Author: jonah.luo

Categories
Math All Contests

The Inequalities Between Sequences

Suppose there are three sequences {an}, {bn}, and {cn}, where n=1, 2, 3 … . These sequences are defined recursively, and with inter-dependence; for example, the definition of bn depends not only on bn-1, but also on an-1 or cn-1. Suppose we are to work out some inequality relations among an, bn, and cn. In what way can we make it?

Here is one such question, as follows.

Let p, q be any two distinct positive real numbers such that pq = 5.

Define A_0 = -1 - p - q, B_0 = p+q+pq, C_0 = - pq <0.

For n \geq 1 , define

An=2Bn−1–(An−1)2,Bn=(Bn−12)–2An−1Cn−1,Cn=−Cn−12A_n = 2 B_{n-1} – (A_{n-1})^2, B_n = (B_{n-1}^2) – 2 A_{n-1} C_{n-1}, C_n = -C_{n-1}^2

Show that 1+B_n > (-C_{n-1} +1)^2 >0 for any n \geq 1 .

Let us first calculate B1. Using the recursive definition, we have that B_1 = (B_0)^2 - 2 A_0 C_0 \\ = (p+q+pq)^2 - 2 (-1-p-q) (-pq) \\ = p^2 + q^2 + p^2 q^2 >=0. Note it is obvious that B1 is non-negative. Continuing in this way, any Bn shall be non-negative.

Let us set the next immediate goal as 1+B_1 > (-C_0 +1)^2 . Bringing the value of C_0 = -pq , and noting that B_1 = p^2 + q^2 + p^2 q^2 , the goal of proof is equivalent to 1 + p^2 + q^2 + p^2 q^2 > (1-pq)^2; which shall be obvious as p^2 + q^2 > 0 > -2pq .

We list the values of {A0, B0, C0; A1, B1, C1}, in terms of p, q, as follows: (previously, we show how B1 can be calculated; the read can try to find A1, C1, similarly)

A_0 = -1-p-q, B_0 = p+q+pq, C_0 = -pq, \\ A_1 = -1 - p^2 -q^2, B_1 = p^2 + q^2+p^2 q^2, C_1 = -p^2 q^2 .

The quick-minded reader must have found that, when advancing the index by 1, the following schema holds true (where p^*, q^* are two distinct positive real numbers), A_{n-1} = -1-p^*-q^*, C_{n-1} = -p^* q^*, \\B_{n-1} = p^*+q^*+p^*q^*, \\ A_{n} = -1 - (p^*)^2 -(q^*)^2, C_{n} = -{p^*}^2 {q^*}^2, \\ B_{n} = {p^*}^2+{q^*}^2+{p^*}^2 {q^*}^2.

So, a straightforward generalization for the proof of 1+B_1 > (-C_0 +1)^2 , leads to the proof of 1+B_n > (-C_{n-1}+1)^2 .

We have two remarks in sequel.

Categories
Math All Contests

Using Factor Theorem (cont.) – selected details for a CSMC question

This posts follows a previous post. For the convenience of readers, the question is repeated:

Suppose that A1=-6, B1 = 10 and C1 = -5. For each positive integer n \geq 2 , let \\ A_n = 2 B_{n-1} – (A_{n-1})^2, \\ B_n = (B_{n-1})^2 – 2 A_{n-1} C_{n-1}, \\ C_n = -(C_{n-1})^2 .

Prove that the polynomial f_{100}(x) = x^3 + A_{100} x^2 + B_{100} x + C_{100} has three distinct positive real roots.Selected Details for the solution of c.

Now we start to work out the problem just-shown.

We find that 1+A_1 + B_1 + C_1 = 0 , then f_1 (x) = x^3 + A_1 x^2 + B_1 x + C_1 = (x-1) (x^2 + (A_1+1) x - C_1) ; (using polynomial division \frac{x^3 + A_1 x^2 + B_1 x + C_1}{x -1} to find the quadratic cofactor). By bringing A1=-6, C1= -5 into the factorized form of f_1(x) :

f_1 (x) = (x-1) (x^2 -5 x +5).

