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The Easter Eggs in the Examinations WONG Hang-chi

教學難點 4: 源於教具不足

7. The Easter Eggs in the Examinations

(b) Find the least value of n such that the sum of the first n terms of the sequence is greater than 8 × 1018 .

The solution of (a) involves some standard techniques of geometric sequences. Let a and r be the 1st term and the common ratio respectively. According to the question, we have two equations ar2 = 144 and ar5 = 486 . After solving, we have

r = 1.5 and a = 64 , provided that the terms in the given sequence

are all real. Since the calculations are quite simple, we omit the detailed steps here.

Part (b) of the question deserves our attention. The question called for “the sum of the first n terms”, which gave us the clue to apply the summation formula of a geometric sequence. We then set up an inequality

64(1.5n – 1)

1.5 – 1 > 8 × 1018 Therefore, we have

1.5n > 6.25 × 1016 + 1

Since the common logarithmic function is increasing, we have log 1.5n > log(6.25 × 1016 + 1)

which implies

n log 1.5 > log(6.25 × 10

16 + 1)

Using a calculator, we obtain n > 95.38167941, and so the

Everything seems to be fine, except for the step that we apply the logarithm. According to the examination regulations, only calculators that are “HKEAA Approved” may be used in the HKDSE. However, these calculators, such as Casio fx-50FH II, usually do not support high-precision arithmetic (HPA). In fact, Casio fx-50FH II can handle at most 15 significant figures in its memory. In other words, the expression

6.25 × 1016 + 1 = 62 500 000 000 000 001

will be rounded off to 62 500 000 000 000 000, correct to 15 significant figures in the calculator memory. For this question, the round off error is tolerable. It is because we can later verify that

64(1.595 – 1)

1.5 – 1 ≈ 6.852981824 × 1018 < 8 × 1018 while

64(1.596 – 1)

1.5 – 1 ≈ 1.027947274 × 1019 > 8 × 1018

Unfortunately, this method does not always work. It might fail in some situations where the round off error cannot be ignored.

Let us approach this problem in the general situation. Suppose that we have a geometric sequence with the 1st term a > 0 and common ratio r > 1 . We aim to find the least value of n such that the sum of the first n terms of the sequence is greater than

some constant b > 0 . Using the method suggested above, we have

n – 1)

and so rn >

a

+ 1 Using logarithm, we have n log r > log



a

+ 1

Thus, the least value of n is log



a

+ 1 log r .

We observe that if

a

is very large, say, over the order of 1015 , then adding 1 to it becomes virtually negligible in the calculator. We may get into trouble if it happens that

< rn

b(r – 1)a

+ 1

It is because when we compute the value of log



a

+ 1 log r

using the calculator, the “approximate” value of log



a

log r might be returned instead.

To demonstrate this situation of the round off error, let us consider the following example.

In the geometric sequence 9 , 90 , 900 , ... , find the least value of n such that the sum of the first n terms of the sequence is at least 1016 .

The question is equivalent to finding the smallest positive integer n such that

9 + 90 + 900 + ... + 9(10)n – 1 ≥ 1016

It is obvious that the common ratio of the geometric sequence is 10, we can then establish the inequality

9(10n – 1)

10 – 1 ≥ 1016 Therefore, we have

10n ≥ 1016 + 1 By trial and error, we can easily see that

1016 < 1016 + 1 and 1017 > 1016 + 1

Since the sequence 10n is monotonic increasing, we conclude that the required least value of n is 17.

We shall see an interesting phenomenon if we try to solve 10n ≥ 1016 + 1

we get n log 10 ≥ log(1016 + 1)

Astonishingly, we arrive at the result n ≥ 16, which gives the least value of 16. How come we have two different solutions to the same question? It seems that we have done each step under logical deduction, but why the calculator returns an obviously erroneous answer? Recall that the HKEAA approved Casio fx-50FH II can handle at most 15 significant figures. The problem here is that 1016 is far larger than 1015. To be precise, we know that

1016 + 1 = 10 000 000 000 000 001

To store this exact value into the calculator, at least 16 significant figures are required, which exceed the memory capacity of Casio fx-50FH II. Therefore, the calculator will automatically round off the number to 1016, and thus give rise to an incorrect result. In fact, we can use computer programs that support HPA to find the approximate value of log(1016 + 1). The following value is found by Wolfram Alpha, an online computational knowledge engine developed by the company Wolfram Research.

log(1016 + 1)

≈ 16.000000000000000043429448190325180593640482375 401514738622407081

Interested readers might visit the Wolfram Alpha website (https://www.wolframalpha.com/) to explore the engine. We thus find that log(1016 + 1) is slightly larger than 16, but a calculator just do not have enough significant figures to display this value to a higher precision.

