2019 AMC 10A Problems/Problem 19
Contents
Problem
What is the least possible value of where is a real number?
Solution 1
Grouping the first and last terms and two middle terms gives , which can be simplified to . Noting that squares are nonnegative, and verifying that for some real , the answer is .
Solution 2
Let . Then the expression becomes .
We can now use the difference of two squares to get , and expand this to get .
Refactor this by completing the square to get , which has a minimum value of . The answer is thus .
Solution 3 (calculus)
Similar to Solution 1, grouping the first and last terms and the middle terms, we get .
Letting , we get the expression . Now, we can find the critical points of to minimize the function:
To minimize the result, we use . Hence, the minimum is , so .
Note: We could also have used the result that minimum/maximum point of a parabola occurs at .
Note 2: This solution is somewhat "lucky", since when we define variables to equal a function, and create another function out of these variables, the domain of such function may vary from the initial one. This is important because the maximum and minimum value of a function is dependent on its domain, e.g:
has no maximum value in the the integers, but once restricting the domain to the maximum value of is .
Also, observe that if we were to evaluate this by taking the derivative of , we would get as the -value to obtain the minimum -value of this expression. It can be seen that is actually an inflection point, instead of a minimum or maximum.
-Note 2 from Benedict T (countmath1)
Solution 4(guess with answer choices)
The expression is negative when an odd number of the factors are negative. This happens when or . Plugging in or yields , which is very close to . Thus the answer is .
Solution 5 (using the answer choices)
Answer choices , , and are impossible, since can be negative (as seen when e.g. ). Plug in to see that it becomes , so round this to .
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.
Solution 6 (also calculus but more convoluted)
We can ignore the and consider it later, as it is a constant. By difference of squares, we can group this into . We pull a factor of into each term to avoid dealing with decimals:
Now, we let . Our expression becomes:
Taking the derivative, we get This is equal to when , and plugging in , we get the expression is equal to and therefore our answer is
~Technodoggo
Video Solutions
https://www.youtube.com/watch?v=Vf2LkM7ExhY by SpreadTheMathLove
https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)
- https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything
- https://www.youtube.com/watch?v=Mfa7j2BoNjI
- https://youtu.be/tIzJtgJbHGc - savannahsolver
- https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14
- https://youtu.be/GmUWIXXf_uk?t=1134 ~ pi_is_3.14
See Also
2019 AMC 10A (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 10 Problems and Solutions |
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