Obviously, x^2 – 5x +5 = 0 has two distinct roots [as the discriminant is (-5)2 – 4(5) = 5 >0; and note for this quadratic, none of the roots is 1]. Thus f_1(x) has three distinct roots.

In general, if 1 + A_n + B_n + C_n = 0 (this is indeed the case), then

f_n (x) = x^3 + A_n x^2 + B_n x + C_n \\ = (x-1) (x^2 + (A_n+1) x – C_n).

To show that 1 + A_n + B_n + C_n = 0 , we proceed as follows: 1 + A_n + B_n + C_n \\ = 1 + (2B_{n-1} - A^2_{n-1}) + (B^2_{n-1} - 2 A_{n-1} C_{n-1}) + (- C^2_{n-1}) \\ = (1+B_{n-1})^2 - (C_{n-1} + A_{n-1})^2 = 0 (the last equality holds if assuming 1 + A_{n-1} + B_{n-1} + C_{n-1} =0 ). Since 1 + A_1 + B_1 + C_1 =0 , from math induction, we obtain for any n, 1 + A_n + B_n + C_n = 0 .

It is clear now that fn(x) always has 1 as one of its roots. Let us denote D_n as the discriminant of the quadratic (x^2 + (A_n+1) x – C_n) . Then we should have (if fn(x) has three distinct roots)

  • (claim 1) D_n >0 ;
  • (claim 2) At x = 1, x^2 + (A_n+1) x - C_n = 2 + A_n - C_n \neq 0 .

Yet the two claims are not confirmed up to here. So more effort is needed! The following table (that lists feature related to function fn(x)) is made:

AnBnCnDnD_n2+An−Cn2+A_n-C_n
n=1-610-55>01
n=2-1640-25125>011
n=3-176800-6251752+4(-625)>0451

By inspection, we find an interesting pattern that A_n <0, B_n >0, C_n <0 , or equivalently -A_n >0, B_n >0, -C_n >0 . From this we can deduce claim 2: 2 + A_n - C_n >2 >0 . We observe another interesting pattern: B_n > - C_n - 2 C_{n-1} >0 which implies B_n + C_n > -2 C_{n-1}>0, \\ -1 - A_n > -2 C_{n-1} >0. Note that (– Cn )= (Cn-1)2, and squaring on both sides of the previous line, to obtain: D_n = (1+A_n)^2 -4 (-C_n) >0 which is exactly Claim 1. (There is one last thing: it remains to be shown formally, the two patterns mentioned above .. ..)

Categories
Bulletin

Training Sessions for 2026 Fall at AA (Access Academy)

From September 26th to November 14th, on most Saturdays (except a couple of Saturdays off), Access Academy welcome those students who desire to challenge math problems to the following sessions, that prepare them to participate in the Math Contests (problems given by University of Waterloo), and to write in the COMC (Canadian Open Math Challenge).

Each training session for COMC is 2.5 hours long. And each training session for CSMC (Canadian Senior Math Contests) or CIMC (Canadian Intermediate Math Contest) is 2 hours long.

While these training sessions are going, relevant posts will be added to the column, and we allow students to download some materials for training (these materials are password-protected). The posts are public accessible. More posts or training materials will be added when they are available.

See the relevant posts:

https://prime.mathatjonahs.com/2026/06/26/proof-diamater-is-the-shortest-curve-that-bisects-circular-area/

See the relevant training material:

Categories
Math All Contests

Using Factor Theorem – work out a question in CSMC

The following question comes from 2024 CSMC (Canadian Senior Math Contest) – Part B, Question 2. (There are three sub-questions, a, b, and c. We present only a. and c.)

a. The quadratic equation x2-2x-1= 0 has solutions x= r and x = s. Determine integers b and c for which quadratic equation x2 + bx + c = 0 has solutions x = 2r + s and x = r + 2s.

c. Suppose that A1=-6, B1 = 10 and C1 = -5. For each positive integer n \geq 2 , let \\ A_n = 2 B_{n-1} – (A_{n-1})^2, \\ B_n = (B_{n-1})^2 – 2 A_{n-1} C_{n-1}, \\ C_n = -(C_{n-1})^2 .

