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Difference between revisions of "2010 AIME II Problems/Problem 5"

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== Solution ==  
 
== Solution ==  
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Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}x)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>.  
+
Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>.  
  
 
Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>.  
 
Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>.  
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After plugging in the values into the equation, we find that <math>\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</math>.  
 
After plugging in the values into the equation, we find that <math>\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</math>.  
  
−
However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{75}</math>.
+
However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{075}</math>.
 +
 
 +
==Solution 2==
 +
Let <math>a=\log_{10}x</math>, <math>b=\log_{10}y,</math> and <math>c=\log_{10}z</math>.
 +
 
 +
We have <math>a+b+c=81</math> and <math>a(b+c)+bc=ab+ac+bc=468</math>. Since these two equations look a lot like Vieta's for a cubic, create the polynomial <math>x^3-81x^2+468x=0</math> (leave the constant term as <math>0</math> to make things easy). Dividing by <math>x</math> yields <math>x^2-81x+468=0</math>.
 +
 
 +
Now we use the quadratic formula:
 +
 
 +
<math>x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}</math>
 +
 
 +
<math>x=\frac{81\pm\sqrt{4689}}{2}</math>
 +
 
 +
<math>x=\frac{81+3\sqrt{521}}{2}</math>, <math>x=\frac{81-3\sqrt{521}}{2}</math>
 +
 
 +
Since the question asks for <math>\sqrt{a^2+b^2+c^2}</math>  (remember one of the values was the solution <math>x=0</math> that we divided out in the beginning), we find:
 +
 
 +
<math>\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}</math>
 +
 
 +
<math>=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}</math>
 +
 
 +
<math>=\sqrt{\frac{11250}{2}}</math>
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 +
<math>=\boxed{075}</math>
 +
 
 +
~bad_at_mathcounts
 +
 
 +
==Video solution==
 +
 
 +
https://www.youtube.com/watch?v=Ix6czB_A_Js&t
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 
{{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 +
{{MAA Notice}}

Latest revision as of 22:49, 5 February 2022

Problem

Positive numbers $x$, $y$, and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$. Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$.

Solution

Using the properties of logarithms, $\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468$ by using the fact that $\log_{10}ab = \log_{10}a + \log_{10}b$.

Through further simplification, we find that $\log_{10}x+\log_{10}y+\log_{10}z = 81$. It can be seen that there is enough information to use the formula $\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, as we have both $\ a+b+c$ and $\ 2ab+2ac+2bc$, and we want to find $\sqrt {a^2 + b^2 + c^2}$.

After plugging in the values into the equation, we find that $\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2$ is equal to $\ 6561 - 936 = 5625$.

However, we want to find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$, so we take the square root of $\ 5625$, or $\boxed{075}$.

Solution 2

Let $a=\log_{10}x$, $b=\log_{10}y,$ and $c=\log_{10}z$.

We have $a+b+c=81$ and $a(b+c)+bc=ab+ac+bc=468$. Since these two equations look a lot like Vieta's for a cubic, create the polynomial $x^3-81x^2+468x=0$ (leave the constant term as $0$ to make things easy). Dividing by $x$ yields $x^2-81x+468=0$.

Now we use the quadratic formula:

$x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}$

$x=\frac{81\pm\sqrt{4689}}{2}$

$x=\frac{81+3\sqrt{521}}{2}$, $x=\frac{81-3\sqrt{521}}{2}$

Since the question asks for $\sqrt{a^2+b^2+c^2}$ (remember one of the values was the solution $x=0$ that we divided out in the beginning), we find:

$\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}$

$=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}$

$=\sqrt{\frac{11250}{2}}$

$=\boxed{075}$

~bad_at_mathcounts

Video solution

https://www.youtube.com/watch?v=Ix6czB_A_Js&t

See also

2010 AIME II (Problems • Answer Key • Resources)
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