MỘT số vấn đề về dạy TOÁN BẰNG TIẾNG ANH ở cấp THCS

vertex vertices đỉnh3.4 Angle góc complement of an angle góc phụ nhau của một góc supplement of an angle góc bù của một góc tangent chord angle góc tiếp tuyến và dây angles on the same a

MỘT SỐ VẤN ĐỀ VỀ DẠY TOÁN BẰNG TIẾNG ANH Ở CẤP THCS

intersection of (sets) giao của(các tập hợp)

rational number số hữu tỉ

canonical factorization biểu diễn chính tắc

LCM (least common multiple) bộ chung lớn nhất

GCD (greatest common divisor) ước chung nhỏ nhất)

congruent modulo (m) cùng đồng dư modun (m)linear congruence system hệ đồng dư tuyến tínhcomplete set of residue classes hệ thặng dư đầy đủ

simplified residue system hệ thặng dư thu gọn

2.1 Kinds Number (các loại số)

finite decimal (number) số thập phân hữu hạninfinite decimal (number) số thập phân vô hạn

periodical infinite decimal số thập phân vô hạn tuần hoàn

multiple (multiple of) bội (bội của)

divisor (divisor of) số chia, ước số (ước của)

2.3 Fraction (phân số)

2.4 Number-digit (số- chữ số)

3.2 Triangle (tam giác)

Congruent Triangles tam giác bằng nhau

midperpendicular (perpendicular bisector) trung trực

equilateral triangle (regular triangle) tam giác đều

orthocenter(orthocentre) trực tâm

circumcentre (circumcenter) tâm ngoại tiếp

excentre (opposite A) tâm bàng tiếp (đỉnh A)

3.3 Polygon (đa giác)

vertex (vertices) đỉnh

3.4 Angle (góc)

complement of an angle góc phụ nhau của một góc

supplement of an angle góc bù của một góc

tangent chord angle góc tiếp tuyến và dây

angles on the same arc các góc chắn cùng một cung

internal common tangent tiếp tuyến chung trong

external common tangent tiếp tuyến chung ngoài

circumscribed circle đường tròn ngoại tiếp

power (of a point) phương tích (của một điểm)

3.6 Cartesian coordinates system (hệ tọa độ đề các)

x-axis (horizontal axis) trục hoành

origin (of coordinates) gốc tọa độ

negative x-axis, y-axis phần âm trục hoành (trục tung)quadrant I, II, III, IV góc phần tư thứ nhất, hai, ba, bốn

3.7 Measure (các đại lượng đo)

3.8 Interrelationship in Geometry (sự tương giao trong Hình học)

go through, pass through (v) đi qua

compose (v) hợp thành

3.9 Grammar of Geometry

perpendicular from X to Y vuông góc từ X tới Y

reflection of X across Y ảnh của X qua Y

tangent line to circles tiếp tuyến với đường tròn

multiple variable function hàm nhiều biến

continue function hàm liên tục

non-trivial solution nghiệm không tầm thường

doubly periodic function hàm tuần hoàn

concave (convex) function hàm lõm (lồi)

degree (of polynomial) bậc (của đa thức)

polynomial of degree n đa thức bậc n

homogeneous polynomial đa thức thuần nhất

5 Other Terminologys (thuật ngữ khác)

5.1 Theorem terminology (Thuật ngữ

6 Words in a proof

From (by) the assumption Từ giả thiết

assume that, suppose that giả sử rằng

It is easy to see (show) dễ thấy, dễ chỉ ra

So vì thế

The first case

trường hợp 1, second case: trường hợp 2

If and only if (iff) khi và chỉ khi, nếu và chỉ nếu

Combining (1) and (2) tổng hợp (1) và (2)

Contradict (v), contradiction (n) vô lí, mâu thuẫn

follow the condition (assumption) theo điều kiện (giả thiết)

Without loss of generality không mất tính tổng quát

Equivalent (adj), equivalently (adv) tương đương

Respectively (corresponding) tương ứng

MẪU CÂU THƯỜNG DÙNG TRONG VĂN BẢN TOÁN

1 Definition.

Let … be … - Let f be a continuous function.

- Let n be a non-negative integer.

- Let p be a prime.