Prove that the polynomial f_{100}(x) = x^3 + A_{100} x^2 + B_{100} x + C_{100} has three distinct positive real roots.

Now we look into how to solve the question above.

Analysis for a.

It shall hold x2 + px + q = (x-r) (x-s) if r, s are two roots of the equation. By expanding the right hand side, we obtain x2 – (r+s)x + rs. Now comparing it to x2 + px + q to find [this is indeed Viete’s Theorem]

r+s = -p, (rs) = q ~---~ -- (*)

Using (*) for the question in 2a), it holds that r+s = -p =2, and (rs) =q = -1; meanwhile we have -b = (2r+s) + (r+2s) = 3(r+s) etc. from which the value of b can be calculated.

Analysis for c.

We intend to show that for any n, equation f_n (x) = x^3 +A_n x^2 +B_n x + C_n has three distinct roots. (This is actually stronger than what the question is asked; yet if this is indeed true, then a clear path can be paved leading all the way to f100 (x). )

If one of the three roots (call it x1) is known (preferably an integer), then we have a linear factor (x – x_1) (by Factor Theorem), and the co-factor (quadratic) can be worked out with polynomial division. For the said quadratic co-factor, we argue it has real distinct roots by checking that the discriminant is positive, and be ensured neither root of the quadratic factor equals x_1 .

Note we find x=1 is a root for fn(x) (so let x1 =1.) And f1(x), f2(x), … each has three distinct roots. [Details to be worked out]

Next step: The readers are trusted to work out the solution to a. (analysis for a. was already given). For question c., in addition to the analysis given above, there are some gaps need to be filled. Please attempt to work them out by yourself, yet if you get stuck or want to check whether you have done it right, check out the next post. Using Factor Theorem (cont.) – selected details for a CSMC question

Categories
Bulletin

Bulletin on Math Contests – Accepting Registration

  • For year 2026, the math contest CSMC/CIMC (Canadian Senior and Intermediate Math Contest) will be held on Wednesday, November 18th.   CSMC is recommended to students in Grade 11 and 12 to participate, while CIMC is for students in Grade 9 and 10. 
  • If you are new to the math contests in Canada, please read this post. 
  • For year 2026, COMC (Canadian Open Math Challenge) will be held on October 29th. Recommended for students in grade 7-12 who love the challenge of math.  — Registration will be opening soon. 
Categories
Bulletin

公告板 – 数学竞赛报名消息

  • 2026 年COMC 加拿大公开数学挑战赛征集报名（九月开始）；CSMC/CIMC 加拿大高级与中级数学竞赛也征求报名。有意参加的学生可以联系 Access Academy。请注意相关的信息更新。
  • 如果您对于加拿大数学竞赛不熟悉，请参考这个帖子。

  • 2026 年的 CSMC/CIMC 将在11月18日（周三）举行。参加CSMC（高级数学竞赛）的应为在11或12 年级就读的学生。参加CIMC 的应是 9或10 年级就读的学生。
  • 今年的COMC 加拿大公开数学挑战赛将在十月的最后一个周四（10月29日）举办。建议热爱数学挑战的学生参加。参赛学生应在 7 – 12 年级就读。COMC 是加拿大数学竞赛中接受公开报名的最高级别的挑战，也是唯一一个参赛者可能加入加拿大数学奥林匹克（CMO）及国际数学奥林匹克（IMO）的比赛（经选拔参加后续竞赛活动）。

  • 今夏在上海举办的国际数学奥林匹克（IMO）上，中国，美国队获冠亚军。加拿大队取得第 13 名。
Categories
Numbers

Prime numbers

Prime numbers are those that have 1 (one) and itself as the only two divisors. Examples of primes are 2, 3, 5, 7, 11. None of 4, 6, 9 is a prime since 4 = 2 × 2, 6 = 2 × 3, and 9 = 3 × 3.

