# Quod Erat Demonstrandum

## 2007/10/26

### [Fun] Gag

Filed under: Fun — johnmayhk @ 4:24 下午

Whenever I’m in F.5C, I must hear someone making jokes. (Broken gag in Chinese!) They ask me some questions for making gags. OK, this time I give a question to revenge,

source : Facebook group “一日一爛gag, I am not a plastic bag."

### Blood donation

Filed under: School Activities — johnmayhk @ 2:14 下午

The blood donation talk for form five students was held on this Monday. The purpose of the talk was for promotion, hoping more students would join the blood donation activity. The speaker was ms. MK. By delivering the talk through the PowerPoint, I wondered how many of students might be interested in the history of blood transfusion centers, the functions of different parts in blood cells, etc. When it closed to the lunch hour, a ‘short’ video was getting started! Students showed ‘ORZ’ when watching that 70s or 80s cartoon show with repeated information on the functions of blood cells and the importance of blood donation. The first blood donation day was held on 2007-10-25 (THU) and the number of blood donors was 58 only. I remembered that there were 108 blood donors once in a blood donation activity. It seems that the HSL was Tsang Chi Fung (?) that year. The main reason was not something about our students, but the lack of manpower from the blood transfusion centers. There were only 8 chairs! (Instead of 12 chairs in early time) Ms. MK asked me to extend the time of using the hall to 5 p.m.! Oh, it was impossible for the full booking. Before that, a red cross member’s mother called me and asked me that why her son was required to do the service in school hours! Oh, my.

## 2007/10/24

### [AL][AM] Solving DE, sorry!

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 5:29 下午

It’s my fault to give the following D.E. to F.6 students for solving.

$\frac{d^2y}{dx^2} + y = \tan(x)$

They can’t use the method of undetermined coefficients as stated in the note to solve. The problem is what $y_p$ we should take? They may use the method of Variation of Parameters by taking

$y_p = L_1y_1 + L_2y_2$

where $L_1$, $L_2$ are functions in $x$ to be determined and $y_1$ and $y_2$ are two functions in the complementary solution $y_c$.

In this example, it’s easy to obtan $y_c = A\sin(x) + B\cos(x)$ and we take $y_1 = \sin(x)$ and $y_2 = \cos(x)$.

To determine $L_1$ and $L_2$, we establish the following:

$L_1'y_1 + L_2'y_2 = 0$
$L_1'y_1' + L_2'y_2' = \tan(x)$

Solving the simultaneous equations above, we have

$L_1' = \sin(x)$
$L_2' = \cos(x) - \frac{1}{\cos(x)}$

Performing integration, omitting the arbitary constants, we have

$L_1 = -\cos(x)$
$L_2 = \sin(x) - \ln|\tan(x) + \sec(x)|$

Hence we obtain the $y_p$ as

$y_p = L_1y_1 + L_2y_2$
$= (-\cos(x))\sin(x) + (\sin(x) - \ln|\tan(x) + \sec(x)|)\cos(x)$
$= -\cos(x)\ln|\tan(x) + \sec(x)|$

And the general solution to the D.E. is

$y = A\sin(x) + B\cos(x) -\cos(x)\ln|\tan(x) + \sec(x)|$

Just replace $\tan(x)$ by $f(x)$ for more general case.

Um, sorry to tell you that the method above is “out of syllabus", time wasting, right?

## 2007/10/23

### [初中] Simple geometry problems

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 6:20 下午

Here’s just an assignment question given to F.2 students.
Refer to the figure, AB // CE. D lies on CE. BC cuts AD and AE at G and F respectively.

Show that $\frac{AF \times EC}{EF} = \frac{AG \times DC}{DG}$.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

In most of the cases, showing something about ratios in lengths, we look for similar triangles. It is easy to show that △ABF ~ △ECF and △ABG ~ △DCG. To help to visualize the situation, it may be better to draw similar triangles in “proper orientations”, that isFor △ABF ~ △ECF, we draw

Now, try to look at the conclusion we need to show, L.H.S. involves AF, EC, EF and all these lengths are shown in the figure above. By similarity of △ABF and △ECF, we have
$\frac{AF}{EF} = \frac{AB}{EC}$

$\frac{AF \times EC}{EF} = AB$ — (1)

Now, we create the expression of L.H.S. now. Similarly, we look at another pair of similar triangles.

