2021 Fall AMC 12A Problems/Problem 19
Contents
Problem
Let be the least real number greater than such that , where the arguments are in degrees. What is rounded up to the closest integer?
Solution 1
The smallest to make would require , but since needs to be greater than , these solutions are not valid.
The next smallest would require , or .
After a bit of guessing and checking, we find that , and , so the solution lies between and , making our answer
Note: One can also solve the quadratic and estimate the radical.
~kingofpineapplz
Solution 2
For choice we have For choice we have For choice we have For choice we have For choice we have Therefore, the answer is as is the closest to
~Steven Chen (www.professorchenedu.com)
Easy Solution with Intermediate Value Theorem
we know that x^2-x=180-2x+360k or x^2-x=360k from looking at the sin values on the unit circle. with these equations, we can just insert the values that are given in the 5 answers. let's try A first. we know that one of the equations should be greater on the left, and another equation we make should be greater on the right for the solution to be in between the values inserted into the equations the two numbers it is in between is 9 and 10 since it is rounded up. 100-10 vs 180-20 81-9 vs 180-18 this does not satisfy what we would like to find now let's try B between 12 and 13 144-12 vs 180-24 169-13 vs 180-26 now this satisfies our hunt for the solution the answer is B
emilyyunhanq@gmail.com solution by Emily Q
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=H0pNJFbV4jE
Video Solution by TheBeautyofMath
Solved both Mentally and by writing things down
~IceMatrix
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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