If a number greater than one is not a prime, then it is a composite number, and can be factored into the product of primes — called prime factorization. We have given the prime factorization of 4, 6, 9 as above. For a couple of more examples:

12 = 2 × 2 × 3

36 = 2 × 3 × 3 × 3

28 = 2 × 2 × 7

So all natural numbers are divided into three classes: the number 1, the prime numbers, and the composite numbers.

A bonus point: π, besides representing in a circle, the ratio of circumference to diameter, also stands for a special function related to prime numbers. This is described as follows.

Function π(x) — for integer x, represents the number of primes less than or equal (i.e. not exceeding) x.

 

For example,

π(2) = 1, π(3) = 2, π(10) = 4, π(20) = 8 etc.
[To find why π(10) = 4, recall the 4 prime numbers not exceeding 10: they are 2,3,5, and 7.]

Categories
Algebra

Irrational Number – Proving the square root of 2 is an irrational number

Yes. We are to claim the irrationality of [pmath size=14] sqrt 2 [/pmath].

We will use proof by contradiction.

Assume that [pmath size=14] sqrt 2 [/pmath] is a rational number. Then by definition of rational numbers, we may write that

[pmath size=12] sqrt 2 = n / m [/pmath]

where n, m are integers.

Squaring on both sides: [pmath size=12] (sqrt 2)^2 = (n / m)^2[/pmath], or

[pmath size=12] 2 = n^2 / m^2 [/pmath]

which leads to [pmath size=12] 2 m^2 = n^2 [/pmath] (*)

Let us show that (*) cannot hold true.

Method 1.

Let [pmath size=12] m = 2^p K, n = 2^q L [/pmath], where p, q are integers, K and L are odd integers.

{If either K or L are even numbers, then factor 2 can be extracted iteratively until an odd number (co-factor) is revealed. }

Bringing m, n in the above forms into [pmath size=12] 2 m^2 = n^2 [/pmath], then:

[pmath size=12] 2 ((2^p) K)^2 = (2^q L)^2 [/pmath]

[pmath size=12] 2^{2p+1} K^2 = 2 ^{2q} L^2 [/pmath] (++)

Let us count the number of 2’s on each side. The l.h.s has (2p+1) of factor 2’s (an odd count of 2), and the r.h.s. has (2q) of 2’s (an even count of 2’s). However, it is well established that for any integer, the prime factorization is unique. Therefore, equation (++) cannot holds. Tracing back, we have to revoke the initial assumption

[pmath size=12] sqrt 2 = n / m [/pmath]

Therefore, number [pmath size=12] sqrt 2 [/pmath] is not a rational number.

Method 2.

Note in [pmath size=12] n^2 = 2 m^2[/pmath], the right hand side is even, so the left side is also even. This implies that n must contain factor 2. Let n = 2 N, then

[pmath size=12] 2 m^2 = (2 N)^2[/pmath] — > [pmath size=12] m^2 = 2 N^2 [/pmath]

where the new equation [pmath size=12] m^2 = 2 N^2 [/pmath] implies that number m must contain factor 2. So each of n, m and N contains factor 2.

This process may be applied in an iterative manner, forever. Consequently, both m and n contains an infinite many factors of 2’s. However, this is not possible as either n or m is a finite number.

You got it?

If you got it, you shall be able to prove that [pmath size=12] sqrt 3 [/pmath] is also irrational (i.e. [pmath size=12] sqrt 3 [/pmath] cannot be written as the ratio of two integers.) How? Just use the same idea, but this time, you need to count the number of 3’s.

Categories
Math Contests - Problems and Discussion

加拿大 数学竞赛 介绍

您听到过加拿大数学竞赛吗？对那些热爱数学，有较强逻辑思考能力，愿意挑战自己的中小学生，这真地是一个很好的舞台来展现他们的智力和才华。

Click here to read the English version of this post.

