Quod Erat Demonstrandum

十一月 3, 2009

不能秒殺的提問之在三角形內

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 12:43 pm
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剛下課,中五生 Woody 問:如何判別一點是否在三角形內?

即已知 A(a,b), B(c,d), C(e,f),如何知道 D(x,y) 是否在三角形 ABC 內?

畫一個準確的圖當然可以,但這不是數學人的答案。

我只想出一個方法: (更多…)

十一月 2, 2009

病中打字

Filed under: Additional / Applied Mathematics, Fun, HKCEE, Junior Form Mathematics — johnmayhk @ 10:44 pm
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一。釘已打

太太生日,想談更多;但這不宜,只記某刻:她去看衫,兒子上課,我去看書。

基礎數學,頗為有趣,暼見如下:

P1130376 (904 x 768)

唯銀根出缺,書釘已打,匆匆離去,滿足。新近一套:《數學文化小叢書》,短小精悍,其中一本,價值 $14,題為《幾何學在文明中所扮演的角色》,著者項武義教授(姑勿論存在關於項教授的負面言論),也吸引了眼目。在起始一段: (更多…)

十月 18, 2009

致 5E 同學

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 1:40 pm

在測驗中,我隨便擬了一道極顯淺的題:

設直線 L 及曲線 C 的方程分別是 x + 4y = 03x + (y - 1)^3 = 4。若 L (的圖像) 切 C (的圖像) 於 P,求 P 的坐標。另設 C 上一點 Q,其 y-坐標為 1,求 CQ 處的法線(normal)之方程。 (更多…)

十月 14, 2009

1,2,3,4 之後是

Filed under: HKCEE, Junior Form Mathematics, NSS, mathematics — johnmayhk @ 7:32 am

為讓 NSS 的同學多一點探究,在下嘗試在數學課引入一些活動,其中一個舊活動是「交通擠塞」,見

http://mathforum.org/alejandre/java/jam/Jam.html

當天和同學探究 n 對「人」和「最少步數」的關係,易知

n = 1,「最少步數」是 3;
n = 2,「最少步數」是 8;

隨即,我著同學「估」:當 n = 3 時,「最少步數」如何? (更多…)

九月 27, 2009

奇異解

Filed under: Additional / Applied Mathematics, HKALE, HKCEE — johnmayhk @ 9:11 pm
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感謝中五的 Carman 回應了上一個 post,讓我也閒聊幾句,高手見諒。

比如

\frac{dy}{dx} = f(x)

那麼 y 其實是什麼?

尋找 y,就是解微分方程的過程,上例不過用積分,得到

y = \int f(x)dx (更多…)

九月 25, 2009

Find dy/dx at a point not on the curve

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 4:50 pm
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When distributing the marked test paper to students, one student, Carman, reminded me that there was a ‘question’ in the following question:

If x^3 - 4x^2y + 3xy^2 - y^5 = 10, find \frac{dy}{dx} at the point (-2,1).

Carman said, ‘the point does NOT lie on the curve.’

Good observation! I had to say thank you to him. Although I’m not the setter, I should bear the responsibility of checking the paper.

But a natural follow-up question turns up: what is the meaning of the number \frac{dy}{dx}|_{(-2,1)} = \frac{31}{33} we are obtaining? Is the number meaningless or standing for something?

Let’s consider a simple example. (更多…)

九月 24, 2009

Differentiation of parametric equations

It is extremely easy to set up questions about differentiation techniques (but good real life application questions are really rare, esp. at secondary school level), apart from tedious computation, when the differentiation involves parameter, students may have difficulties, like mistaking:

\frac{d^2y}{dx^2} = 1/\frac{d^2x}{dy^2} (wrong!)
\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2}/\frac{d^2x}{dt^2} (wrong!)

Here is a question in recent quiz, which involves parametric equations: (更多…)

九月 14, 2009

類似地?

小心,對一些運算法則,我們定要正本清源,不能單以一句「類似地」便隨便進行「類似」運算。

e.g. 1 循環小數

0.3 \times 0.4 = 0.12 正確,但不是「類似地」得到:

0.\dot 3 \times 0.\dot 4 = 0.\dot 1\dot 2(錯!) (更多…)

八月 30, 2009

Finding general term by generating function

It is extremely easy to set up questions on number pattern, like

1, 3, 8, 19, 42, 89, ?

for more details, we may tabulate the question as:

the question is, when n = 6, what is the value of a_n?

