Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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==Video Solutions== | ==Video Solutions== |
Latest revision as of 20:22, 19 August 2024
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 since at the minimum, two of the numbers are less than
, but two are between
and
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.