Tag: The Season of Math Contest

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

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.

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