My first reply to such kind of question is

“no need to do” (更多…)

八月 23, 2009

等闊曲線 2

Filed under: HKCEE, mathematics — johnmayhk @ 12:28 pm
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上回介紹了其中一類等闊曲線:curvilinear polygon(曲線多邊形)的構作方法。依循該作法而得的曲線多邊形,其角(corner)的數目必為奇數,為何?以下圖顯示的七角曲線多邊形為例:

可以看到,

「每一個角對面對應著一個圓弧;同樣,每一個圓弧對面也對應著一個角」 – - – - – - (*)

任意由一個角出發,比方說,由角 A 出發, (更多…)

八月 10, 2009

等闊曲線 1

Filed under: Fun, HKCEE, mathematics — johnmayhk @ 9:32 pm
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以前看小學 ETV,其中有討論:為何車輪的形狀要是圓形?因為圓形車輪可使車輛行駛時平穩,為說明這事,ETV 中虛擬(比方說)三角形車輪的情況,結果車子行駛時上下嚴重搖晃,非常不平穩云云。所謂平穩,乃指在行車時車身和平地距離保持一致,即 w = 常數(見下圖)。

諸君莫見笑,車,豈會如此簡單?平穩與否,起碼和懸掛系統有關。抱歉,我講數而已。

車輪圍繞固定車軸轉動,那麼要保持(所謂的)平穩,車輪必為圓形;但若沒有車軸,純粹好像原始人把重物放置在一排圓木上推動般,那麼除了圓形,還有沒有別的形狀,可以達至「w = 常數」的效果? (更多…)

七月 2, 2009

十進制轉二進制

Filed under: HKCEE, Junior Form Mathematics, mathematics — johnmayhk @ 12:13 pm
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同事在中二的數學考卷擬了一道題:把 101.5_{(10)} 表達成二進制的數字。

整數部分,同學易求,現在的關注點是

0.5_{(10)} 如何轉成二進數?這個也簡單,

0.5_{(10)} = \frac{1}{2} = 0.1_{(2)}

對卷後,梁同學問我,那麼

0.4_{(10)} 如何轉成二進數? (更多…)

六月 26, 2009

閒談一些基本東西:導數符號,函數,解釋

1. 高階導數的符號

同學問,為何 D 兩次(即求二階導數)的符號是

\frac{d^2y}{dx^2}

而不是

\frac{dy^2}{dx^2} 或 \frac{d^2y}{d^2x}(更多…)

六月 16, 2009

橢圓規

Filed under: Additional / Applied Mathematics, HKALE, HKCEE, Pure Mathematics — johnmayhk @ 5:57 pm

不知有沒有授課員用過橢圓規這個教具?(實情我不知這名稱是否正確,網上找到 Ellipsograph 這個字,不知是否橢圓規的正確英文名稱。)

20090608-ellipsograph

我「靜靜雞」用科組錢買了一個,操作見下: (更多…)

六月 4, 2009

溫書題

Filed under: HKALE, HKCEE, Pure Mathematics, mathematics — johnmayhk @ 4:28 pm
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這是溫書時期。

1. 有關數列的題目

這是校內 2007-2008 年度純數期終試其中一題:
=======================================
Let {a_n} be a sequence of positive integers. Define sequences {b_n} and {c_n} as
b_1 = a_1, b_2 = a_1a_2 + 1, b_{n+2} = a_{n+2}b_{n+1} + b_{n}. (n \in \mathbb{N})
c_1 = 1, c_2 = a_2, c_{n+2} = a_{n+2}c_{n+1} + c_{n}. (n \in \mathbb{N})
Let x_n = \frac{b_n}{c_n}. (n \in \mathbb{N})

Show that x_1 \le \lim_{n \rightarrow \infty}x_n \le 1 + x_1.
======================================= (更多…)

五月 23, 2009

點數差的期望值

Filed under: Additional / Applied Mathematics, HKALE, HKCEE, mathematics — johnmayhk @ 3:03 pm

擲兩顆公平骰子一次,求兩個點數差的期望值。

學生甲:

嗯,擲一顆骰子,點數的期望是 \frac{1}{6}(1+2+3+4+5+6) = 3.5。無論擲兩顆,三顆,它們的點數期望值皆是 3.5,故此,兩個點數差的期望值就是 3.5- 3.5= 0(更多…)

五月 12, 2009

用 logarithm 計出頭幾個位

Filed under: HKCEE, mathematics — johnmayhk @ 12:05 pm
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昨天匆匆出了份中四 logarithm 的習作,其中一題:

Find the first digit of 1997^{2009}.

留意,我們不是找「個位」,而是找「第一個位」是什麼。 (更多…)

五月 6, 2009

2009 CE Additional Mathematics Paper Section B

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 6:04 pm

重點題目:

以下是 2009 CE Additional Mathematics Q.18,同學,看看你可否在 16 分鐘內正確地完成它:

The following figure shows a park AED on a horizontal ground. The park is in the form of a right-angled triangle surrounded by a walking path with negligible width. Henry walks along the path at a constant speed. He starts from point A at 7:00 am. He reaches points B, C and D at 7:10 am, 7:15 am and 7:30 am respectively and returns to A via point E. The angles of elevation of H, the top of a tower outside the park, from A and D are 45^o and 30^o respectively. At point B, Henry is closest to the point K which is the projection of H on the ground. Let HK = h m.

