Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Adding problem section)
(→‎Solution 2)
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
−
 
−
==Problem==
 
 
== Problem ==
 
== Problem ==
 
Positive numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy <math>xyz = 10^{81}</math> and <math>(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468</math>. Find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>.
 
Positive numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy <math>xyz = 10^{81}</math> and <math>(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468</math>. Find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>.
Line 14: Line 12:
  
 
==Solution 2==
 
==Solution 2==
−
<math>a = \log{x}</math>
+
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>b = \log{y}</math>
+
<math>x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}</math>
  
−
<math>c = \log{z}</math>
+
<math>x=\frac{81\pm\sqrt{4689}}{2}</math>
  
−
<math>xyz = 10^{81}</math>
+
<math>x=\frac{81+3\sqrt{521}}{2}</math>, <math>x=\frac{81-3\sqrt{521}}{2}</math>
  
−
<math>\log{xyz} = 81</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>\log{x} + \log{y} + \log{z} = 81</math>
+
<math>\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}</math>
  
−
<math>a + b + c = 81</math>
+
<math>=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}</math>
  
−
<math>\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468</math>
+
<math>=\sqrt{\frac{11250}{2}}</math>
  
−
<math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math>
+
<math>=\boxed{075}</math>
  
−
<math>a^2 + b^2 + c^2 = 5625 = 75^2</math>
+
~bad_at_mathcounts
  
−
<math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math>
+
==Video solution==
  
−
~MathleteMA
+
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}}
 
{{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)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_5&oldid=171010"