Difference between revisions of "2010 AIME II Problems/Problem 5"
Mathletema (talk | contribs) (→Solution 2) |
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<math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math> | <math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math> | ||
− | <math>a^2 + b^2 + c^2 = | + | <math>a^2 + b^2 + c^2 = 5625 = 75^2</math> |
<math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math> | <math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math> |
Revision as of 19:57, 21 February 2018
Contents
[hide]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
~MathleteMA
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.