(a) Express DK in terms of h. (1 mark)
(b) Show that AB = h\sqrt{\frac{2}{3}} m. (3 marks)
(c) Find the angle of elevation of H from C correct to the nearest degree. (3 marks)
(d) Henry returns to A at 8:10 am. It is known that the area of the park is 9450 m^2.
(i) Find h.
(ii) A vertical pole of length 3 cm is located such that it is equidistant from A, D and E. Find the angle of elevation of H from the top of the pole correct to the nearest degree. (5 marks) (更多…)

四月 28, 2009

有關圓形方程和軌跡的無聊討論

Filed under: HKCEE, mathematics — johnmayhk @ 4:28 pm
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【其一】 上課材料

To 4E students: you may download the worksheet (2009-04-24) at

f4-additional-mathematics-worksheet-20090424

together with some exploration at

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877&current_page=&i=1247383&t=1247383

【其二】 笑到喊

形如下式的二元二次方程

x^2 + y^2 + Dx + Ey + F = 0 – - – - – - (*)

中四的同學知道,其圖像代表著圓形(指實圓圓周 circumference of real circle)。

現在反過來,代表著圓形的方程式,是否一定形如 (*)?(即二元二次)

不一定,以下是一條二元四次方程方程

x^4 + 2x^3 + (2y^2 - 2y - 1)x^2 + (2y^2 - 2y - 2)x + y^4 - 2y^3 - y^2 + 2y + 1 = 0

其圖像,也是圓形。為何? (更多…)

四月 27, 2009

2009 CE Mathematics Paper

Filed under: HKCEE, mathematics — johnmayhk @ 3:38 pm

今天考會考數學,同事看卷後,感到題目很淺,考生出來,紛紛說「太淺了,怕取不到 A」云云。

聊舉數例

2009 CE Mathematics Paper I Q.17

image002

Given

A, B, C and D lie on the horizontal ground. CE is perpendicular to the plane ABE.
AC = 28 cm, BC = 25 cm, BD = 6 cm, BE = 24 cm, \angleABC = 57^o.

Find the shortest distance from E to the horizontal ground. (更多…)

四月 1, 2009

Discussion of F.4 General Mathematics Test Paper (2009-03-26)

Filed under: HKCEE, mathematics — johnmayhk @ 9:22 am
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To F.4E students

Please visit my forum and do some exploration on some easy geometry questions in the previous quiz by clicking the following.

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877&current_page=&i=1247268&t=1247268

三月 26, 2009

Just answer an easy additional mathematics question

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 12:09 am

A F.5 student asked me the following question some days ago, reply now.

A(-3,0) and B(-1,0) are two points and P(x,y) is a variable point such that PA = \sqrt{3}PB. Let C be the locus of P.

(a) Show that the equation of C is x^2 + y^2 = 3.

(b) T(a,b) is a point on C. Find the equation of the tangent to C at T.

(c) The tangent from A to C touches C at a point S in the second quadrant. Find the coordinates of S.

(d) L is a straight line which passes through point A and makes an angle \theta with the positive x-axis, where -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}. Q(x,y) is a point on L such that AQ = r. (See the figure below)

(i) Write down the coordinates of Q in terms of r and \theta.

(ii) L cuts C at two distinct points H and K. Let AH = r_1, AK = r_2.

(1) Show that r_1, r_2 are roots of the quadratic equation r^2 - 6r\cos\theta + 6 = 0.

(2) Find the range of possible values of \theta, giving your answers correct to three significant figures.

(HKCEE 1999) (更多…)

二月 24, 2009

Minor point in differentiation

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 1:43 pm
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This is a simple question in differentiation.

Let x^2 = \sqrt{y^6} for any real number y, determine \frac{dy}{dx} at (1,-1). (更多…)

二月 20, 2009

Some F.4 textbook questions

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 4:55 pm
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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.

f4-trigo-compound-angle

Show that \tan\alpha = \frac{2m - 1}{\sqrt{3}}. (更多…)

二月 19, 2009

Just an old question about F.4 trigonometry

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 9:55 pm
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It is glad that F.4 students asked me how to evaluate

\sin1^o \times \sin2^o \times \sin3^o \times \dots \times \sin90^o

Here is a way. (更多…)

二月 3, 2009

出現 A 先於 B 的機會

常見題目:擲一顆公平骰子若干次,求得到 1 點先於得到 6 點的概率。

此例容易,因 1 點和 6 點在地位上無異,易知要求概率為 (更多…)

一月 9, 2009

錯在哪裡系列:零是不定義?

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 4:01 pm
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When giving the following basic trigonometry question in F.4 additional mathematics lesson:

Given A + B + C = 90^o, prove that \tan A\tan B + \tan B\tan C + \tan C\tan A = 1.

It should be extremely easy, (更多…)

十二月 7, 2008

做數雕蟲小技系列:看整體

看整體有時比看局部好。舉例

1. Factorize (x + y)^3 + (x - y)^3. (更多…)

十一月 21, 2008

Exist or does not exist

Is giving hints a good way to help students in solving mathematics problems? Urm, sometimes it may not.