For △ABG ~ △DCG,

At a look at the conclusion, R.H.S. involves AG, DC and DG, and all these lengths are shown in the figure above. By similarity of △ABG and △DCG., we have,

$\frac{AG}{DG} = \frac{AB}{DC}$

$\frac{AG \times DC}{DG} = AB$ — (2)

By comparing (1) and (2) , we can draw the conclusion that
$\frac{AF \times EC}{EF} = \frac{AG \times DC}{DG}$

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Here is another textbook question, but I think not many students could manage to do it. Thus I should teach them here for saving my time. Refer to the following figure. △ACE and △CBF are equilateral. D lies on AB. $\angle$BCA and $\angle$CDB are right angles. Prove that △ADE ~ △CDF. [Please do it on you own first, then try to compare your method with mine.]

It is very easy to show $\angle EAD = \angle FCD$. Since $\angle EAC = \angle FCB = 60^o$ and $\angle CAD = \angle BCD$ just consider the right angles.

Then, what’s next? Can I find another pair of equal corresponding angles? It seems that it is not easy. Hence we need to switch to consider ratio of lengths.

If $\frac{AD}{CD} = \frac{AE}{CF}$, it is done by stating the reason that “ratio of 2 sides, inc. $\angle$”. But how to show $\frac{AD}{CD} = \frac{AE}{CF}$?

The ratio $\frac{AD}{CD}$ makes me consider △ABC, we come across this type of setting many times in the topic of similar triangles. It is easy to find three triangles in △ABC such that they are similar to each other. Namely, △ADC ~ △CDB ~ △ABC.

Consider the similarity of △ADC and △ABC, it is easy to obtain

$\frac{AD}{AC} = \frac{CD}{CB}$
$\frac{AD}{CD} = \frac{AC}{CB}$

But AC = AE and CB = CF (look at equilateral triangles), hence we have

$\frac{AD}{CD} = \frac{AE}{CF}$

And the story ends.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

## 2007/10/18

### Happy blogging

Filed under: Life — johnmayhk @ 6:48 下午

1. The most frequently seen post in xanga or blog etc seems to be something like “long time no update la” (Um, I can see the date wor.)
2. Avoid saying something like “I’m busy” in blog, someone must give comment quickly that “then why you have time to write blog?” (People don’t believe that some blog writers just spend several minutes to finish writing)
3. Minimize the “copy and paste”, try to write something original.

### [AL][CE][PM][AM] C + kL = 0

Filed under: Additional / Applied Mathematics,HKALE,HKCEE,Pure Mathematics — johnmayhk @ 6:46 下午

Suppose a line $L$ cuts a circle $C$ at two distinct points, then, the family of circles passing through the interesting points are given by $C + kL = 0$. (Um, we all know that $C$ is a circle and $L$ is a straight line, so it is not good to say $C = 0$, $L = 0$ are equations, just for convenience in presentation onwards.) How about when $L$ touches $C$?

What is the meaning of $C + kL = 0$ in this case? This question and its further questions had been discussed some times ago. Suppose $L$ touches $C$ at $P$, it is not difficult to ‘guess’ that $C + kL = 0$ represents the family of circles touching $C$ at the $P$. Well it’s true, but why?
In AL pure mathematics textbook, we have the following question.Let $C: x^2 + y^2 + 2gx + 2fy + c = 0$
Suppose $P(x_1,y_1)$ lies on $C$.
Show that
$(x^2 + y^2 + 2gx + 2fy + c) + k(x_1x + y_1y + g(x + x_1) + f(y + y_1) + c) = 0$
represents a circle touches $C$ at $P$ for any real value of $k$.

It is given in the solution that,

since $L : x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0$ is the tangent to $C$ at $P$, $C + kL = 0$ represents a circle touches $C$ at $P$.

First of all, it’s unlikely to come across this type of question in the present public examination, so students, just be calm. For discussion purpose only, do you think that the one-sentence solution given is a proof? (“What is a proof?” or “What is a valid proof” may induce many further mathematical or philosophical discussions, I hope I have time to share it later.)

I think that we need to explain why $C + kL = 0$ touches $C$.

How to show two circles touching each other? There should be many ways. Intuitively, it is not difficult to see. Just think about “common tangent is the limiting case of common chord”. But is it a “proof”? Seems to be not satisfactory. One commonly used method in additional mathematics is to show the distance between centres of two circles is equal to either the sum or the difference of radii. But it’s a nightmare to apply this to the previous question. I tried another way in class and was buried in complicated algebraic operations very soon. Firstly, it’s easy to show $P$ lies on $C + kL = 0$. Then I wrote down the tangent $T$ to $C + kL = 0$ at $P$. Next, write down the tangent $T_1$ to $C$ at $P$. Finally, I’d like to show $T$ is the same as $T_1$ by showing the corresponding coefficients of $x$ and $y$ and the constant term are in proportion. But, it’s another nightmare.

