BC =6rad2. If y =BP then PC= 6-rad2-y. PT = TC = (6rad2-y/rad2).PTC is an isosceles rt triangle
Oh gosh, I didn't realise I'll come down to having something like 0 = 0 if I used TC in terms of y...
Anyway, maybe a way avoiding the calculator using my previous method:
Let $\displaystyle \sin\theta_1 = \frac{1}{\sqrt5}$
($\displaystyle \theta_1$ is angle HAM after cancelling the x that I introduced.)
Let $\displaystyle \sin\theta_2 = \frac{2}{\sqrt5}$
($\displaystyle \theta_2$ is angle HAB)
Angle HBP is then $\displaystyle \frac{\pi}{4} - \theta_1$
Angle APB = $\displaystyle \pi - \frac{\pi}{4} - \theta_2 = \frac{3\pi}{4} - \theta_2$
Angle APC = $\displaystyle \frac{\pi}{4} + \theta_2$
then, using the sine rule:
Okay, let AB = 1 now for the sake of simplicity and BP = x
$\displaystyle \frac{x}{\sin \left(\theta_2\right)} = \frac{1}{\sin \left(\frac{3\pi}{4} - \theta_2\right)}$
$\displaystyle x = \frac{\frac{2}{\sqrt5}}{\sin \left(\frac{3\pi}{4}\right)\cos(\theta_2) - \cos\left(\frac{3\pi}{4}\right)\sin (\theta_2)}$
$\displaystyle x = \frac{\frac{2}{\sqrt5}}{\left(\frac{1}{\sqrt2} \right)\left(\frac{1}{\sqrt5}\right) - \left(-\frac{1}{\sqrt2}\right)\left(\frac{2}{\sqrt5} \right)}$
$\displaystyle x = \frac{\frac{2}{\sqrt5}}{\left(\frac{3}{\sqrt{10}} \right)}$
$\displaystyle x = \frac{2\sqrt2}{3}$
Now, the same with PC, which I call y.
$\displaystyle \frac{y}{\sin \left(\theta_1\right)} = \frac{1}{\sin \left(\frac{\pi}{4} + \theta_2\right)}$
$\displaystyle y = \frac{\frac{1}{\sqrt5}}{\sin \left(\frac{\pi}{4} \right)\cos\left(\theta_2\right) + \cos\left(\frac{\pi}{4} \right)\sin\left(\theta_2\right)}$
$\displaystyle y = \frac{\frac{1}{\sqrt5}}{\left(\frac{1}{\sqrt2} \right)\left(\frac{1}{\sqrt5}\right) + \left(\frac{1}{\sqrt2}\right)\left(\frac{2}{\sqrt5 } \right)}$
$\displaystyle y = \frac{\frac{1}{\sqrt5}}{\left(\frac{3}{\sqrt{10}} \right)}$
$\displaystyle y = \frac{\sqrt2}{3}$
You immediately see that x:y = 2:1
Well, if you don't want to use trigonometry, maybe a sketch will do:
When I complete the sketch some image of the initial triangle, I get a square, with two parallel lines running across the diagonal. These lines start at corners and then the next line start right where the previous line started (in this case, the black line parallel to the initial blue line starts at the left, directly opposite to where the blue line ended). This divides the diagonal into three pieces of the same length.
This applies to rectangles too (second picture) and even a greater number of lines drawn across the diagonal. In the last diagram, the diagonal is equally divided into 4 parts by the three lines.
Thanks bjhopper,
But I've been trying to solve this using your x and y method. And I come up with such messy solutions for triangle APT. Something like (12-6rad2-6/2)^2 + (6rad2-y/2)^2 = (rad(171/5) + x)^2 and when your try to simplify it, it becomes HELL. There are like 4 questions like these for a 60 min exam so I don't think it's practical. And this exam is meant for Graduating High school students :/
Now to try Unknow008's solution!