Here is a common m.i. question in recent F.4 additional mathematics regular test:

Show that n^3 - n + 3^n is divisible by 3 for any positive integer n. (更多…)

十一月 18, 2008

Create an m.i. question

It is not difficult to create questions like:

Prove by mathematical induction that

\frac{3^3\times1}{4!} + \frac{3^4\times2}{5!} + \frac{3^5\times3}{6!} + \dots + \frac{3^{n+2}\times n}{(n+3)!} = \frac{9}{2} - \frac{3^{n+3}}{(n+3)!} (更多…)

十一月 14, 2008

錯在哪裡系列:切於一點

Filed under: HKCEE, mathematics — johnmayhk @ 12:25 pm
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在下每年幾乎都要教四個課程,同工一定了解,到時到侯就是測驗「旺季」,同時段出三四份卷 (更多…)

十一月 10, 2008

正割餘割

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 9:26 pm
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中四開新課,同學認識一下新「朋友」吧:

正弦 = sine (\sin)
餘弦 = cosine (\cos)
正切 = tangent (\tan)
餘割 = cosecant (\csc) 定義:\csc\theta \equiv \frac{1}{\sin\theta}
正割 = secant (\sec) 定義:\sec\theta \equiv \frac{1}{\cos\theta}
餘切 = cotangent (\cot) 定義:\cot\theta \equiv \frac{1}{\tan\theta}

我今天才知「正割」「餘割」這兩個譯名,慚愧。

十一月 2, 2008

不能秒殺的提問:餘式定理

Filed under: HKCEE — johnmayhk @ 4:53 pm
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教了餘式定理,隨便亂出題:設 f(x) 是多項式,已知

f(x) \div (x - 1),餘式為 3
f(x) \div (x - 2),餘式為 5

那麼,當 f(x) \div (x - 1)(x - 2) 時,餘式是什麼?

同學秒殺之,答案是 2x + 1,毫無難度。但有 4E 同學提出他的解法如下: (更多…)

十月 17, 2008

不能秒殺的提問

在堂上,一般情況下,學生提問,在下多半手起刀落,秒殺解之。今天 4E 同學問了兩個問題,在下不能秒殺:

1. 誰發明 _nC_r 這個符號?
2. 畢氏數組中是否必然存在 3 或 5 的倍數? (更多…)

十月 15, 2008

Assuming step in mathematical induction

Just share a minor point in the presentation of M.I.

To prove that a proposition P(n) is true for all positive integers n by using M.I.

We need ‘4′ steps, namely (更多…)

十月 7, 2008

論商餘(二)

有同學問到綜合除法(synthetic division)之原理,讓我詳述之。

本文共分四部份:
一、算法
二、原理
三、推廣
四、應用
(更多…)

九月 22, 2008

論商餘 (一)

Filed under: HKCEE, Junior Form Mathematics, Pure Mathematics — johnmayhk @ 10:34 pm
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所謂「商」「餘」乃 quotient 及 remainder 是也。這個 post 只是一些有關除法的簡單討論。 (更多…)

九月 3, 2008

Prerequisite of F.4 algebra

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 5:17 pm

As usual, we start with basic algebraic computation drilling at the beginning of F.4 mathematics lessons. But, I just did some useless mathematical chatting and gave something about infinity (e.g. the discussion of 1 – 1 + 1 – 1 + … ) and they showed their excitment with hands clapping. Well, of course, I switched to serious matter very soon. As expected, most of the students did not know the fact that “zero is an even number” (even, they did not know that zero is an integer). I went through some ‘prerequisite’ questions with them, one is:

(a) \sqrt{9} = \pm3  [True or false?]
(b) If x^2 = 9, then x = 3. [True or false?]
(c) The square roots of 9 are 3 and -3. [True or false?]
(d) If x^2 = 9, then x = 3 and x = -3. [True or false?]

F.4 students, here are the answers. (更多…)

八月 4, 2008

[FW][News] 19支火箭之後……

Filed under: HKCEE, Life — johnmayhk @ 12:24 pm

(明報)8月4日 星期一 05:10

【明報專訊】每年會考放榜 ,鎂光燈總是集中在十優狀元身上,聽他們講自己不必苦讀便取得佳績,希望進入醫學院或環球金融系,舉起勝利手勢讓記者拍照。另一邊廂則請來一堆明星名人,勉勵失意考生不必氣餒,會考零分不等於永遠失敗,順便重溫今天賺大錢的明星,當日慘不忍睹的會考分數。

今天又是放榜天,又會產生一班狀元和一班零分生。A有助找工作嗎?A會幫你賺大錢嗎?到底A的意義是什麼?31歲的許正宇(Chris)和40歲的薛俊良(Isaac),在會考高考合共拿到19個A,就由他們來說19個A之後的故事。 (更多…)

七月 31, 2008

正多邊形作圖法

Filed under: Additional / Applied Mathematics, HKCEE, Junior Form Mathematics — johnmayhk @ 11:01 pm

我在工業學校渡過中學生活。已往在工業學校,繪圖是必修科(現在好像稱為『圖象傳意科』)。從中一開始,我們便手拿大大塊的繪圖木版,插著 T 尺上課,每星期也要寫 lettering(功效類似我年代幼稚園練字用的 copy book,不知現在還有嗎?)以打好寫字的基本功。 (更多…)

七月 18, 2008

錯在哪裡之數學歸納法

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 10:11 pm

對於任意自然數 n,證明

n = \sqrt{1 + (n - 1)\sqrt{1 + n\sqrt{1 + (n + 1)\sqrt{1 + (n + 2)\sqrt{1 + (n + 3)\dots}}}}} (更多…)

七月 7, 2008

[FW] 如何造假資料?

