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

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

@@ Line 12: / Line 12: @@
 ==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>.
-<math>b = \log{y}</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>.
-<math>c = \log{z}</math>
+Now we use the quadratic formula:
-<math>xyz = 10^{81}</math>
+<math>x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}</math>
-<math>\log{xyz} = 81</math>
+<math>x=\frac{81\pm\sqrt{4689}}{2}</math>
-<math>\log{x} + \log{y} + \log{z} = 81</math>
+<math>x=\frac{81+3\sqrt{521}}{2}</math>, <math>x=\frac{81-3\sqrt{521}}{2}</math>
-<math>a + b + c = 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{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468</math>
+<math>\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}</math>
-<math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math>
+<math>=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}</math>
-<math>a^2 + b^2 + c^2 = 5621 = 75^2</math>
+<math>=\sqrt{\frac{11250}{2}}</math>
-<math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math>
+<math>=\boxed{075}</math>
-~MathleteMA
+~bad_at_mathcounts
+==Video solution==
+https://www.youtube.com/watch?v=Ix6czB_A_Js&t
 == See also ==
 {{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 {{MAA Notice}}

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