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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 3"

(Created page with "==Problem== Suppose that <math>a, b, c</math> are real numbers such <cmath> \begin{align*} a + b - c &= 4 \\ a^2 + b^2 + c^2 &= 14 \\ a^3 + b^3 - c^3 &= 34 \\ \end...")
 
 
(2 intermediate revisions by the same user not shown)
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</cmath>
 
</cmath>
 
Find the sum of all possible values of <math>a+b+c</math>.
 
Find the sum of all possible values of <math>a+b+c</math>.
 +
 
==Solution==
 
==Solution==
 +
 +
Let <math>x = -c.</math> Therefore, we have
 +
 +
<cmath>\begin{align}
 +
a+b+x&=4,\\
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a^2+b^2+x^2&=14,\text{ and }\\
 +
a^3+b^3+x^3 &= 34.
 +
\end{align}</cmath>
 +
 +
Subtracting equation (2) from the square of (1), we have  <cmath>2(ab+ax+bx) = 2\implies ab+ax+bx = 1.</cmath>
 +
Multiplying this by the first equation, we have <cmath>(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+3abx = 4.</cmath>
 +
Cubing the first equation, we have
 +
 +
\begin{align*}
 +
(a^3+b^3+x^3)+3(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+6abx &= \\
 +
(a+b+x)^3\implies34+3(4-3abx)+6abx = 64\implies abx=6.
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\end{align*}
 +
 +
Thus, <math>a,b,x</math> are the roots of the polynomial <math>P(y) = y^3-4y^2+y-6,</math> which factors <math>(y-3)(y-2)(y+1).</math> So, the possible values of <math>a+b+c</math> are
 +
 +
<cmath>\begin{align*}
 +
a+b+c = a+b-x = 3+2-(-1) = 6,\\
 +
a+b+c = a+b-x = 3+(-1)-(2) = 0,\text{ and }\\
 +
a+b+c = a+b-x = 2+(-1)-(3) = -2.
 +
\end{align*}</cmath>
 +
 +
Thus, the answer is <math>6+0+(-2) = \boxed{4}.</math>
 +
 +
~SMO_Team

Latest revision as of 14:51, 9 September 2025

Problem

Suppose that $a, b, c$ are real numbers such \begin{align*} a + b - c &= 4 \\ a^2 + b^2 + c^2 &= 14 \\ a^3 + b^3 - c^3 &= 34 \\ \end{align*} Find the sum of all possible values of $a+b+c$.

Solution

Let $x = -c.$ Therefore, we have

\begin{align} a+b+x&=4,\\ a^2+b^2+x^2&=14,\text{ and }\\ a^3+b^3+x^3 &= 34. \end{align}

Subtracting equation (2) from the square of (1), we have \[2(ab+ax+bx) = 2\implies ab+ax+bx = 1.\] Multiplying this by the first equation, we have \[(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+3abx = 4.\] Cubing the first equation, we have

\begin{align*} (a^3+b^3+x^3)+3(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+6abx &= \\ (a+b+x)^3\implies34+3(4-3abx)+6abx = 64\implies abx=6. \end{align*}

Thus, $a,b,x$ are the roots of the polynomial $P(y) = y^3-4y^2+y-6,$ which factors $(y-3)(y-2)(y+1).$ So, the possible values of $a+b+c$ are

\begin{align*} a+b+c = a+b-x = 3+2-(-1) = 6,\\ a+b+c = a+b-x = 3+(-1)-(2) = 0,\text{ and }\\ a+b+c = a+b-x = 2+(-1)-(3) = -2. \end{align*}

Thus, the answer is $6+0+(-2) = \boxed{4}.$

~SMO_Team

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