrt3}2\end{align}$$

Splitting the larger square into trapeziums
The area of each trapezium is $\frac12\left(\left(x+y\right)+x\right)y$

Green square occupies a third of the area
$\therefore$ the three shapes have equal area


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.

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