如果一个学生想要参加数学竞赛，而他/她 又在加拿大的全日制中小学校学习（公立，教会学校，或私立，都可以）那么要做的第一步是问询他所在的学校有否参赛机会。您也可以问询我们 –Jonah’s Math Corner：我们是登记过的考点之一，可以和竞赛组织者合作监考数学竞赛–具体竞赛项目和时间安排，欢迎您来咨询。

报名参赛时，请注意年级资格： 一般规则是低年级可以参加高年级的竞赛（只要有足够自信，能说服学校或者报名点），但是高年级不可以参加低年级的竞赛，规则不允许。比方说九年级竞赛，十年级以上学生是不被允许参加的。不用解释，这是为着公平。

想参加竞赛的学生，或者他们的家长，需要了解一下竞赛的种类，出题形式和难度状况。Math Kangaroo 是蛮有人气的竞赛，参加者遍及1 – 12 年级学生。除此而外，多数竞赛是面向中学生的。本篇就是为这一目的给大家做一个概括的介绍：有哪些竞赛，参赛年级和形式；相比而言难度如何？

1）Math Kangaroo，起源于欧洲，现在在全世界越来越有人气。比赛分成6 个年龄组，每两年级是一组。

（我们目前暂未组织这项竞赛）

2）初中数学竞赛（参赛 限于 9 年级以下学生）其中以为7-8 年级而设的高中数学竞赛最为著名。有不少学校组织全体学生参加。如果你住在 Alberta 省，卡加利大学和阿尔伯塔大学每年都会各自主办由他们命题的初中生竞赛。

3）中级和高中数学竞赛 从九年级开始，数学竞赛不仅区分年级，而且有不同难度等级。中级是指九十年级，处于初中到高中过渡的阶段；再往上是高中11-12 年级竞赛。（顺便说一句，12 年级参赛的学生较少，因为不少学生在准备 大学/学院的入学申请，或者联系工作；竞赛成绩突出的学生才会考虑参加，一般参加的都是高水平的竞赛 – 即选拔性竞赛）。难度等级在下面结合各个竞赛点明。

下面介绍由滑铁卢大学 (University of Waterloo) 主办的三个系列的竞赛。

系列一 – PCF 竞赛：Pascal – 九年级，Cayley – 十年级，Fermat – 十一年级。多项选择题，难度中等。

系列二 – FGH 竞赛：Fryer – 九年级，Galois – 十年级，Hypatia – 十一年级。没有多项选择题；部分问题要求给出答数，另一些问题要求写出全部计算和推理过程。答题时间为 75 分钟。

系列三 – 加拿大中级数学竞赛（CIMC，九十年级）和高中数学竞赛（CSMC，十一年级以上）。竞赛形式类似 FGH 竞赛，综合性更强，难度略高。

4）由加拿大数学会主办的选拔性竞赛。数学会有一系列选拔性竞赛，其中第一项 加拿大公开数学挑战赛 （Canadian Open Math Challenge）公开接受报名。接下来的加拿大奥数赛（CMO）和国际数学奥林匹克 （IMO）只是邀请在历次其他竞赛中表现突出的学生参加，不公开接受报名。最终获邀参加 IMO 的将组成加拿大代表队，代表加拿大国家参赛。

5）其他未列的竞赛。除上面提到的竞赛项目外，我们还组织参加过加拿大或者美国的区域性竞赛。有的竞赛可以上网参加：需要老师组织监考，按时完成。

Jonah’s math corner 是正式登记的考点，可以组织参加竞赛。我们也有面向竞赛的辅导项目（当然也有补习提高的辅导）。注意，竞赛辅导和考试登记是分开的。在参加了竞赛辅导后，学生可以自主决定是否参加，和在哪里参加竞赛。选择在本考点参赛的，是否参加竞赛辅导皆可。学生或家长有这方面的需求，欢迎联系，欢迎咨询我们。

