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...") |
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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,\\ | ||
| + | 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. | ||
| + | \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. | ||
| + | </cmath> | ||
| + | Thus, the answer is <math>6+0+(-2) = \boxed{4}.</math> | ||
Revision as of 14:50, 9 September 2025
Problem
Suppose that 
are real numbers such
Find the sum of all possible values of 
.
Solution
Let 
Therefore, we have
Subtracting equation (2) from the square of (1), we have ![\[2(ab+ax+bx) = 2\implies ab+ax+bx = 1.\]](http://latex.artofproblemsolving.com/8/9/4/894b8156d33d13a503c363c01fdb211d87c9a89e.png)
Multiplying this by the first equation, we have ![\[(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+3abx = 4.\]](http://latex.artofproblemsolving.com/7/d/4/7d4fce5b2e9388625ccce4ca9510e8ac4a73abae.png)
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, 
are the roots of the polynomial 
which factors 
So, the possible values of 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. (Error compiling LaTeX. Unknown error_msg)
Thus, the answer is 
