Difference between revisions of "2021 WSMO Accuracy Round Problems/Problem 3"
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==Solution== | ==Solution== | ||
If <math>3+\sqrt{5}</math> is a root of the polynomial, then <math>3-\sqrt{5}</math> is too. Similarly, if <math>4-\sqrt{7}</math> is a root of the polynomial, then so is <math>4+\sqrt{7}.</math> Thus, <math>f(x)=d(x)\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right),</math> for some polynomial <math>d.</math> To minimize the degree of <math>f,</math> we can set the degree of <math>d</math> to 0, which means that <math>d(x)=c.</math> This means that the leading coefficient of <math>f(x)=c.</math> Since <math>f</math> is monic, <math>c=1,</math> which means that <cmath>f=\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right)=</cmath><cmath>\left(\left(x-3\right)-\sqrt{5}\right)\left(\left(x-3\right)+\sqrt{5}\right)\left(\left(x-4\right)+\sqrt{7}\right)\left(\left(x-4\right)-\sqrt{7}\right)=</cmath><cmath>\left(\left(x-3\right)^{2}-\left(\sqrt{5}\right)^{2}\right)\left(\left(x-4\right)^{2}-\left(\sqrt{7}\right)^{2}\right)=</cmath><cmath>\left(x^{2}-6x+4\right)\left(x^{2}-8x+9\right)=x^{4}-14x^{3}+61x^{2}-86x+36.</cmath> We conclude that <cmath>|f(1)|=|1-14+61-86+36|=|-2|=\boxed{2}.</cmath> | If <math>3+\sqrt{5}</math> is a root of the polynomial, then <math>3-\sqrt{5}</math> is too. Similarly, if <math>4-\sqrt{7}</math> is a root of the polynomial, then so is <math>4+\sqrt{7}.</math> Thus, <math>f(x)=d(x)\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right),</math> for some polynomial <math>d.</math> To minimize the degree of <math>f,</math> we can set the degree of <math>d</math> to 0, which means that <math>d(x)=c.</math> This means that the leading coefficient of <math>f(x)=c.</math> Since <math>f</math> is monic, <math>c=1,</math> which means that <cmath>f=\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right)=</cmath><cmath>\left(\left(x-3\right)-\sqrt{5}\right)\left(\left(x-3\right)+\sqrt{5}\right)\left(\left(x-4\right)+\sqrt{7}\right)\left(\left(x-4\right)-\sqrt{7}\right)=</cmath><cmath>\left(\left(x-3\right)^{2}-\left(\sqrt{5}\right)^{2}\right)\left(\left(x-4\right)^{2}-\left(\sqrt{7}\right)^{2}\right)=</cmath><cmath>\left(x^{2}-6x+4\right)\left(x^{2}-8x+9\right)=x^{4}-14x^{3}+61x^{2}-86x+36.</cmath> We conclude that <cmath>|f(1)|=|1-14+61-86+36|=|-2|=\boxed{2}.</cmath> | ||
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| + | ~pinkpig | ||
Latest revision as of 11:29, 23 December 2021
Problem
If 
is a monic polynomial of minimal degree with rational coefficients satisfying 
and 
find the value of 
.
Solution
If 
is a root of the polynomial, then 
is too. Similarly, if 
is a root of the polynomial, then so is 
Thus, 
for some polynomial 
To minimize the degree of 
we can set the degree of 
to 0, which means that 
This means that the leading coefficient of 
Since is monic, 
which means that ![\[f=\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right)=\]](http://latex.artofproblemsolving.com/b/d/f/bdf068f758b60b7b3e752258348c48a577eb9433.png)
![\[\left(\left(x-3\right)-\sqrt{5}\right)\left(\left(x-3\right)+\sqrt{5}\right)\left(\left(x-4\right)+\sqrt{7}\right)\left(\left(x-4\right)-\sqrt{7}\right)=\]](http://latex.artofproblemsolving.com/c/2/3/c2330863df10250ffa68de5d7889952920312124.png)
![\[\left(\left(x-3\right)^{2}-\left(\sqrt{5}\right)^{2}\right)\left(\left(x-4\right)^{2}-\left(\sqrt{7}\right)^{2}\right)=\]](http://latex.artofproblemsolving.com/f/2/f/f2f3c4f6f483a62b372a6a25cd5d4566dcd86bbf.png)
![\[\left(x^{2}-6x+4\right)\left(x^{2}-8x+9\right)=x^{4}-14x^{3}+61x^{2}-86x+36.\]](http://latex.artofproblemsolving.com/2/6/e/26ea652ac0d802eb5896fabdef6befe30e626353.png)
We conclude that ![\[|f(1)|=|1-14+61-86+36|=|-2|=\boxed{2}.\]](http://latex.artofproblemsolving.com/a/e/4/ae425db8a64104984be3c5843b8e1b33dc463642.png)
~pinkpig
