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| | ==Problem== | | ==Problem== |
| − | The roots of <math>x^3 + 2x^2 - x + 3</math> are <math>p, q,</math> and <math>r.</math> What is the value of <cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)?</cmath><math>\textbf{(A) } 64 \qquad \textbf{(B) } 75 \qquad \textbf{(C) } 100 \qquad \textbf{(D) } 125 \qquad \textbf{(E) } 144</math>
| + | Suppose <math>p</math> and <math>q</math> are real numbers for which <cmath>\log_{p}(q) - \log_{2p}(q) =\frac{1}{3} \qquad \operatorname{and} \qquad \log_{2p}(q) - \log_{4p}(q) =\frac{1}{4}.</cmath> What is the value of <math>\log_{4p}(q) - \log_{8p}(q)</math>? |
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| − | ==Solution 1==
| + | <math>\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{7}{40}\qquad\textbf{(C)}~\frac{8}{45}\qquad\textbf{(D)}~\frac{7}{36}\qquad\textbf{(E)}~\frac{1}{5}</math> |
| − | You can factor <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math> as <math>(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)</math>.
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| − | For any polynomial <math>f(x)</math>, you can create a new polynomial <math>f(x+2)</math>, which will have roots that instead have the value subtracted.
| + | ==Solution== |
| − | | + | Let <math>x = \log_{2}p</math> and <math>y = \log_{2}q</math>. Then note <cmath>\log_{p}(q) - \log_{2p}(q) =\frac{\log_{2}(q)}{\log_{2}(p)} -\frac{\log_{2}q}{\log_{2}(2p)} =\frac{y}{x} -\frac{y}{x + 1} =\frac{y}{x(x + 1)} =\frac{1}{3} \implies x(x + 1) = 3y.</cmath> Similarly, <cmath>\log_{2p}(q) - \log_{4p}(q) =\frac{\log_{2}(q)}{\log_{2}(2p)} -\frac{\log_{2}(q)}{\log_{2}(4p)} =\frac{y}{x + 1} -\frac{y}{x + 2} =\frac{y}{(x + 1)(x + 2)} =\frac{1}{4} \implies (x + 1)(x + 2) = 4y.</cmath> Dividing the two equations, <math>\tfrac{x}{x + 2} = \tfrac{3}{4}</math> which solves to <math>x = 6</math>. Substituting, <math>y = \tfrac{1}{3} \cdot 6 \cdot 7 = 14</math> and <cmath>\log_{4p}(q) - \log_{8p}(q) =\frac{y}{x + 2} -\frac{y}{x + 3} = \frac{14}{8} - \frac{14}{9} = \boxed{\textbf{(D)}~\frac{7}{36}}.</cmath> |
| − | Substituting <math>x-2</math> and <math>x+2</math> into <math>x</math> for the first polynomial, gives you <math>10i-5</math> and <math>-10i-5</math> as <math>c</math> for both equations. Multiplying <math>10i-5</math> and <math>-10i-5</math> together gives you <math>\boxed{\textbf{(D) }125}</math>.
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| − | -ev2028
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| − | ~Latex by eevee9406
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| − | ==Solution 2== | |
| − | Let <math>f(x)=x^3 + 2x^2 - x + 3</math>. Then | |
| − | <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)=(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)=f(2i)f(-2i)</math>.
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| − | We find that <math>f(2i)=-8i-8-2i+3=-10i-5</math> and <math>f(-2i)=8i-8+2i+3=10i-5</math>, so <math>f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}</math>.
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| − | Select D
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| − | ~eevee9406
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| − | ==Solution 3==
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| − | First, denote that
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| − | <cmath>p+q+r=-2,
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| − | pq+pr+qr=-1,
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| − | pqr=-3</cmath>
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| − | Then we expand the expression | |
| − | <cmath>(p^2+4)(q^2+4)(r^2+4)</cmath> | |
| − | <cmath>=(pqr)^2+4((pq)^2+(pr)^2+(qr)^2)+4^2(p^2+q^2+r^2)+4^3</cmath>
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| − | <cmath>=(-3)^2+4((pq+pr+qr)^2-2pqr(p+q+r))+4^2((p+q+r)^2-2(pq+pr+qr))+4^3</cmath>
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| − | <cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath>
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| − | <cmath>=\fbox{(D) 125}</cmath>
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| − | ~lptoggled
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| − | ==Solution 4 (Reduction of power)==
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| − | The motivation for this solution is the observation that <math>(p+c)(q+c)(r+c)</math> is easy to compute for any constant c, since <math>(p+c)(q+c)(r+c)=-f(-c)</math> (*), where <math>f</math> is the polynomial given in the problem. The idea is to transform the expression involving <math>p^2, q^2, r^2</math> into one involving <math>p, q, r</math>.
