# Quod Erat Demonstrandum

## 2009/02/20

### Some F.4 textbook questions

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 4:55 下午
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Students may find the following textbook questions difficult.

Question 1

Refer to the figure below, given that $DE \perp AB$ and $DF \perp AC$; $ED = a$; $AB : AC = m$. Show that $\tan\alpha = \frac{2m - 1}{\sqrt{3}}$.

Solution

To use the given information $AB : AC = m$, we express $AB$ and $AC$ in terms of $\alpha, \beta$ and $a$. $AB = AE + EB = \sqrt{3}a + a\tan\alpha$ $AC = AF + FC = \sqrt{3}a + a\tan\beta$ (note: $DF = a$)

Now $m = \frac{AB}{AC} = \frac{\sqrt{3}a + a\tan\alpha}{\sqrt{3}a + a\tan\beta} = \frac{\sqrt{3} + \tan\alpha}{\sqrt{3} + \tan\beta}$ – – – – – – (*)

Then, many students stop there and don’t know how to find $\tan\alpha$. May be, students find that, there are, at least, two different approaches.

One is geometry. $\tan\alpha$, the direct meaning in the figure is $\frac{BE}{ED}$, but, it is not that easy to express $BE$ explicitly.

Another one is algebra, that is using trigonometric identities to do some algebraic computation so as to figure out the relation between $\tan\alpha$ and $m$.

We select the latter.

To find $\tan\alpha$, we need to get rid of $\beta$, hence all we need is replacing $\beta$ by something involves $\alpha$. Here is the crucial key: $\alpha + \beta = 60^o\Rightarrow \beta = 60^o - \alpha$, and (*) will be $m = \frac{\sqrt{3} + \tan\alpha}{\sqrt{3} + \tan(60^o - \alpha)}$

Well, it is not a must that students can obtain the required results, because, algebraic computation is still nightmares to some students. $m(\sqrt{3} + \frac{\sqrt{3} - \tan\alpha}{1 + \sqrt{3}\tan\alpha}) = \sqrt{3} + \tan\alpha$ $\frac{2m(\sqrt{3} + \tan\alpha)}{1 + \sqrt{3}\tan\alpha} = \sqrt{3} + \tan\alpha$

Urm, what’s next? Some students may expand further and come up with monsters. But, once we discover that there are similar non-zero terms ( $\sqrt{3} + \tan\alpha$) on both sides, cancel it and the expression above can be further simplied as $\frac{2m}{1 + \sqrt{3}\tan\alpha} = 1$ $\therefore \tan\alpha = \frac{2m - 1}{\sqrt{3}}$

Question 2

This is actually a past-paper question (1992 paper 2), just highlight one of the parts. A regular pentagon inscribing in a circle with radius $r$. Prove that $PA^2 + PB^2 + PC^2 + PD^2 + PE^2 = 10r^2$

Solution

The previous result showing that $PD^2 = 2r^2 - 2r^2\cos(\theta + \frac{6\pi}{5})$.

According to the marking scheme, we have But, as I’d mentioned in class, it seems that the presentation above is not clear enough. Something is missing in the marking scheme. Let me explain.

Refer to the following figure. By cosine law, it is easy to show that $(PA')^2 = 2r^2 - 2r^2\cos\phi$ – – – – – – (**)

Imagine that the point $A'$ is moving along the circumference, like The formula (**) is still valid.

How about when $A'$ moves to some position such that $\phi > \pi$? Like Is (**) still valid? Well, we need to show it out! Just refer to the figure above, by cosine law $(PA')^2$ $= 2r^2 - 2r^2\cos\angle POA'$ $= 2r^2 - 2r^2\cos(2\pi - \phi)$ $= 2r^2 - 2r^2\cos\phi$

Yes, (**) is still valid for $\phi > \pi$. Hence, we have 