Trying again, for $P$ is a common point of $C$ and $C + kL = 0$, I’d like to show that the slope of tangent to $C + kL = 0$ at $P$ is exactly the slope of tangent to $C$ at $P$. Then the amount of calculation is reduced. Here is the sketch.

The tangent to $C$ at $P$ is $x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0$ and hence the slope of tangent at $P$ is $-\frac{y_1 + f}{x_1 + g}$. Besides, to find the slope of tangent to $C + kL = 0$ at $P$, we differentiate $C + kL = 0$ with respect to $x$, yield $2x + 2y\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} + k(x_1 + y_1 + g + f\frac{dy}{dx}) = 0$, hence, putting $(x,y) = (x_1,y_1)$, we have $\frac{dy}{dx} = -\frac{(k + 2)(y_1 + f)}{(k + 2)(x_1 + g)} = -\frac{y_1 + f}{x_1 + g}$ which is the same as the slope of tangent to $C$ at $P$.

Is it a proof? Or a good proof?

Then, I think another way just after finishing the above proof: method of contradiction. We know that $P$ is a common point of $C$ and $C + kL = 0$. Suppose $C + kL = 0$ does not touch $C$, then $C + kL = 0$ will cut $C$ at another point $Q$ (other than $P$) (See the figure below). Let $L_1$ be the line $PQ$, obviously, $L_1$ is NOT $L$. Now consider the family of circles passing through $P$ and $Q$, i.e. $C + k_1L_1 = 0$. Note that $C + kL = 0$ is one of the member of that family, hence $C + kL = C + k_1L_1$ for certain $k_1$ (here, I must emphasize that $C = 0$ and $L = 0$ stand for equations of circle and line), resulting $kL = k_1L_1$ which is implying that the line $L$ is the same as (or equivalent to) $L_1$, contradicting to the fact that $L_1$ is NOT $L$. In this proof, there’s nearly no calculation is involved.

Is it a proof? Or a good proof? Students, could you give another proof?

The remaining case should be, when there is no intersection between $C$ and $L$, what is $C + kL = 0$? Together with the question asked by Zuson, “what is the meaning of $x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0$ if $(x_1,y_1)$ lies insides the circle?”, I need more time to study and present. Just end my utterance here.

## 2007/10/16

### [初中] 對角線經過多少正方形？

Filed under: Junior Form Mathematics — johnmayhk @ 2:44 下午

## 2007/10/15

### [AL][PM] Be careful with the fundamental theorem of integral calculus

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:50 下午

Fundamental theorem of integral calculus tells us that, for continuous function $f(t)$, we have

$\frac{d}{dx}\int_a^xf(t)dt = f(x)$

Now, let’s consider the function $f(t) = (x - t)^n$, by using the above, we have

$\frac{d}{dx}\int_0^x(x - t)^ndt = f(x) = (x - x)^n = 0$

Fine? Um, it is not the correct answer. We could obtain the correct answer $x^n$ by the following way:

Let $u = x - t$, $du = -dt$
$\frac{d}{dx}\int_0^x(x - t)^ndt = \frac{d}{dx}\int_x^0u^n(-du) = \frac{d}{dx}\int_0^xu^ndu = x^n$.

Or, more elementary,

$\frac{d}{dx}\int_0^x(x - t)^ndt$
$=\frac{d}{dx}\int_0^x\sum_{r=0}^nC_r^nx^{n-r}(-1)^rt^rdt$
$=\frac{d}{dx}\sum_{r=0}^nC_r^n(-1)^rx^{n-r}\int_0^xt^rdt$
$=\frac{d}{dx}\sum_{r=0}^nC_r^n(-1)^rx^{n-r}(\frac{x^{r+1}}{r+1})$
$=\frac{d}{dx}\sum_{r=0}^n\frac{C_r^n(-1)^r}{r+1}x^{n+1}$
$= (n+1)\sum_{r=0}^n\frac{C_r^n(-1)^r}{r+1}x^n$
$= x^n$
(Note: F.6 students, try to verify the last equality)

Then why we obtain the incorrect answer in using the first method?

## 2007/10/05

### [初中] Correct answer is the most important?

Filed under: Junior Form Mathematics — johnmayhk @ 5:42 下午

I gave the following question to my F.2 class.
Refer to the following figure.

