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Difference between revisions of "2007 iTest Problems/Problem 46"

(Created page with "== Problem == Let <math>(x,y,z)</math> be an ordered triplet of real numbers that satisfies the following system of equations: <cmath>\begin{align*}x+y^2+z^4&=0,\\y+z^2+x^4&=0,...")
 
(Solution to Problem 46 (credit to RagvaloD) — insane system)
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== Problem ==
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==Problem==
  
 
Let <math>(x,y,z)</math> be an ordered triplet of real numbers that satisfies the following system of equations:  
 
Let <math>(x,y,z)</math> be an ordered triplet of real numbers that satisfies the following system of equations:  
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If <math>m</math> is the minimum possible value of <math>\lfloor x^3+y^3+z^3\rfloor</math>, find the modulo <math>2007</math> residue of <math>m</math>.
 
If <math>m</math> is the minimum possible value of <math>\lfloor x^3+y^3+z^3\rfloor</math>, find the modulo <math>2007</math> residue of <math>m</math>.
  
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== Solution ==
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==Solution==
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Rearrange the terms to get
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<cmath>y^2 + z^4 = -x</cmath>
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<cmath>z^2 + x^4 = -y</cmath>
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<cmath>x^2 + y^4 = -z</cmath>
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Since the left hand side of all three equations is greater than or equal to 0, <math>x,y,z \le 0</math>.  Also, note that the equations have symmetry, so [[WLOG]], let <math>0 \ge x \ge y \ge z</math>.  By substitution, we have
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<cmath>y^2 + z^4 \le z^2 + x^4 \le x^2 + y^4</cmath>
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Note that <math>0 \le x^2 \le y^2 \le z^2</math> and <math>0 \le x^4 \le y^4 \le z^4</math>. That means <math>x^2 + y^4 \le x^2 + z^4</math>.  Since <math>y^2 + z^4 \le x^2 + y^4</math>,
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<cmath>y^2 + z^4 \le x^2 + z^4</cmath>
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<cmath>y^2 \le x^2</cmath>
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Since <math>x^2 \le y^2</math>, then <math>x^2 = y^2</math>.  Because <math>x</math> and <math>y</math> are nonpositive, <math>x = y</math>.
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<br>
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Using substitution in the original system,
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<cmath>x^2 + z^4 = z^2 + x^4</cmath>
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<cmath>(z^2 + x^2)(z^2 - x^2) - (z^2 - x^2) = 0</cmath>
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<cmath>(z^2 - x^2)(x^2 + z^2 - 1) = 0</cmath>
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To find the real solutions, we use [[casework]] and the Zero Product Property.
 +
 
 +
<br>
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'''Case 1: <math>z^2 = x^2</math>'''
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If <math>z^2 = x^2</math>, then since <math>z</math> and <math>x</math> are nonpositive, then <math>z = x</math>.  Substitution results in
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<cmath>x+x^2+x^4 = 0</cmath>
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<cmath>x(1+x+x^3) = 0</cmath>
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That means <math>x = 0</math> or <math>x^3 + x + 1 = 0</math>.  For the first equation, <math>m = 0</math>.  For the second equation, note that <math>x^3 = -x-1</math>, and since <math>x = y = z</math>, <math>m = \lfloor -3x-3 \rfloor</math>, where <math>x</math> is a real number.  Since <math>-\tfrac{1}{3}^3 - \tfrac13 + 1 = \tfrac{16}{27}</math> and <math>-\tfrac{2}{3}^3 - \tfrac23 + 1 = \tfrac{1}{27}</math>, the root of <math>x</math> is less than <math>-\tfrac23</math> but more than <math>-1</math>, so
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<cmath>0 > -3x-3 > -1</cmath>
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<cmath>m = \lfloor -3x-3 \rfloor = -1</cmath>
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 +
<br>
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'''Case 2: <math>x^2 + z^2 = 1</math>'''
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Because <math>x^2 \le z^2</math>, <math>x^2 \le \frac12</math>.  From one of the original equations,
 +
<cmath>z^2 + x^4 = -y</cmath>
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<cmath>1 - x^2 + x^4 + x = 0</cmath>
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Using the [[Rational Root Theorem]],
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<cmath>(x+1)(x^3 - x^2 + 1) = 0</cmath>
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Note that if <math>x = -1</math>, then <math>x^2 \ge \tfrac12</math>, so that won’t work.  Let <math>x = -a</math> (where <math>a \ge 0</math> since <math>x \le 0</math>), so
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<cmath>a^3 + a^2 = 1</cmath>
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If <math>a \le \tfrac{\sqrt2}{2}</math>, then
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<cmath>a^3 + a^2 \le \frac{\sqrt2}{4} + \frac12</cmath>
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<cmath>a^3 + a^2 \le \frac{\sqrt2 + 2}{4} < 1</cmath>
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Thus, there are no solutions in this case.
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<br>
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From the two cases, the smallest possible value of <math>m</math> is <math>-1</math>, so the [[modulo]] <math>2007</math> residue of <math>m</math> is <math>\boxed{2006}</math>.
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==See Also==
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{{ITest box|year=2007|num-b=45|num-a=47}}
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[[Category:Olympiad Algebra Problems]]

