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 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 removing the exponent again gives: Substituting this back into the first equation gives: Therefore,
~IAmTheHazard
Solution 5
We can relate the logarithms as follows:
Now we can convert all logarithm bases to using the identity :
We can solve for as follows:
We get . Verifying that the common ratio is positive, we find the answer of .
~QIDb602
Solution 6
If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as and , respectively. Therefore: Let . We can rewrite the expression as: Zero does not work in this case, so we consider : . Therefore, .
~Bowser498
Solution 7
Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let be the exponent of , then we have Also, we can divide the first equation by two to have the right side be . Also (again), . Setting this equal to , we can divide the two equations to get . Therefore, . After that, we can raise to the th power to get . We then get our sum of .
Solution 8 (Official MAA)
By the Change of Base Formula the common ratio of the progression isHence must satisfyThis is equivalent to . Hence and . The requested sum is . See here for a video solution:
Video Solutions
https://youtu.be/mgRNqSDCvgM?t=281s
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.