Okay, let's try this.

If you came here looking for new problems, then the answer is probably no. (Except if you've never solved these before, then don't read the notes after them.) But this note is an experimental test of this format (and me rambling about these problems without actually solving them).

(This was inspired by Varsha Dani. Thank you!)

Problem 2:

What is the minimum value of qq such that there are always three elements xmx_m, xnx_n, xpx_p of the set A={x1,x2,,xq1,xq}{1;2;;1233;1234}\mathbb A = \{x_1, x_2, \cdots, x_{q - 1}, x_q\} \in \{1; 2; \cdots; 1233; 1234\} such that xn22xpxm<xp2+xm2<xn2+2xpxmx_n^2 - 2x_px_m < x_p^2 + x_m^2 < x_n^2 + 2x_px_m (with every possible combination of the set A\mathbb A)?

We have that

xn22xpxm<xp2+xm2<xn2+2xpxm    xpxm<xn<xp+xm\large x_n^2 - 2x_px_m < x_p^2 + x_m^2 < x_n^2 + 2x_px_m \implies |x_p - x_m| < x_n < x_p + x_m

That means xmx_m, xnx_n, xpx_p are three sides of a triangle. A similar property of a triangle is that if aa is the longest side of the triangle side aa, bb and cc then a<b+ca < b + c.

We can find out the minimum value of qq using induction with the Fibonacci sequence. Why the Fibonacci sequence, you ask? Because first of all, the Fibonacci numbers are the terms of a sequence of integers in which each term is the sum of the two previous terms with or Fn=Fn1+Fn2F_n = F_{n - 1} + F_{n - 2} (nNn \in \mathbb N and n2n \ge 2). And secondly, the few Fibonacci numbers match the first few numbers from 1 to 1234.

More specifically, we'll the set {1;2;3;5;8;;987}\{1; 2; 3; 5; 8; \cdots; 987\}, which has 15 elements.

Now we have to prove that set A\mathbb A with a minimum of 16 elements must have three numbers aa, bb and cc such that a>b>ca > b > c and a<b+ca < b + c. Those 16 elements are x1<x2<x15<x16x_1 < x_2 < \cdots x_{15} < x_{16}. It is true that

We have that

{x4x3+x22x2+x1x5x4+x33x2+2x1x15x14+x13377x2+233x1x16x15+x14610x2+377x1\large \left\{ \begin{aligned} x_4 &\ge x_3 + x_2 \ge 2x_2 + x_1\\ x_5 &\ge x_4 + x_3 \ge 3x_2 + 2x_1\\ \cdots\\ x_{15} &\ge x_{14} + x_{13} \ge 377x_2 + 233x_1\\ x_{16} &\ge x_{15} + x_{14} \ge 610x_2 + 377x_1\\ \end{aligned} \right.

Plugging in x11x_1 \ge 1 and x22x_2 \ge 2, x16610x2+377x11597Ax_{16} \ge 610x_2 + 377x_1 \ge 1597 \notin A, which is incorrect.

So the answer is 16. Check Mark Henning's more detailed explanation. It's better.

Problem 3:

Option 8 Option 4 Option 3 Option 7 Option 2 Option 6 Option 1 Option 5

There was a statement on a page on the book.

"x1x_1 and x2x_2 are the roots of the equation 1x2mx+2=0\square_1 x^2 - mx + \square_2 = 0, where 1{m,n}\square_1 \in \{m, n\} and 2{m,n}\square_2 \in \{m, n\} (with mn>0mn > 0 and x1>x2x_1 > x_2). It is true that x1x2+1x2x1+2mn=0\sqrt{\dfrac{x_1}{x_2}} + \triangle_1\sqrt{\dfrac{x_2}{x_1}} + \triangle_2\sqrt{\dfrac{m}{n}} = 0, where 1{1,1}\triangle_1 \in \{-1, 1\} and 2{1,1}\triangle_2 \in \{-1, 1\}."

Which of the following options is correct?

Option 1: 1=m,2=n,1=1,2=1\square_1 = m, \square_2 = n, \triangle_1 = 1, \triangle_2 = -1

Option 2: 1=m,2=n,1=1,2=1\square_1 = m, \square_2 = n, \triangle_1 = -1, \triangle_2 = 1

Option 3: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = 1, \triangle_2 = -1

Option 4: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = 1, \triangle_2 = 1

Option 5: 1=n,2=m,1=1,2=1\square_1 = n, \square_2 = m, \triangle_1 = -1, \triangle_2 = 1

Option 6: 1=n,2=m,1=1,2=1\square_1 = n, \square_2 = m, \triangle_1 = 1, \triangle_2 = -1

Option 7: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = -1, \triangle_2 = 1

Option 8: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = -1, \triangle_2 = -1

With elimination, you can work out can Option 6 is the answer. I can't really go into detail of why Option 6 is the answer. But what I can do is prove the statement is correct.

We have that x1+x2=x1x2=ba=aba2>0x_1 + x_2 = x_1x_2 = \dfrac{b}{a} = \dfrac{ab}{a^2} > 0, which means x1x_1 and x2x_2 are positives, or x1x2\sqrt{\dfrac{x_1}{x_2}}, x2x1\sqrt{\dfrac{x_2}{x_1}} and ba\sqrt{\dfrac{b}{a}} are defined.


x1x2+x2x1=x1+x2x1x2=baba=ba    x1x2+x2x1ba=0\large \sqrt{\dfrac{x_1}{x_2}} + \sqrt{\dfrac{x_2}{x_1}} = \dfrac{x_1 + x_2}{\sqrt{x_1x_2}} = \dfrac{\dfrac{b}{a}}{\sqrt{\dfrac{b}{a}}} = \sqrt{\dfrac{b}{a}} \implies \sqrt{\dfrac{x_1}{x_2}} + \sqrt{\dfrac{x_2}{x_1}} - \sqrt{\dfrac{b}{a}} = 0

And that's all I have to say! Thanks again to Varsha Dani. Here's her discussion: "AVL Trees meet Fibonacci Numbers".

Note by Thành Đạt Lê
2 years, 5 months ago

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