Area of a triangle in a diagram

Question

I am currently taking the Scottish Mathematical Council's Mathematical Challenge 2016/17. I am unable to solve problem S5.

I need to show that the area of the triangle $UQR$ is $$\frac {\sqrt 3-1}4$$ square units given that the sides of the square are $1$ unit and triangle $PQT$ is an equilateral.

Answer 1

Drop the perpendicular from $U$ to $QR$. Let $X\in QR$ be the foot of that perpendicular and let $h$ be the length of $UX$. Then, of course, the area of $\Delta QUR$ is $\frac 12h$.

To compute $h$: routine angle chasing shows that $<RUX=45$ so $\Delta RUX$ is isosceles and $h=RX$ which implies that $QX=1-h$. We also easily see that $<QUX=60$ so $\Delta QUX$ is a $30-60-90$ triangle. It follows that $$\frac {1-h}h=\sqrt 3\implies h=\frac 1{\sqrt 3+1}=\frac {\sqrt 3-1}2$$ And the desired result follows immediately.

Answer 2

the most straightforward way to me just putting this on a co-ordinate plane. Hopefully somebody will post something more elegant.

Let this be the unit square with bottom left corner at $(0,0)$. Find a line that defines $TQ$: the slope is $\tan(60)=\sqrt 3$ and it goes through $1,1$, so we get: $y-1=\sqrt 3(x-1)$. Similarly, $PR$ is given by $y=1-x$. We find that the intersection is $x=\sqrt 3/(1+\sqrt 3)$ whose distance to the line $x=1$ is exactly $1-\sqrt 3/(1+\sqrt 3)=1/(\sqrt3+1)$ and so we can multiply the numerator and denominator to get $(\sqrt 3-1)/2$

So the area is: $\frac{1}{2} 1 \cdot (\sqrt 3-1)/2=(\sqrt 3-1)/4$.

Quite ugly, but it gets the job done.