Filed under: Additional / Applied Mathematics, HKCEE, Junior Form Mathematics — johnmayhk @ 5:37 pm

http://chanlikhangnick.googlepages.com/writing20080703.htm

簡單有趣的介紹,推薦大家一看!

六月 30, 2008

Something about 2008 CE (additional) mathematics

Filed under: HKCEE, Report — johnmayhk @ 5:18 pm
Tags:

會考數學和附加數學已是『夕陽工業』,下面談的,不到兩三年便成垃圾。 (更多…)

六月 18, 2008

附加數題之最大四邊形

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 4:03 pm

這是一道中四附加數學題。(相信這樣的好題,在將來的公開試會漸漸式微。)它的根本要問的是:一個固定四邊長度的四邊形,在什麼情況下面積最大?題目在教科書可找,詳表如下:
(更多…)

五月 3, 2008

Wrong method, but correct answer

Filed under: Additional / Applied Mathematics, HKCEE, Junior Form Mathematics — johnmayhk @ 7:11 am

Obviously, the methods throughout the following sets are completely wrong in general; however, correct answers will be obtained. Try to verify them and explain why it happens. You may read the explanation at the end for confirmation.

(Set 1) Erase the indices

\frac{5^3 + 2^3}{5^3 + 3^3} = \frac{5 + 2}{5 + 3}
\frac{7^3 + 3^3}{7^3 + 4^3} = \frac{7 + 3}{7 + 4}

(Set 2) Erase the logarithm

\frac{\log 2}{\log 4} = \frac{2}{4}
\log \frac{9}{4}\div \log \frac{27}{8} = \frac{9}{4} \div \frac{27}{8}

(Set 3) Product to sum

\frac{8}{7} \times 8 = \frac{8}{7} + 8
\frac{11}{10} \times 11 = \frac{11}{10} + 11

(Set 4) Out of radical sign

\sqrt{5\frac{5}{24}} = 5\sqrt{\frac{5}{24}}
\sqrt{7\frac{7}{48}} = 7\sqrt{\frac{7}{48}}

(Set 5) Exchange indices

(\frac{5}{7})^2 + \frac{2}{7} = \frac{5}{7} + (\frac{2}{7})^2
(\frac{\pi}{4})^2 + \frac{4 - \pi}{4} = \frac{\pi}{4} + (\frac{4 - \pi}{4})^2

(Set 6) Sum to product

\sec^238^{o} + \csc^238^{o} = \sec^238^{o}\csc^238^{o}
\sec^254^{o} + \csc^254^{o} = \sec^254^{o}\csc^254^{o}

Explanation
Try to prove the following identities to see the reasons why we have equations from set 1 to set 6. (更多…)

四月 26, 2008

有關某圓形特性的逆命題

Filed under: HKCEE — johnmayhk @ 8:33 pm

中四五的同學應該知道:『圓心角是圓周角的兩倍』(angle at centre twice angle at circumference),即是說,若 O 是下圖圓形的圓心,則有 \angle BOC = 2\angle BAC

反過來說,參下圖,假如知道 \angle BOC = 2\angle BAC,是否表示,存在圓心是 O 的圓形,穿過 A, B, C

(更多…)

二月 28, 2008

Evaluate cube roots

Filed under: Additional / Applied Mathematics, HKCEE, Junior Form Mathematics — johnmayhk @ 4:02 pm

A F.5 student, Yan, asked me to do the following without using calculator

\sqrt[3]{10 + 6\sqrt{3}} + \sqrt[3]{10 - 6\sqrt{3}}

(更多…)

二月 23, 2008

Quadratic formula program @ CASIO 3950p/3650p

Filed under: HKCEE, Junior Form Mathematics — johnmayhk @ 11:17 pm
Tags:

驚恐中。我班的中五同學不懂做

Factorize 4x^2 - 4xy + y^2.

(不變金句:教了,不等於學了)

在中二的課,談到 cross method,我教同學用計算機。Casio 3950p (or 3650p) 較以往好,是它顯示的根(roots)是分數,不是點數,這樣可方便同學寫出答案。中五的同學,若你的 Casio 3950p 還未有 quadratic formula,快快看:

http://intranet.sfxc.edu.hk/it-school/homepage/nwc/casio_%20fx_3950p_quadratic_formula_program.doc

(更多…)

二月 20, 2008

Additional Mathematics 補底

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 11:01 am

更新日期:2008-02-21 (希望不斷更新。)
免插聲明:本文為修附加數學科的中五同學純粹補底之用,高手勿插,謝謝。

1. 口訣:一冇 trigo,即寫 general

有關 general solution,考生常犯毛病見下

\sin(2\theta) = \sin30^o
2\theta = 30^o
\theta = 15^o
\therefore \theta = 180^on + (-1)^n(15^o)

錯!最後一步才寫 general solution 的形式是錯的!

