2022 AMC 10A Problems/Problem 16
- The following problem is from both the 2022 AMC 10A #16 and 2022 AMC 12A #15, so both problems redirect to this page.
Contents
Problem
The roots of the polynomial are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by units. What is the volume of the new box?
Solution 1 (Vieta's Formulas)
Let , , be the three roots of the polynomial. The lengthened prism's volume is By Vieta's formulas, we know that a cubic polynomial with roots , , satisfies: We can substitute these into the expression, obtaining
- phuang1024
Solution 2 (Guessing Roots)
From the answer choices, we can assume the roots are rational numbers, and therefore this polynomial should be easily factorable. The coefficients of must multiply to , so these coefficients must be or in some order. We can try one at a time, and therefore write the factored form as follows: Note that have to multiply to , so they must be either or in some order. Again, we can try one at a time in different positions and see if they multiply correctly. We try and multiply the terms, and sure enough they add up to . You can try to add up the terms and they add up to . Therefore the roots are , and . Now if you add to each root, you get the volume is .
~KingRavi
Solution 3 (Rational Root Theorem Bash)
We can find the roots of the cubic using the Rational Root Theorem, which tells us that the rational roots of the cubic must be in the form , where is a factor of the constant and is a factor of the leading coefficient . Therefore, is and q is
Doing Synthetic Division, we find that is a root of the cubic:
Then, we have a quadratic Using the Quadratic Formula, we can find the other two roots: which simplifies to
To find the new volume, we add to each of the roots we found: Simplifying, we find that the new volume is
-MathWizard09
Solution 4
Let , and let be the roots of . The roots of are then so the product of the roots of is the area of the desired rectangular prism.
has leading coefficient and constant term .
Thus, by Vieta's Formulas, the product of the roots of is .
-Orange_Quail_9
Solution 5
Let . This can be factored m as , where , , and are the roots of . We want .
"Luckily" . , giving .
-Oxymoronic15
(It's not just lucky. If has roots , has roots . By Vieta, the product of the roots is the negation of the constant term divided by the leading coefficient , which is , which is . -oinava )
Solution 6 (Desperate Final Effort - Estimation Guess)
By Vieta's, we can see that . Using this, we can see that if each side is the same length, the length is between () and (). Adding to these numbers would give us three numbers that are close to . Rounding up, we will just assume they are all three. If we multiply all of them, it gives us . The closest answer choice is as all of the other choices are far from this number (the second closest answer choice being away).
is a lower bound for the answer (if the roots are more spread out then adding to a smaller root stretches the product more than adding 2 to a larger root shrinks the product), but a different with the same product of roots could have roots that lead to a much larger answer (but not exactly 48, it turns out). Going by this bound alone, only answers A, B, and C can be eliminated, leaving a guess between D and E.
-oinava
Video Solution (Quick and Simple)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=08YkinzFcCc
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Very nice Trick)
https://youtu.be/7yAh4MtJ8a8?si=URJLqRFDNbizEVbM&t=3280
~Math-X
See also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.