Category: Math All Contests

Everything about the math contests.

The Inequalities Between Sequences

Post author By jonah.luo
Post date September 30, 2026
Sticky post

Suppose there are three sequences {a_n}, {b_n}, and {c_n}, where n=1, 2, 3 … . These sequences are defined recursively, and with inter-dependence; for example, the definition of b_n depends not only on b_n-1, but also on a_n-1 or c_n-1. Suppose we are to work out some inequality relations among a_n, b_n, and c_n. 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

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

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 B_n 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.

Tags The Season of Math Contest

Math All Contests

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

Post author By jonah.luo
Post date September 23, 2026

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 f_n(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 f_n(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:

	A_n	B_n	C_n	$D_n$	$2+A_n-C_n$
n=1	-6	10	-5	5>0	1
n=2	-16	40	-25	125>0	11
n=3	-176	800	-625	175²+4(-625)>0	451

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 (– C_n )= (C_n-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 .. ..)

Math All Contests

Using Factor Theorem – work out a question in CSMC

Post author By jonah.luo
Post date September 19, 2026
Sticky post

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 f_n(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

Tags The Season of Math Contest

Math All Contests Math Contests - Problems and Discussion

Introduction to Canadian Math Contest

Post author By jonah.luo
Post date August 31, 2018
Sticky post

(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.

Tags The Season of Math Contest