2018 AIME I Problems/Problem 5
For each ordered pair of real numbers satisfying there is a real number such that Find the product of all possible values of .
Using the logarithmic property , we note that That gives upon simplification and division by . Factoring by Simon's Favorite Factoring Trick gives Then, From the second equation, If we take , we see that . If we take , we see that . The product is .
Do as done in Solution 1 to get Do as done in Solution 1 to get If then If then Hence our final answer is -vsamc -minor edit:einsteinstudent
-style edit: yeaboi
Solution 3 (Official MAA)
Because the right side of the first equation is real. It follows that the left side of the equation is also real, so and Thus which implies that Therefore either or and because must be positive and Similarly, If then when If then when The requested product is
|2018 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|