USAMO 2014 Day 1

by liberator, Aug 18, 2014, 9:07 PM

Even though I am not an American citizen, I decided to try several USAMO questions for practice. Here is my work for day 1.

Problem 1: Let $a$ , $b$ , $c$ , $d$ be real numbers such that $b-d \ge 5$ and all roots $x_1, x_2, x_3,x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.

[b]My solution[/b]

\[ \begin{align*}\prod_{\text{cyc}} (x_1^2+1) = \prod_{\text{cyc}} (x_1-i)(x_1+i) &= \left[\prod_{\text{cyc}} (i-x)\right] \left[\prod_{\text{cyc}} (-i-x)\right] \\ &= P(i)P(-i) \\  &= (b-d-1)^2+(c-a)^2 \\ &\ge (5-1)^2+0^2 \\ &= 16.\end{align*} \]

Equality occurs iff $b-d = 5$ and $c=a$ ; e.g. $P(x) = x^4 + 4x^3 + 6x^2 + 4x + 1 = (x+1)^4$ .

Problem 2: Let $\mathbb{Z}$ be the set of integers. Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that $xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))$ for all $x, y \in \mathbb{Z}$ with $x \neq 0$ .

[b]My solution[/b]

Problem 3: Prove that there exists an infinite set of points $\dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots$ in the plane with the following property: For any three distinct integers $a,b,$ and $c$ , points $P_a$ , $P_b$ , and $P_c$ are collinear if and only if $a+b+c=2014$ .

[b]My solution[/b]

We show that the set of points $P_n = \left(t,t^3 \right)$ , where $t = n - \tfrac{2014}{3}$ satisfies the conditions.

Shift the indices by $- \tfrac{2014}{3}$ , so that the collinearity condition for $P_a, P_b, P_c$ is $a+b+c=0$ . $P_a, P_b, P_c$ are collinear iff $\left[P_aP_bP_c \right] = 0$ , so by the shoelace determinant

\[ \begin{align*}\begin{vmatrix} a^3 & b^3 & c^3 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = -\sum_{\text{cyc}} a^3(b-c) &= -\sum_{\text{cyc}} \left[a^3(b-c) + a^2(b^2 - c^2) \\ &= -\sum_{\text{cyc}} a^2(b-c)(a+b+c) \\ &= (a-b)(b-c)(c-a)(a+b+c). \end{align*} \]

Since $a,b,c$ are pairwisely distinct, $\left[P_aP_bP_c \right] = 0 \iff a+b+c = 0$ , and the result follows.

This post has been edited 2 times. Last edited by liberator, Aug 21, 2014, 10:04 PM

algebra

complex numbers

Determinants

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That solution to 3 ...

by jlammy, Aug 18, 2014, 10:07 PM

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for #3 what was your motivation for establishing $(n-\frac{2014}{3},(n-\frac{2014}{3})^3)$

by bcp123, Aug 19, 2014, 8:22 PM

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@bcp123

My motivation was to change the collinearity condition to $a+b+c=0$ , so that I could attempt to show that $[P_aP_bP_c] = 0 \iff a+b+c=0$ , which I succeeded in doing. Furthermore, since there was nothing important about $a,b,c \in \mathbb{N}$ , it was perhaps implicit that the solution would be $P_n = (f(n), g(n))$ , where $f,g$ are functions of $n$ . Clearly, since the condition involves three variables, the function would have to have degree of at least 3. To have the value of the shoelace equivalent to zero iff $a+b+c=0$ , I tried to make it equal to an expression that contained a factor of $a+b+c$ , and also with the other factors guaranteed to be non-zero; e.g. $(a-b)$ . Hence the simplest method I could think about was to see whether I could reverse-construct a shoelace determinant from the expression $(a-b)(b-c)(c-a)(a+b+c)$ , (note that this expression has degree 3), which I accomplished.

This post has been edited 1 time. Last edited by liberator, Aug 21, 2014, 1:37 PM

by liberator, Aug 19, 2014, 8:43 PM

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I was wondering if you could explain the thought process behind having it as (t,t^3), as opposed to (t,t^2)? Being that it is third degree, shouldn't the sum of the degrees equal to 3?

Also, what was the logic behind constructing the shoelace determinant? I still don't really understand how you came up with that.

by Taussig, Dec 25, 2014, 1:00 AM

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@Taussig

Taussig wrote:

Being that it is third degree, shouldn't the sum of the degrees equal to 3?

The degree of a polynomial is defined as the greatest degree of any of its individual terms: see http://en.wikipedia.org/wiki/Degree_of_a_polynomial.

v_Enhance wrote:

Taussig wrote:

What was the motivation behind the algebraic construction? This seems like a very geometric problem but I would've never came up with an answer like yours.

pythag011's explanation makes this more evident: the only "geometry" in the problem is collinearity conditions. That's not especially geometric (compare this with something like #5). Also, the sum condition is also very non-geometric. So it's some combination of (a) wishful thinking (hoping that you can get something that is as nice as the above) and (b) realizing that the geometric phrasing is rather artificial.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3691140#p3691140

by liberator, Dec 27, 2014, 8:02 PM

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thx bcp123, I guess its because I only have 10 posts, not much to brag on about
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USAMO 2014 Day 1

by liberator, Aug 18, 2014, 9:07 PM

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