Đặt, địnhnghĩa …

If … then … - If  0 then the equation has two distinct roots

- If a b and b c thena c

- If m and n are relative prime then

m

n isirreducible

- We define a to be multiple of b if b divides a.

It is easy to see/show

that … (clause) - It is easy to see that m and n are not relativeprime

- It is easy to show that the inequality is hold for

all x.

- It is easy to see that the equation has two

distinct roots for all m.

From … we have… - From (1) we have the following inequality

- From the hypothesis (hypotheses) we have theequality

- From (1) and (2) we have expression…

the … Let us denote by A the set of event integers.

Denote by … a/the… Denote by A a set of event integers.

Let … denote a/the… Let A denote the set of event integers.

3 Assumption, condition

- We will make the following assumption:

- From now on we make the assumption:

- The following assumption will be neededthroughout the proof

Give the main ideas of the proof

Examine the first equation

But A B

To prove this, let f 

This is proved by writing g 

We prove this as follows:

Our proof starts with the observation that… Quan sát

Trong chứng minh qui

nạp Assume the formula hold for the degree k, we willprove it for k 1

a similar argument, …the lemma above, …uniquely, …

- As n is odd we conclude that n 1 is event

6 Proof consecutive steps

Let us evaluate/

compute/ apply the

formula to/ regard s as

fixed and …

- Let us evaluate the left side first

- Let us compute the value of the secondexpression

- Let us apply the formula (1) to the followingequation

- Let us regard s as fixed and let n tend to infinity.

Đánh giáTính toán

Áp dụngXem (coi)

7 Proof: it is sufficient to…

It is sufficient to

Show thatMake the following observationUse (1) together with the observation that

We need only to consider 3 cases: firstly… secondly … thirdly…

We need to show that…

What is left is to show that

The proof is completed by showing that

8 Proof: It is easily seen that

It is

Easily seen that…

Easy to check that…

It follows easily that…

We see at once that…

It follows immediately that…

An easy computation shows that…

(2) makes it obvious that …

9 Proof conclusion and remarks

…, which

Is impossibleProves the problemCompletes the proof

mong muốn

…, and

The proof is complete

The lemma follows(2) is proved

This contradicts our assumption

This contradicts the facts that, …

CẤU TRÚC CÂU TRONG HÌNH HỌC

I Một số quan hệ trong quan hệ Hình học

A- Cấu trúc đối tượng

Gọi các đối tượng là X, Y (đường thẳng, đường tròn, điểm, tứ giác….)

Kiểu 1: Gọi tên đối tượng, cho đối tượng

Let X be (đặt X là), given X (cho X), X is called (X được gọi là), denote X by Y, denote by (X) +(Y) (kí hiệu X là Y)

Exam: Let ABC be an isosceles and M , N, P be the midpoints of BC, CA, AB

respectively Denote G by (denote by G) the intersection of AM, BN, CP, then G is called centroid of triangle ABC

Kiểu 2: Quan hệ song song hai đối tượng X, Y

X and Y + are + adj

Exam: Given a triangle ABC Let M, N be the points lying on segments AB, AC respectively such that AM/AB=AN/AC It is easy to show that MN and BC are parallel

Kiểu 3: Quan hệ một chiều của đối tượng X với đối tượng Y

X +is +adj +to Y

Exam: Given a right triangle ABC Let D be the foot of altitude at vertex A Prove that BA

is tangent to circumcircle of triangle ACD

Kiểu 4: Điều kiện (giải thích) cho đối tượng

Clause A +so that, such that+ clause B

Exam: On diagonal AC of parallelogram ABCD points P and Q are taken so that AP =

CQ Let M be a point on AD Let us draw through point M line EF such that EF is

parallel to CD (points E and F lie on lines BC and AD)

Kiểu 5: Dựng đối tượng chưa có: Draw, take (kẻ, vẽ, lấy)

Exam: Given a triangle ABC We draw a line d passing through A and parallel to BC, and take E, F on d (A is between E and F) such that AE=AF Drawing a circle with center A and radius AE Two tangents draw from E and F respectively intersect at M

B- Quan hệ liên thuộc và tương giao trong Hình học

1- Quan hệ liên thuộc

Kiểu 1: lie on (nằm trên), is on (nằm trên), lie in (nằm trong), belong to (thuộc về), contain (chứa)

