2020 AIME I Problems/Problem 2
Contents
Problem
There is a unique positive real number such that the three numbers , , and , in that order, form a geometric progression with positive common ratio. The number can be written as , where and are relatively prime positive integers. Find .
Solution
Since these form a geometric series, is the common ratio. Rewriting this, we get by base change formula. Therefore, the common ratio is 2. Now
. Therefore, .
~ JHawk0224
Solution 2
If we set , we can obtain three terms of a geometric sequence through logarithm properties. The three terms are In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: which can be solved to reveal . Therefore, , so our answer is .
-molocyxu
Solution 3
Let be the common ratio. We have Hence we obtain Ideally we change everything to base and we can get: Now divide to get: By change-of-base we obtain: Hence and we have as desired.
~skyscraper
Solution 4 (Exponents > Logarithms)
Let be the common ratio, and let be the starting term (). We then have: Rearranging these equations gives: Deal with the last two equations first: Setting them equal gives: Using LTE results in: Using this value of , substitute into the first and second equations (or the first and third, it doesn't really matter) to get: Changing these to a common base gives: Dividing the first equation by 2 on both sides yields: Setting these equations equal to each other and applying LTE again gives: Substituting this back into the first equation gives: Therefore,
~IAmTheHazard
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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