2012 AIME I Problems/Problem 9
Let and be positive real numbers that satisfy The value of can be expressed in the form where and are relatively prime positive integers. Find
Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that Then Solving these equations, we quickly see that and then Finally, our desired value is and thus
Notice that , and .
From this, we see that is the geometric mean of and . So, for constant : Since are in an arithmetic progression, so are .
Therefore, is the geometric mean of and We can plug in to any of the two equal fractions aforementioned. So, without loss of generality:
Thus and .
Since we are given that , we may assume that , and are all powers of two. We shall thus let , , and . Let . From this we get the system of equations:
Plugging equation into equation yields . Plugging equation into equation and simplifying yields , and substituting for and simplifying yields . But , so , so .
Knowing this, we may substitute for in equations and , yielding and . Thus, we have that . We are looking for . and , so . The answer is .
Video Solution by Richard Rusczyk
|2012 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|