Difference between revisions of "2010 AIME II Problems/Problem 5"
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==Solution 2== | ==Solution 2== | ||
− | <math>a = \ | + | Let <math>a=\log_{10}x</math>, <math>b=\log_{10}y,</math> and <math>c=\log_{10}z</math>. |
− | <math>b = | + | 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> | + | <math>x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}</math> |
− | <math>\ | + | <math>x=\frac{81\pm\sqrt{4689}}{2}</math> |
− | <math>\ | + | <math>x=\frac{81+3\sqrt{521}}{2}</math>, <math>x=\frac{81-3\sqrt{521}}{2}</math> |
− | <math>a + b + c = | + | 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>\ | + | <math>\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}</math> |
− | <math> | + | <math>=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}</math> |
− | <math> | + | <math>=\sqrt{\frac{11250}{2}}</math> |
− | <math> | + | <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}} | {{MAA Notice}} |
Latest revision as of 22:49, 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.