不要到最後一步,而是要一開始便寫,即是

\sin(2\theta) = \sin30^o
2\theta = 180^on + (-1)^n(30^o)

即口訣的:一冇 trigo function 符號(本例是 sin),立即 (係立即!!!) 要寫 general solution 的形式。於是,我們有

\sin(2\theta) = \sin30^o
2\theta = 180^on + (-1)^n(30^o)
\therefore \theta = 90^on + (-1)^n(15^o)

才是正確答案。

2. 又 degree 又 radian;個樣衰過 Edit 神

(更多…)

二月 19, 2008

A beautiful question in binomial theorem

Filed under: Additional / Applied Mathematics, HKALE, HKCEE, Pure Mathematics — johnmayhk @ 5:02 pm

證明:對任何正整數 m,n 恒有

\frac{(m + n)!}{(m + n)^{m + n}} < \frac{m!}{m^m}\frac{n!}{n^n}

如何證?MI?

(更多…)

一月 25, 2008

[CE][Math] Statistics: More dispersed?

Filed under: HKCEE — johnmayhk @ 2:52 pm

In CE mathematics, we know that there are at least 3 measurements for dispersion, namely, range, inter-quartile range and standard derivation.

Describing or comparing the dispersion of different sets may be a very important application.

About the standard derivation, you may read my old messages for your reference
http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1159940&t=1159940&v=t
http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=965435&t=959956

Given two numerical data sets A and B.

If we come across the following situations

range of A > range of B and
inter-quartile range of A < inter-quartile range of B

How to compare which set is more dispersed?

Well, you may say, just look at the standard derivation.

OK. If we have the following

range of A > range of B and
inter-quartile range of A > inter-quartile range of B and
standard derivation of A < standard derivation of B

Is it true to draw the conclusion that set B is more dispersed?

Further, how about when (set A \ne set B) such that

range of A = range of B and
inter-quartile range of A = inter-quartile range of B and
standard derivation of A = standard derivation of B

Urm, let me give the following as an example

Set A ={1,10,10,10}
Set B = (1,1,1,10}

Is it true to draw the conclusion that both sets have the same dispersion?

The problem may be something about the vagueness of the concept of “dispersion”. How to resolve it?

一月 12, 2008

[軟件][教學] Scilab 在中學數學的應用初探

Filed under: HKALE, HKCEE, Junior Form Mathematics, Teaching — johnmayhk @ 6:01 pm

以往介紹過一些網上的數學工具,諸如 integrator 及 calc101 等。鑑於部分教師及同學沒有上網習慣,讓我介紹一個離線作業且完全免費的數學工具:Scilab,希望大家用得著。關於這個法國研發的開放軟件,只要大家 google 一下,不難找到更詳盡的資訊,在此不贅。感謝有關研究員之努力及慷慨!幾個月前,我才知這軟件存在,這裡我主要提及它在中學數學的一些應用(再進深的我不懂了):

大家先下載(現時)軟件的最先版本
http://www.scilab.org/download/4.1.2/scilab-4.1.2.exe

安裝後,執行便出現以下版面:

看到閃動著的游標(cursor)嗎?可以輸入東西了。為方便大家『唔駛打』,可下載以下的 text file,之後的例子都在這 file 中,大家可以 copy and paste 來試試例子的效果。

http://johnng.inscyber.net/mathgif2/Scilab-johnmayhk.txt

操作小提示:輸入後,按 Enter 顯示結果。打錯了或想修改某些之前輸入的東西,只要按向上箭咀鍵若干次,重新輸入便可。

數字(Numbers)

以下是一般基本計算機也可處理的運算工作。

這是一些初等函數(elementary functions)的運算,留意其輸入的方式。

更多的函數,比如 arcsin 是 asin(),可參考軟件中的 help file。

多項式(Polynomials)
這裡處理的是一元(single variable)的多項式。對多項式的輸入,我們不能直接輸入諸如 x+1 之類,因為電腦不知道何謂 x,我們要先定義清楚,才可繼續運算。這裡介紹 3 種定義方法:

(1) 直接輸入

(2) 透過系數(coefficients)
當多項式的項數多,直接輸入頗麻煩,我們可以透過系數定義,詳見如下:

留意,不一定用 x,其他字母作 variable 也可,但 coeff 這個就字不能修改。

(3) 透過根 (roots)

方便呀,forming equation 的題目,立即得到結果!

好了,定義了多項式,我們可以進行有關多項式的運算。

1. 基本運算

2. 有關除法

3. 解多項式方程 (solve polynomial equations)

4. 因式分解(factorization)

頗有用的功能!不過,對一些根為『不漂亮』的有理數之多項式,比如 – 7 + 26*x – 19*x^2 + 12*x^3,它的因式分解的答案也同樣『不漂亮』,大家不妨試試及想一想解決方法。(注,我知配合 Maxima.exe 或用 Mathematica 可處理多元的多項式。但不太懂,希望以後再研究一下。)

5. 求最大公因式(GCD)及解丟番圖方程(Diophant equation)
讀 AL Pure Mathematics 的同學注意了,這是 algebra 中輾轉相除法 (Euclidean algorithm) 的題目,感覺如何呢?

