Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
− | Grouping the first and last terms and two middle terms gives <math>(x^2+5x+4)(x^2+5x+6)+2019</math>, which can be simplified | + | |
+ | Grouping the first and last terms and two middle terms gives <math>(x^2+5x+4)(x^2+5x+6)+2019</math>, which can be simplified to <math>(x^2+5x+5)^2-1+2019</math>. Noting that squares are nonnegative, and verifying that <math>x^2+5x+5=0</math> for some real <math>x</math>, the answer is <math>\boxed{\textbf{(B) } 2018}</math>. | ||
==Solution 2== | ==Solution 2== | ||
− | Let <math>a=x+\tfrac{5}{2}</math>. Then <math>(x+1)(x+2)(x+3)(x+4)</math> becomes <math>(a-\tfrac{3}{2})(a-\tfrac{1}{2})(a+\tfrac{1}{2})(a+\tfrac{3}{2} | + | Let <math>a=x+\tfrac{5}{2}</math>. Then the expression <math>(x+1)(x+2)(x+3)(x+4)</math> becomes <math>\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)</math>. |
− | |||
− | |||
− | + | We can now use the difference of two squares to get <math>\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)</math>, and expand this to get <math>a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}</math>. | |
− | - | + | Refactor this by completing the square to get <math>\left(a^2-\tfrac{5}{4}\right)^2-1</math>, which has a minimum value of <math>-1</math>. The answer is thus <math>2019-1=\boxed{\textbf{(B) }2018}</math>. |
− | ==Solution 3 ( | + | ==Solution 3 (calculus)== |
Similar to Solution 1, grouping the first and last terms and the middle terms, we get <math>(x^2+5x+4)(x^2+5x+6)+2019</math>. | Similar to Solution 1, grouping the first and last terms and the middle terms, we get <math>(x^2+5x+4)(x^2+5x+6)+2019</math>. | ||
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Letting <math>y=x^2+5x</math>, we get the expression <math>(y+4)(y+6)+2019</math>. Now, we can find the critical points of <math>(y+4)(y+6)</math> to minimize the function: | Letting <math>y=x^2+5x</math>, we get the expression <math>(y+4)(y+6)+2019</math>. Now, we can find the critical points of <math>(y+4)(y+6)</math> to minimize the function: | ||
− | <math>\frac{d}{ | + | <math>\frac{d}{dy}(y^2+10y+24)=0</math> |
<math>2y+10=0</math> | <math>2y+10=0</math> | ||
− | <math>2y | + | <math>2y=-5</math> |
<math>y=-5,0</math> | <math>y=-5,0</math> | ||
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To minimize the result, we use <math>y=-5</math>. Hence, the minimum is <math>(-5+4)(-5+6)=-1</math>, so <math>-1+2019 = \boxed{\textbf{(B) }2018}</math>. | To minimize the result, we use <math>y=-5</math>. Hence, the minimum is <math>(-5+4)(-5+6)=-1</math>, so <math>-1+2019 = \boxed{\textbf{(B) }2018}</math>. | ||
− | ( | + | ''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>. |
+ | |||
+ | ''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: | ||
+ | |||
+ | <math>f(x)=x^2</math> has no maximum value in the the integers, but once restricting the domain to <math>(-5, 5)</math> the maximum value of <math>f(x)</math> is <math>25</math>. | ||
+ | |||
+ | Also, observe that if we were to evaluate this by taking the derivative of <math>(x+1)(x+2)(x+3)(x+4)+2019</math>, we would get <math>-5</math> as the <math>x</math>-value to obtain the minimum <math>y</math>-value of this expression. It can be seen that <math>-5</math> 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 <math>-2 < x < -1</math> or <math>-4 < x < -3</math>. Plugging in <math>x = -\frac32</math> or <math>x = -\frac72</math> yields <math>-\frac{15}{16}</math>, which is very close to <math>-1</math>. Thus the answer is <math>-1 + 2019 = \boxed{\textbf{(B) }2018}</math>. | ||
+ | |||
+ | ==Solution 5 (using the answer choices) == | ||
+ | |||
+ | Answer choices <math>C</math>, <math>D</math>, and <math>E</math> are impossible, since <math>(x+1)(x+2)(x+3)(x+4)</math> can be negative (as seen when e.g. <math>x = -\frac{3}{2}</math>). Plug in <math>x = -\frac{3}{2}</math> to see that it becomes <math>2019 - \frac{15}{16}</math>, so round this to <math>\boxed{\textbf{(B) }2018}</math>. | ||
+ | |||
+ | We can also see that the limit of the function is at least <math>-1</math> since at the minimum, two of the numbers are less than <math>1</math>, but two are between <math>1</math> and <math>2</math>. | ||
+ | |||
+ | ==Solution 6 (also calculus but more convoluted)== | ||
+ | We can ignore the <math>2019</math> and consider it later, as it is a constant. By difference of squares, we can group this into <math>\left((x+2.5)^2-0.5^2\right)\left((x+2.5)^2-1.5^2\right)</math>. We pull a factor of <math>4</math> into each term to avoid dealing with decimals: | ||
+ | |||
+ | <cmath>\dfrac{\left((2x+5)^2-1\right)\left((2x+5)^2-9\right)}{16}.</cmath> | ||
+ | |||
+ | Now, we let <math>a=2x+5</math>. Our expression becomes: | ||
+ | |||
+ | <cmath>\dfrac{(a-1)(a-9)}{16}=\dfrac{a^2-10a+9}{16}.</cmath> | ||
+ | |||
+ | Taking the derivative, we get <math>\dfrac{2a-10}{16}=\dfrac{a-5}8.</math> This is equal to <math>0</math> when <math>a=5</math>, and plugging in <math>a=5</math>, we get the expression is equal to <math>-1</math> and therefore our answer is <math>2019-1=\boxed{\text{(B)}~2018}.</math> | ||
− | + | ~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== | ==See Also== | ||
{{AMC10 box|year=2019|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2019|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:21, 2 September 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.