Difference between revisions of "2010 AIME II Problems/Problem 5"
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− | == 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>. | ||
== Solution == | == Solution == | ||
− | 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} | + | 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>. | ||
+ | |||
+ | 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{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> | ||
+ | |||
+ | <math>=\boxed{075}</math> | ||
+ | |||
+ | ~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}} |
Latest revision as of 23:49, 5 February 2022
Problem
Positive numbers ,
, and
satisfy
and
. Find
.
Solution
Using the properties of logarithms, by taking the log base 10 of both sides, and
by using the fact that
.
Through further simplification, we find that . It can be seen that there is enough information to use the formula
, as we have both
and
, and we want to find
.
After plugging in the values into the equation, we find that is equal to
.
However, we want to find , so we take the square root of
, or
.
Solution 2
Let ,
and
.
We have and
. Since these two equations look a lot like Vieta's for a cubic, create the polynomial
(leave the constant term as
to make things easy). Dividing by
yields
.
Now we use the quadratic formula:
,
Since the question asks for (remember one of the values was the solution
that we divided out in the beginning), we find:
~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.