想进一步了解的可以参考以下网站：（英文网站）

Centre of Education for Mathematics and Computing Science, Math Contests

— 数学和计算科学中心，数学竞赛

— 上述竞赛的说明，准备和相关材料

Canadian Math Society, Canadian Open Math Challenge

加拿大数学会，数学公开挑战赛

Math Kangaroo Company

Math Kangaroo 数学竞赛

我们的中英文网站 Jonah’s Math Corner在建设中。

www.mathatjonahs.com

www.mathatjonahs.com/eduforum/

Categories
Math All Contests Math Contests - Problems and Discussion

Introduction to Canadian Math Contest

(Revised in August 20, 2026)

Have you heard about the Canadian Math Contests? It is certainly for those talented in math and logic to show their ability, and a good opportunities for those who love the challenges.

For a student to participate in a contest, talk to the school one attends, or talk to us (Jonah’s Math Corner). Be sure to have a current enrollment in a secondary school (junior, high school) — or sometimes elementary school, located in Canada.

Check your grade too — if it says grade 9, then all students in grade 10 and higher cannot write that contest. A students in lower grade are allowed to participate in contests for higher grade (only if he or she feels confident), but not the other way round.

If you are looking at very selective competitions, then COMC (Canadian Open math challenge) is the contest to participate (typically held in October). The very best performed students can earn an entry to represent Canada for the IMO (International Math Olympiad) through further selective process.

Students in grade 9/10/11 may look for CIMC (Canadian Intermediate Contest) and CSMC (Canadian Senior Math Contest) — here Senior means “Senior High School” — Both are held in November on the same time, but no one is allowed to participate in both (just choose one of them).

Keep reading for information about different Contest series, detailed as follows: 

1) Junior math contests (have to be under grade 9 to participate): the most popular is the Gauss contest for grades 7 & 8, organized by U. of Waterloo. Gauss is very popular that some school teachers enrolled all of their students to participate.

There are some grade 9 contest – see the mid-level and high-school level contests, as the challenge is already close to high levels.

If you live in Alberta, then both U. of Calgary and U. of Alberta have set up challenges for Junior students.

2) Mid-level and High-school level Contests — the main stage for the middle school math challenge. University of Waterloo has organized 3 series respectively for grade 9, 10 and 11. The contests questions appears typically in multiple-choice format.

Series I: PCF contest: Pascal for grade 9, Cayley for grade 10 and Fermat – for grade 11.

Questions in PCF appear in multiple-choice format.

Series II: FGH contest: Fryer – for grade 9, Galois – for grade 10 and Hypatia – for grade 11.

No multiple-choice question in FGH series. All answers have to be worked out by participants. In any of these series, there will be two types of questions, one will ask participants for answers only, and the other type will ask students to write full solutions. Each contests is 75 minutes long.

Series III: CIMC – Canadian Intermediate Math Contests (for grade 9/10) and CSMC – Canadian Senior Math Contests (for grade 11/12). The format is similar to FGH, however, the level of challenge is in general higher, and students are provided with 2 hours time to answer.

4) Selective Contests for those best talents in math: Canadian Math Society have a challenge series for those who want to participate in Math Olympiad. The first step is the COMC (Canadian Open Math Challenge). It is selective, so only COMC is open to students; all following contests in the series, like CMO (Canadian Math Olympiad) and IMO (International Math Olymiad), are by-invitation ONLY.

5) Popular Math Contests: Math Kangaroo, started in Europe and becoming popular across world, is organized by Math Kangaroo company. Their are 6 age groups and each group bracket two grades, like grades 1-2, grades 3-4, .. .. According to the organizer’s claim, the purpose of the contest is to challenge students in a playful setting; instead of academic, it aims at developing mental powers and flexibility.

5) Other Contests: besides those mentioned, we occasionally will register students in Regional Math League of Canada or US – some contests are online only – save students the cost and time to travel.

Jonah’s math corner can serve as a site to supervise students to write the math contests. If you have any question, just talk to us.