The moral of this question is that checking the answer is always important. A calculator is a merely an aid when solving problems. Not only should we make suitable use of the calculator, but also justify our answers by appropriate logical reasoning.

PQRS is a quadrilateral paper card, where PQ = 60 cm, PS = 40 cm,

PQR = 30°, PRQ = 55° and

QPS = 120°. The paper card is held with QR lying on the horizontal ground as shown in Figure 1.

Figure 1

(a) Find the length of RS.

(b) Find the area of the paper card.

(c) It is given that the angle between the paper card and the horizontal ground is 32°.

(i) Find the shortest distance from P to the horizontal ground.

(ii) A student claims that the angle between RS and the horizontal ground is at most 20°. Is the claim correct?

Parts (a) and (b) merely involved some relatively simple techniques in trigonometry and mensuration, and so we shall omit the detailed calculations here. But note that the result of (a),

P

Q

R

S

RS ≈ 16.90879944 cm ≈ 16.9 cm will be used in the later parts

of the question.

Concerning (c)(i), it is helpful for us to draw some auxiliary lines in the figure in order to help us understand the situation. It is clear that the point G on the horizontal ground closest to P must be vertically below P. In other words, G is the projection of P onto the horizontal ground. We then join G and P. Now, the required distance is GP. It is given that the angle between the paper card and the horizontal ground is 32°. This information lead us to think of the angle between the two planes PQRS and

GQR, with the line of intersection QR. It is natural to drop the

foot of perpendicular H from P to QR, and draw the lines GH and HP, as shown in Figure 2 below.

P

Q

R

S

H G

K

T

Thus, we have GHP = 32°. As we know that HP = 60 sin 30°

= 30 cm , we have GP = HP sin 32° = 30 sin 32° ≈ 15.9 cm.

But wait! In Mathematics, it is crucial to justify that our answer is correct. By definition, the two arms of the angle between the planes PQRS and GQR must be both perpendicular to QR. By construction, we can make sure that PHQ = 90°. Nevertheless, how do we know that GHQ also equals 90° ?

Actually there is a little trick here, namely, the theorem of three perpendiculars, which will be added into the revised curriculum, as mentioned in Learning Objective 14.8 in the Curriculum and

Assessment Guide (Secondary 4 - 6). It is recommended that the

schools should implement the revised senior secondary Mathematics curriculum for the Compulsory Part at Secondary 4 to 6 progressively from Secondary 4 with effect from the school year 2023/24.

The theorem of three perpendiculars states that if GP is perpendicular to the plane GHQ and PHQ = 90°, then

GHQ = 90°. The marking scheme produced by the HKEAA took this theorem for granted in this question, and did not include a proof of this fact. However, we found it beneficial for teachers and students to understand the theorem more

thoroughly. Two different proofs of this theorem will be presented here, the first using the elementary geometric approach, and the second using vectors.

Let us investigate the first method. As GP is perpendicular to the plane GHQ , GP must also be perpendicular to any straight line on plane GHQ , in particular GQ and GH. By Pythagoras’

Theorem, we have

GP

2 + GQ2 = PQ2 ... (1) and

GP

2 + GH2 = HP2 ... (2) Also, since PHQ = 90°, we deduce that

HP

2 + HQ2 = PQ2 ... (3) Substituting equations (1) and (2) into (3), we get

(GP2 + GH2) + HQ2 = GP2 + GQ2

GH

2 + HQ2 = GQ2

The GP2 term amazingly canceled out, and using the converse of Pythagoras’ Theorem, we conclude that GHQ = 90°.

The second proof is very elegant and it involves the techniques of vectors in the Extended Part Module 2. In the foregoing arguments, the dot product of any two vectors a and b is denoted by a · b. The term “inner product” or “scalar product” are also sometimes used interchangeably in some textbooks, with the

notation ‹a, b› standing for a · b. It is well known that a · b = 0 if and only if a and b are orthogonal, that is, either a and b are perpendicular to each other, or at least one of the vectors a and

b equals 0.

We know that GP is perpendicular to the plane GHQ, and so

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