6. 補充一下:有關整數求 HCF 及 LCM 的方法

7. 有理函數化成部分分式(resolving into partial fractions)
Partial fractions 這類熱門的 pure mathematics 題目,也可以 Scilab 輕易處理。

顯示的三個ans就是答案了,即

矩陣(Matrices)
Scilab 最強大的功能是處理矩陣,諸君請看!開始時,當然要定義矩陣。方法是相當簡單。

好了,定義了矩陣,我們可以進行有關矩陣的運算。

注意:矩陣的 entries,可以是多項式,運算如舊!

這是有關矩陣『自身』的運算,修 Pure Mathematics 的同學,感覺良好嗎?

其他功能

Scilab 可做的遠超上述,除了有關統計數據資料之處理,還有下列一些。但因為別的,常用的軟件也有類似功能,我略舉以下數例作結,圖收拋磚引玉之效。

一月 9, 2008

[CE][AM] 錯在哪裡 – general solutions to trigonometric equations

Filed under: Additional / Applied Mathematics, HKCEE — johnmayhk @ 12:01 pm

Given that a,b are non-zero constants such that |a| < 1 and |b| < 1.

Solve the following simultaneous equations

\sin(x) = a - – - (1)
\cos(x) = b – - – (2)

Solution (so-called)

divide (1) by (2), yield

\tan(x) = \frac{a}{b}

Hence

x = 180^{o}n + \tan^{-1}(\frac{a}{b}) where n is any integer

上述的答案是錯的,為何?

十一月 7, 2007

[CE] A typical curve sketching question

Filed under: HKCEE — johnmayhk @ 9:40 pm

Just busy with the cheering team practice. For a break, write something quick…

This is a part of the question of 2000-CE-A.Math-Q.10, a typical question.Let y = f(x) = \frac{7 - 4x}{x^2 + 2}.
Show that the maximum and minimum values of f(x) are 4 and -\frac{1}{2} respectively.

When it is given to F.5C students, some of them puzzled and asked, ‘is that something wrong in the question?’ They added, ‘-\frac{1}{2} is maximum and 4 is minimum.’ How come?

In fact, students’ concept is so vague to give that comment. Saying ”-\frac{1}{2} is maximum and 4 is minimum.” is not clear! They should say something like:

y attains its maximum value when x = -\frac{1}{2}. Hence,
to find maximum value of y,
we put x = -\frac{1}{2} into y = \frac{7 - 4x}{x^2 + 2},
that is, the maximum value of y = \frac{7 - 4(-\frac{1}{2})}{(-\frac{1}{2})^2 + 2} = 4.

Also, they should say
y attains its minimum value when x = 4. Hence,
the minimum value of y = \frac{7 - 4(4)}{4^2 + 2} = -\frac{1}{2}.

The designer of this question played little tricks on numbers. It may be quite interesting to set up similar question by choosing ’suitable’ numbers a and b satisfying

a < b;
f(a) = b = maximum value of y;
f(b) = a = minimum value of y; and
It is not too easy (or too difficult) to find derivatives of f(x).

Students try to find such function to help me set up A. Math Examination Paper ^_^.

十一月 6, 2007

[CE] Exploring on curve sketching

Filed under: HKCEE — johnmayhk @ 3:31 pm

1.

Could you guess that equation of the graph of the curve in red (as shown)? You may say it looks like y = x^3. Well, it is actually y = x|x|. Justin discovered that the graphs of y = x^3 and y = x|x| are quite similar in shape. Further, the following shows the graph of y = x^3|x| which looks similar to the graph of y = x^5.

Can we generalize the above as the graph of y = x^{n-1}|x| looks quite ‘similar’ (quite a vague idea actually) to y = x^{n+1} for n > 1? How to explain that finding? I just gave comment briefly that y = x^{n + 1} = x^{n - 1}\times x^2 ‘behaves similarly’ to y = x^{n - 1}\times |x| in the way that x^{n - 1} are multiplied by a non-negative factor. Could you give further comments and explanations?

2.

Many interesting matters have rooms for further discussions. Um, just give you something to ‘play’ with. Can you draw a square? What a silly question. Um, let me refine, could you use one single equation to plot a square with curve plotting software (e.g. Winplot.exe)? I had done one, see

Could you guess the equation of the graph above? Strictly speaking, the above graph is NOT a square (see the trouble corners). How to make the curve above? Just consider:

We all know that the equation of circle is something like

x^2 + y^2 = 1

Instead of making the power of 2, how about

x^3 + y^3 = 1 or
x^4 + y^4 = 1

You may use curve plotting software to see the difference of the graphs of the above curves (could you explain the difference?). When the power of x and y is getting larger and larger, what will be the changes? Try to use winplot.exe to check. In fact, the figure I posted is the graph of x^{100} + y^{100} = 1. It looks like a square except the situation at corners. Students, can you explain, without using any software, why the shape is something like a square? More, if we consider when the powers of x and y are getter smaller, what will be the changes of the shape of graph? Could you imagine the shape of the graphs of the following

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 1
x^{\frac{1}{3}} + y^{\frac{1}{3}} = 1

Finally, when the powers are different, any further observation you may discover? As for example

x^2 + y^3 = 1

We still have a lot of elementary mathematics stuff for exploring, think more!