Exam: Given a triangle ABC Let (O; R) be circumcircle of triangle ABC and M be

intersection of line AO and (O) Let I be reflection of O across line BC Prove that

a, The orthocenter of triangle ABC lies on the line connecting M and midpoint of BC

b, The orthocenter of triangle is on the circle with center I and radius R

Kiểu 2: go through; pass through (đi qua)

Exam: The line goes through the centroid and the circumcentre of the triangle ABC is

called Euler line Euler line also passes through orthocenter

Kiểu 3: inscribed (nội tiếp); circumscribed, tangential (ngoại tiếp), cyclic (đồng

viên)

Exam: A quadrilateral ABCD is inscribed in a circle iff the sum of opposite angles is

180 We say that points A, B, C, D are cyclic or ABCD is inscribed quadrilateral If there is a circle which is tangent to four sides of quadrilateral ABCD then ABCD is

circumscribed quadrilateral

2- Quan hệ tương giao

Kiểu 1 (vị trí): bisect (chia đôi), separate (tách đôi, chia rẽ), same side (cùng phía), other side (khác phía), toward (về phía), lie closer (nằm ở), lie away (nằm ở xa), move away (di chuyển ra xa), move closer (di chuyển đến gần), inside (inward) (bên trong), outside (outward) (bên ngoài)

Exam 1: if a point A lies on the other side of d from B, then d separates the points A and B if A and B lie on the same side of d, then segment AB does not meet d

Exam 2: If we move F towards A along the angle bisector of angle ACD at A, the

intersection of DF with l moves away from C, but the intersection of the parallel to CF through A with l moves closer to C.

Exam 3: Equilateral triangles ABK, BCL, CDM, DAN are constructed inside thesquare ABCD and squares ABPQ, BCRS, CDTU and DAVW are constructed outsidesquare ABCD

Kiểu 2 (so sánh): collinear (thẳng hàng), perpendicular (vuông góc), concurrent (đồng quy), similar (đồng dạng), congruent (bằng nhau, trùng nhau)

Exam: Given a triangle ABC Let M, N, P be the feet perpendiculars from A, B, C to

BC, CA, AB, respectively Assume that NP meet BC at X; PM meet CA at Y and MN meet AB at Z Prove that

a, The triangles ABC and ANP are similar

b, The points X, Y, Z are collinear

Kiểu 3 (giao điểm): meet; intersect

Exam: The altitude from vertex A of triangle ABC intersects again the circumcircle of

this triangle at A’ Then, the segment AA’ meets BC at D

Kiểu 4 (tiếp xúc): tangent (tiếp xúc), touch (tiếp xúc), contact (tiếp điểm)

Exam: AB is tangent to circle (O) We say that circle (O) touches AB Two circle (O) and (O’) touch at M, then M is called the point of contact

II Một số lưu ý trong khi đọc, viết câu Hình học bằng tiếng Anh

1, Chia động từ số ít một số từ đặc biệt hay dùng:

Động từ: touch (touches), go (goes), pass (passes)…

Danh từ: Radii (Radius), focus (foci)…

2, Sử dụng mạo từ: “a”, “an” và “the” trong câu.

+ “an” đứng trước một số từ như: isosceles, in center, inradius, excenter, arbitrary, + “the” được đặt ở trước hầu hết trong mô tả các yếu tố xuất hiện duy nhất, họ các đốitượng

hoặc các danh từ được nhắc tới trước đó

Exam: (IMO 2009, G6): Let the sides AD and BC of the quadrilateral ABCD (suchthat AB

is not parallel to CD) intersect at point P Points O1 and O2 are the circumcenters andpoints H1 and H2 are the orthocenters of the triangles ABP and DCP, respectively.Denote the midpoints of segments O1H1 and O2H2 by E1 and E2, respectively Provethat the perpendicular from E1 on CD, the perpendicular from E2 on AB, and the line

H1H2 are concurrent

(IMO2 2008, G2): Given a trapezoid ABCD with parallel sides AB and CD, assumethat there exist points E on line BC outside the segment BC, and F inside the segment

AD, such that ∠DAE = ∠CBF Denote by I the intersection point of CD and EF and

by J the intersection point of AB and EF Let K be the midpoint of the segment EF.Assume that K does not lie on the lines AB and CD Prove that I belongs to thecircumcircle of △ ABK if and only if K belongs to the circumcircle of △ CDJ

3, Giới từ hay sử dụng: to, from, on, in, of, through, by, inside, outside

Exam: Draw the perpendicular from E2 on AB; we construct the perpendicular from B

to AD

Exam: A point M lies on the segment BC of triangle ABC Let d be the line passing through M and perpendicular to BC such that points A and B lie on the same side of d Suppose that H is foot of the altitude from M to d

Exam: Prove that the midpoint of the segment EF lies on the line through the twointersection points of ω1 and ω2.

