Difference between revisions of "2021 WSMO Accuracy Round Problems/Problem 3"
(Created page with "==Problem== If <math>f</math> is a monic polynomial of minimal degree with rational coefficients satisfying <math>f\left(3+\sqrt{5}\right)=0</math> and <math>f\left(4-\sqrt{7}...") |
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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 12:29, 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
We conclude that
~pinkpig