Also read
http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1231449&t=1231383&v=t

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=358012&t=358012

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=445770&t=445770

十一月 1, 2007

[CE] Linear programming – system of inequalities

Filed under: HKCEE — johnmayhk @ 3:58 pm

1.

Just start teaching the solving of system of 2-variable linear inequalities. One F.5C student, Yeung, told me that “no solution for parallel lines”, Oops, his comment was too brief but I understood him immediately. Let me refine his observation, as follows.

Consider a system of two 2-variable linear inequalities (x and y are real with no extra restriction), the system MAY have no solution only when two different corresponding straght lines are parallel to each other.

As for example,

x + 2y < 1
2x + 4y > 4

The two corresponding lines x + 2y = 1 and 2x + 4y = 4 are parallel, and the system above has no solution.

In other words, the (two 2-variable linear inequalities) system MUST be solvable if the system is ‘corresponding to’ two non-parallel lines. Although it’s not something very striking, I appreciate the discovery, because it is totally orignated from a student himself. Now, could you show me the proof of Yeung’s observation?

[But, most of F.5C students were out of mood of attending the lesson that day, may be they thought that the content is TOO EASY.]

2.

To solve Ax + By + C < 0,
we draw L: Ax + By + C = 0, and
L divides the x-y plane into 2 regions, namely R_1 and R_2.
Knowing that P(x_1,y_1) lies on R_1 and
P satisfies Ax + By + C < 0, that is
Ax_1 + By_1 + C < 0, then
R_1 is the required region.
That is, R_1 represents the solution to the inequality Ax + By + C < 0.

The question is: why we just try one point? Should we try another? Is it possible that there is another point Q lying on R_1 but Q does not satisfy the inequality Ax + By + C < 0? How to prove that “one point is enough for representing the whole region”?

Well, the first thing in my mind is the signed distance (有向距離) from a point to a line.

Additional Mathematics tells us that the distance between a point P(x_1,y_1) and a line L : Ax + By + C = 0 is |\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}|. The absolute sign is for ensuring non-negative distance. How about getting rid of the absolute sign? The numerator Ax_1 + By_1 + C may be positive, zero or negative. What is the significant geometric meaning of the positive and negative values obtained?

Just tell you the following.

If Ax_1 + By_1 + C is positve, then P and O are at the different sides of L (as shown)

If Ax_1 + By_1 + C is negative, then P and O are at the same sides of L (as shown)

Back to our questions, if O does not lie on L, O must lie on one of the regions R_1 and R_2. Suppose O lies on R_1, then all the points on R_1 and O are at the same side of L. Hence any point Q(a,b) lying on R_1 will lead to the result A(a) + B(b) + C < 0. Hence one point is enough! It is impossible to have another point T lying on R_1 such that T does not satisfy Ax + By + C < 0.

Please complete the discussion when O lies on L.

3.

F.5C students asked, ‘how about the case for 3 or more variables?’ To the best of my memory, we may use simplex method.

十月 18, 2007

[AL][CE][PM][AM] C + kL = 0

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


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.

九月 23, 2007

[CE][AM][教學] Common tangents to two circles

Filed under: Additional / Applied Mathematics, HKCEE, Teaching — johnmayhk @ 4:24 pm

Reply to http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1241115&t=1241115

Refer to the following figure, L is a common tangent to C_1 and C_2, find the equation of L.
 

There are, at least, two methods to cope with this question. Let me sketch one.

Let y = mx + c be the equation of L. Since L is a tangent to both circles, the distances between centres and L will be radii. Hence we have |\frac{c}{\sqrt{m^2 + 1}}| = 1 and |\frac{8m + c}{\sqrt{m^2 + 1}}| = 3

By squaring the above equations and eliminating c^2, we have 64m^2 + 16mc - 8(m^2 + 1) = 0 — (*) 

Then c = \frac{-7m^2 + 1}{2m}

Hence (*) can be converted into45m^4 - 18m^2 + 1 = 0

By solving and rejecting, we come up with the conclusion that y = -\frac{x}{\sqrt{15}} - \frac{4}{\sqrt{15}} is the equation of L.

八月 24, 2007

RE M.I. 的 “必然性”?

Filed under: Additional / Applied Mathematics, HKCEE, Teaching — johnmayhk @ 12:19 am

Reply to http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1240450&t=1240450

The proof quoted by Ricky is a problem (not M.I. itself). In the usual procedure in M.I., we need to guarantee P(k) \Rightarrow P(k+1) (say) by valid logic. However in the so-called proof, on assuming the general election to be held in the k^{th} year, we had nothing to say about the situation in the (k + 1)^{th} year, and it is problematic to say證畢 because the so-called proof was incomplete and it could not be a counter-example to establish the statement『我們可以證明 M.I. 並不是經常性成立。』For validity of M.I., you may refer to the following post:http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=926996&t=926996

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