Given that $AB = CD$ and $\angle A = \angle D = 36^o$, find the size of $\angle B$.
Many of them could obtain correct answer $144^o$, but only one student could give correct method.

Well, it’s not bad to give $144^o$, right? Could I appreciate their “number sense" or “spatial sense" by the way? Or, next time, I should draw the figures ugly enough to remind students that “seeing is believing" may not come first, at least, the proofs are at the higher priority in mathematics.

As you may guess, many students claimed that $\angle B = \angle C$ or $AD // BC$ without any reason.

My suggested solution is extending $AB$ and $CD$, and suppose two lines intersect each other at a point $E$. (Well, how to ensure the existence of $E$ anyway?) Then, we try to show that $EB = EC$ and hence $\angle EBC = \angle ECB$; therefore we have $\angle ABC = \angle DCB$.

Many students hate to do problems in geometry, because it requires clear and logical presentation, sometimes, the process is more important than the conclusions (answers); however, people tend to pay attention to the conclusion (outcome) only without enjoying or appreciating the process. Well, it may be my fault to set the expectation too high. I need adjustment.

### [AL][PM] Past Paper 1998 Paper II Q.11

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 5:01 下午

(a)

Let $f$ be a non-negative continuous function on [a , b]. Define $F(x) = \int_a^xf(t)dt$ for all $x\in [a,b]$.

Show that $F$ is an increasing function on [a , b]. Hence deduce that if $\int_a^bf(t)dt = 0$, then $f(x)\:=\:0$ for all $x\in [a,b]$.

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$F$ 的遞增性，得 $F(a)\:\le\:F(x)\:\le\:F(b)$，即 $\int_a^af(t)dt \le F(x) \le \int_a^bf(t)dt$，亦即 $0\le F(x)\le 0$，故 $F(x) = 0$ on [a , b]，意思是 $F(x)$ 在 [a , b] 區間內恆等於零，即 $F(x)$ 是 constant function 是也。那麼，對 $F(x)$ 求導，得 $F'(x) = f(x) = 0$，證畢。

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(b)

Let $g$ be a continuous function on [a , b]. If $\int_a^bg(x)u(x)dx = 0$ for any continuous function $u$ on [a , b], show that $g(x) = 0$ for all $x \in [a,b]$.

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Part (a) 的結論是 $f(x) = 0$，part (b) 的結論是 $g(x) = 0$，我們自然想到 $f(x)$$g(x)$ 有沒有什麼關聯？Part (a) 中，$f(x)$ 是一個非負函數；Part (b) 中，$g(x)$ 不一定是非負函數，所以我們斷不能直接 put $g(x) = f(x)$ 的。看看 $g$ 的條件：對任何連續函數 $u(x)$，恒有 $\int_a^bg(x)u(x)dx = 0$。Pure Mathematics 的題目常有這樣的技巧：既然對任意的 $u$ 也成立，那麼我們可否代入某個特別的 $u$，使我們做出一些結果？如果要運用 part (a)，我們要代入一個特別的 $u$，以產生一個非負函數。不妨設 $u(x) = g(x)$（這不是唯一選擇，$g^{2n-1}(x)$ 也可），則，由 $g(x)$ 的條件，有 $\int_a^bg(x)g(x)dx = 0$$\int_a^bg^2(x)dx = 0$，看，我們製作了一個非負函數 $g^2(x)$，我們便可運用 part (a)，設所謂的 $f(x)$$g^2(x)$，得 $g^2(x) = 0$，即 $g(x) = 0$.

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(c)

Let $h$ be a continuous function on [a , b]. Define $A = \frac{1}{b-a}\int_a^bh(t)dt$.

(i) If $v(x) = h(x) - A$ for all $x \in [a,b]$, show that $\int_a^bv(t)dt = 0$.

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(ii) If $\int_a^bh(x)w(x)dx = 0$ for any continuous function $w$ on [a , b] satisfying $\int_a^bw(x)dx = 0$, show that $h(x) = A$ for all $x \in [a,b]$.

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1. $\int_a^bg(x)u(x)dx = 0$ for any continuous function $u$ on [a , b].

2. $\int_a^bh(x)w(x)dx = 0$ for any continuous function $w$ on [a , b] satisfying $\int_a^bw(x)dx = 0$.

$\int_a^bh(x)v(x)dx = 0$

$\int_a^b(A + v(x))v(x)dx = 0$

$\int_a^bAv(x)dx + \int_a^bv(x)v(x)dx = 0$

$\int_a^bv^2(x)dx = 0$ [因為 $\int_a^bv(x)dx = 0$]

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