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| − | Since <math>p</math> is a root of <math>f</math>,
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| − | <cmath>p^3+2p^2-p+3=0\implies p^3+2p^2=p^2(p+2)=p-3,</cmath>
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| − | which gives us that <math>p^2=\frac{p-3}{p+2}</math>. Then
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| − | <cmath>p^2+4=\frac{p-3}{p+2} + 4 = \frac{5p+5}{p+2}=5\cdot \frac{p+1}{p+2}.</cmath>
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| − | Since <math>q</math> and <math>r</math> are also roots of <math>f</math>, the same analysis holds, so
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| − | \begin{align*} | |
| − | (p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\ | |
| − | &= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\
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| − | &= 125 \frac{-f(-1)}{-f(-2)} \\
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| − | &= 125\cdot 1\\
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| − | &=\boxed{\textbf{(D) }125}.
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| − | \end{align*}
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| − | (*) This is because
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| − | <cmath>(p+c)(r+c)(q+c)=(-1)^3(-c-p)(-c-r)(-c-q)=-f(-c),</cmath>
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| − | since
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| − | <cmath>f(x)=(x-p)(x-q)(x-r)</cmath>
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| − | for all <math>x</math>.
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| − | ~tsun26
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| − | ~KSH31415 (final step and clarification)
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| − | ==Solution 5 (Cheesing it out)==
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| − | Expanding the expression
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| − | <cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)</cmath>
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| − | gives us
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| − | <cmath>(pqr)^2+4p^2q^2+4p^2r^2+4q^2r^2+16p^2+16q^2+16r^2+64</cmath>
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| − | Notice that everything other than <math>(pqr)^2</math> is a multiple of <math>4</math>. Solving for <math>(pqr)^2</math> using vieta's formulas, we get <math>9</math>. Since <math>9</math> is <math>1\pmod4</math>, the answer should be as well. The only answer that is <math>1\pmod4</math> is <math>\boxed{\textbf{(D) }125}</math>.
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| − | ~callyaops
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| − | ==Solution 6==
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| − | Suppose <math>y = x^2 + 4</math>
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| − | then <math>x =\pm \sqrt{y - 4}</math>. Substitute <math>x = \sqrt{y - 4}</math> into <math>x^3 + 2x^2 - x + 3 = 0</math> (It is same for <math>x = -\sqrt{y - 4}</math> because the squares in <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math>)
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| − | <math>(\sqrt{y - 4})^3 + 2(\sqrt{y - 4})^2 - \sqrt{y - 4} + 3 = 0 \implies (y - 5)^2(y - 4) = (-2y + 5)^2</math> whose constant is 125
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| − | according to Vieta's theorem, <math>y_1y_2y_3 = 125</math>
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| − | <math>y_1y_2y_3 = 125 \implies (x_1^2 + 4)(x_2^2 + 4)(x_3^2 + 4) = 125 \implies (p^2 + 4)(q^2 + 4)(r^2 + 4) = \boxed{\textbf{(D) }125}</math>.
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| − | ~JiYang
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| | ==See also== | | ==See also== |
| | {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | | {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
and 
are real numbers for which ![\[\log_{p}(q) - \log_{2p}(q) =\frac{1}{3} \qquad \operatorname{and} \qquad \log_{2p}(q) - \log_{4p}(q) =\frac{1}{4}.\]](http://latex.artofproblemsolving.com/8/1/3/813cdbfad6dc9278751c7c8e487b83ddc95eb3f5.png)
What is the value of 
?
and 
. Then note ![\[\log_{p}(q) - \log_{2p}(q) =\frac{\log_{2}(q)}{\log_{2}(p)} -\frac{\log_{2}q}{\log_{2}(2p)} =\frac{y}{x} -\frac{y}{x + 1} =\frac{y}{x(x + 1)} =\frac{1}{3} \implies x(x + 1) = 3y.\]](http://latex.artofproblemsolving.com/b/1/6/b16fd641df70c288d26c9674aba5d2994627f0e6.png)
Similarly, ![\[\log_{2p}(q) - \log_{4p}(q) =\frac{\log_{2}(q)}{\log_{2}(2p)} -\frac{\log_{2}(q)}{\log_{2}(4p)} =\frac{y}{x + 1} -\frac{y}{x + 2} =\frac{y}{(x + 1)(x + 2)} =\frac{1}{4} \implies (x + 1)(x + 2) = 4y.\]](http://latex.artofproblemsolving.com/0/f/0/0f027b5c42205e60a9f4c57b05758709fe0af7c0.png)
Dividing the two equations, 
which solves to 
. Substituting, 
and ![\[\log_{4p}(q) - \log_{8p}(q) =\frac{y}{x + 2} -\frac{y}{x + 3} = \frac{14}{8} - \frac{14}{9} = \boxed{\textbf{(D)}~\frac{7}{36}}.\]](http://latex.artofproblemsolving.com/3/8/3/38333aa04376a219fa60a3baab678d30b0dfbe31.png)
