Difference between revisions of "2007 iTest Problems/Problem 46"
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− | == Problem == | + | ==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: | ||
Line 5: | Line 5: | ||
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>. | ||
− | == Solution == | + | ==Solution== |
+ | |||
+ | Rearrange the terms to get | ||
+ | <cmath>y^2 + z^4 = -x</cmath> | ||
+ | <cmath>z^2 + x^4 = -y</cmath> | ||
+ | <cmath>x^2 + y^4 = -z</cmath> | ||
+ | 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 | ||
+ | |||
+ | <cmath>y^2 + z^4 \le z^2 + x^4 \le x^2 + y^4</cmath> | ||
+ | |||
+ | 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>, | ||
+ | <cmath>y^2 + z^4 \le x^2 + z^4</cmath> | ||
+ | <cmath>y^2 \le x^2</cmath> | ||
+ | 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>. | ||
+ | |||
+ | <br> | ||
+ | Using substitution in the original system, | ||
+ | <cmath>x^2 + z^4 = z^2 + x^4</cmath> | ||
+ | <cmath>(z^2 + x^2)(z^2 - x^2) - (z^2 - x^2) = 0</cmath> | ||
+ | <cmath>(z^2 - x^2)(x^2 + z^2 - 1) = 0</cmath> | ||
+ | To find the real solutions, we use [[casework]] and the Zero Product Property. | ||
+ | |||
+ | <br> | ||
+ | '''Case 1: <math>z^2 = x^2</math>''' | ||
+ | |||
+ | 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 | ||
+ | <cmath>x+x^2+x^4 = 0</cmath> | ||
+ | <cmath>x(1+x+x^3) = 0</cmath> | ||
+ | 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 | ||
+ | <cmath>0 > -3x-3 > -1</cmath> | ||
+ | <cmath>m = \lfloor -3x-3 \rfloor = -1</cmath> | ||
+ | |||
+ | <br> | ||
+ | '''Case 2: <math>x^2 + z^2 = 1</math>''' | ||
+ | |||
+ | 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> | ||
+ | <cmath>1 - x^2 + x^4 + x = 0</cmath> | ||
+ | Using the [[Rational Root Theorem]], | ||
+ | <cmath>(x+1)(x^3 - x^2 + 1) = 0</cmath> | ||
+ | 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 | ||
+ | <cmath>a^3 + a^2 = 1</cmath> | ||
+ | If <math>a \le \tfrac{\sqrt2}{2}</math>, then | ||
+ | <cmath>a^3 + a^2 \le \frac{\sqrt2}{4} + \frac12</cmath> | ||
+ | <cmath>a^3 + a^2 \le \frac{\sqrt2 + 2}{4} < 1</cmath> | ||
+ | Thus, there are no solutions in this case. | ||
+ | |||
+ | <br> | ||
+ | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{ITest box|year=2007|num-b=45|num-a=47}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 17:23, 22 July 2018
Problem
Let be an ordered triplet of real numbers that satisfies the following system of equations:
If
is the minimum possible value of
, find the modulo
residue of
.
Solution
Rearrange the terms to get
Since the left hand side of all three equations is greater than or equal to 0,
. Also, note that the equations have symmetry, so WLOG, let
. By substitution, we have
Note that and
. That means
. Since
,
Since
, then
. Because
and
are nonpositive,
.
Using substitution in the original system,
To find the real solutions, we use casework and the Zero Product Property.
Case 1:
If , then since
and
are nonpositive, then
. Substitution results in
That means
or
. For the first equation,
. For the second equation, note that
, and since
,
, where
is a real number. Since
and
, the root of
is less than
but more than
, so
Case 2:
Because ,
. From one of the original equations,
Using the Rational Root Theorem,
Note that if
, then
, so that won’t work. Let
(where
since
), so
If
, then
Thus, there are no solutions in this case.
From the two cases, the smallest possible value of is
, so the modulo
residue of
is
.
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 45 |
Followed by: Problem 47 | |
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