Exam: The each sentence in following paragraph has one or two mistakes Let repair them!

(1) Suppose that circles with diameters AB and CD meet at points E and F in thequadrilateral

(2) Let ωE be the circle through feet of the perpendiculars from E to the lines AB, BC,and CD

(3) Let ωF be circle through the feet of the perpendiculars at F to the lines CD, DA, and

AB

(3) Prove that midpoint of the segment EF lies on the line through the two intersectionpoints of ωE and ωF

(3) Let ABC be a acute triangle with circumcircle

(4) Let D be the foot of the altitude at A, and let G be the centroid in the triangle ABC.(5) Let ω be a circle through B0 and C0 that is tangent with the circle at a point X  A.Prove that the points D, G, and X are collinear

(6) Let D and E be the second intersection points of ω and the lines AI and BI,respectively The chord DE meets AC at a point F, and BC at a point G

(7) Let P be the intersection point of the line pass F parallel to AD and the line pass Gparallel to BE

(8) Suppose that the tangents of ω at A and at B meet at a point K

(3) Let ωF be the circle through the feet of the perpendiculars from F to the lines CD,

DA, and AB

(4) Prove that the midpoint of the segment EF lies on the line through the twointersection points of ωE and ωF

(5) Let ABC be an acute triangle with circumcircle

(6) Let D be the foot of the altitude from A , and let G be the centroid of the triangleABC

(7) Let ω be a circle through B0 and C0 that is tangent to the circle at a point X  A.Prove that the points D, G, and X are collinear

(8) Let D and E be the second intersection points of ω with the lines AI and BI,respectively The chord DE meets AC at a point F, and BC at a point G

(9) Let P be the intersection point of the line through F parallel to AD and the linethrough G parallel to BE

(10) Suppose that the tangents to ω at A and at B meet at a point K

MỘT SỐ BÀI ĐỌC HIỂU

PERMUTATIONS AND COMBINATIONS

1 Two basics counting principles

Addition and Multiplication Principles are two simple examples of counting problemsrelated to what are called “Permutations” and “Combinations” Before constructing

“permutations” and “combinations”, we talk about two principles counting that isAddition and Multiplication Principles:

Definition 1 (Addition Principle:AP): Assume that there are: n1 ways for the event1

E to occur, n2ways for the event E2to occur, …n k ways for the event E k to occur,where k 1 If these ways for the different events to occur are pairwise disjoint, thenthe number of ways for at least one of the events E ,1 E2, ,E k1, or E kto occur is

Example1: One can reach city Hanoi from city Newyork by sea, air and road Suppose

that there are 20 ways by sea, 10 ways by air and 2 ways by road How many waystravel from P to Q?

Solution: Using (AP), the total number of ways from P to Q by sea, air or road is 20 +

10 + 2 = 32

Note 1: Using ternomilogy of set: Addition Principle is given below:

Let A A1, , ,2 A k be any kfinite sets, where k 1 If the given sets are pairwise disjoint,

i.e., A iA j  fori, j 1, 2, ., k , ij, then: 1 1 2 1

i i

Definition 2 (Multiplication Principle:MP): Assume that an Even E can decomposed

in to k order events E ,1 E2, ,E k(separate events) and that there are n1 ways for theevent E1to occur, n2ways for the event E2to occur, …n k ways for the event E k tooccur, where k 1 Then the total number of ways for the even E occur is given by:

Example 2: In order to reach city D from city A, We have pass through B and C city.

There are 5 ways from A to B, 6 ways from B to C and 10 ways from C to D Howmany ways travel from A to D?

Solution: If there are 5 ways to travel from A to B, 6 ways from B to C, and 10 waysfrom C to D, then by (MP), the number of ways from A to D via B and C is given by 5

x 6 x 10 = 300

2 Permutations and combinations