Revision as of 16:22, 22 July 2018

Problem

Let $(x,y,z)$ be an ordered triplet of real numbers that satisfies the following system of equations: \begin{align*}x+y^2+z^4&=0,\\y+z^2+x^4&=0,\\z+x^2+y^4&=0.\end{align*} If $m$ is the minimum possible value of $\lfloor x^3+y^3+z^3\rfloor$, find the modulo $2007$ residue of $m$.

Solution

Rearrange the terms to get \[y^2 + z^4 = -x\] \[z^2 + x^4 = -y\] \[x^2 + y^4 = -z\] Since the left hand side of all three equations is greater than or equal to 0, $x,y,z \le 0$. Also, note that the equations have symmetry, so WLOG, let $0 \ge x \ge y \ge z$. By substitution, we have

\[y^2 + z^4 \le z^2 + x^4 \le x^2 + y^4\]

Note that $0 \le x^2 \le y^2 \le z^2$ and $0 \le x^4 \le y^4 \le z^4$. That means $x^2 + y^4 \le x^2 + z^4$. Since $y^2 + z^4 \le x^2 + y^4$, \[y^2 + z^4 \le x^2 + z^4\] \[y^2 \le x^2\] Since $x^2 \le y^2$, then $x^2 = y^2$. Because $x$ and $y$ are nonpositive, $x = y$.


Using substitution in the original system, \[x^2 + z^4 = z^2 + x^4\] \[(z^2 + x^2)(z^2 - x^2) - (z^2 - x^2) = 0\] \[(z^2 - x^2)(x^2 + z^2 - 1) = 0\] To find the real solutions, we use casework and the Zero Product Property.


Case 1: $z^2 = x^2$ If $z^2 = x^2$, then since $z$ and $x$ are nonpositive, then $z = x$. Substitution results in \[x+x^2+x^4 = 0\] \[x(1+x+x^3) = 0\] That means $x = 0$ or $x^3 + x + 1 = 0$. For the first equation, $m = 0$. For the second equation, note that $x^3 = -x-1$, and since $x = y = z$, $m = \lfloor -3x-3 \rfloor$, where $x$ is a real number. Since $-\tfrac{1}{3}^3 - \tfrac13 + 1 = \tfrac{16}{27}$ and $-\tfrac{2}{3}^3 - \tfrac23 + 1 = \tfrac{1}{27}$, the root of $x$ is less than $-\tfrac23$ but more than $-1$, so \[0 > -3x-3 > -1\] \[m = \lfloor -3x-3 \rfloor = -1\]


Case 2: $x^2 + z^2 = 1$ Because $x^2 \le z^2$, $x^2 \le \frac12$. From one of the original equations, \[z^2 + x^4 = -y\] \[1 - x^2 + x^4 + x = 0\] Using the Rational Root Theorem, \[(x+1)(x^3 - x^2 + 1) = 0\] Note that if $x = -1$, then $x^2 \ge \tfrac12$, so that won’t work. Let $x = -a$ (where $a \ge 0$ since $x \le 0$), so \[a^3 + a^2 = 1\] If $a \le \tfrac{\sqrt2}{2}$, then \[a^3 + a^2 \le \frac{\sqrt2}{4} + \frac12\] \[a^3 + a^2 \le \frac{\sqrt2 + 2}{4} < 1\] Thus, there are no solutions in this case.


From the two cases, the smallest possible value of $m$ is $-1$, so the modulo $2007$ residue of $m$ is $\boxed{2006}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 45 		Followed by:
Problem 47
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