Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12B Problems/Problem 16"

(→‎Solution 5 (good at guessing))
(→‎Solution 5 (Good at Guessing))
Line 32: Line 32:
  
 
==Solution 5 (Good at Guessing)==
 
==Solution 5 (Good at Guessing)==
−
The function <math>g(1) = \text{sum of coefficients}</math>. If it's <math>(x-r)(x-s)(x-t)</math>, then it becomes <math>(x-1/r)(x-1/s)(x-1/t).</math>
+
The function <math>g(1) = \text{sum of coefficients}</math>. If it's <math>(x-r)(x-s)(x-t)</math>, then it becomes <math>(x-\dfrac{1}{r})(x-\dfrac{1}{s})(x-\dfrac{1}{t}).</math>
−
So, <math>-rst</math> becomes <math>-1/rst</math>, so <math>c</math> becomes <math>1/c</math>. Also, there is a <math>x^3</math> so the answer must include <math>1</math>. The only answer having both of these is <math>A</math>.
+
So, <math>-rst</math> becomes <math>-\dfrac{1}{rst}</math>, so <math>c</math> becomes <math>\dfrac{1}{c}</math>. Also, there is a <math>x^3</math> so the answer must include <math>1</math>. The only answer having both of these is <math>A</math>.
  
 
~smellyman
 
~smellyman
  
−
-Extremelysupercooldude (minor latex edits and grammar)
+
-Extremelysupercooldude (Minor Latex Edits and Grammar)
  
 
==Solution 6==
 
==Solution 6==

Revision as of 18:44, 29 October 2022

Problem

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$

$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}$

Solution 1

Note that $f(1/x)$ has the same roots as $g(x)$, if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] so we can see that \[g(x) = \frac{x^3}{c}f(1/x)\] Therefore \[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]

Solution 2 (Vieta's bash)

Let the three roots of $f(x)$ be $d$, $e$, and $f$. (Here e does NOT mean 2.7182818...) We know that $a=-(d+e+f)$, $b=de+ef+df$, and $c=-def$, and that $g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}$ (Vieta's). This is equal to $\frac{def-de-df-ef+d+e+f-1}{def}$, which equals $\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}$. -dstanz5

Solution 3 (Fakesolve)

Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take $f(x) = (x+5)^3 = x^3+15x^2+75x+125$. Then $f(x)$ has a triple root of $x = -5$. Then $g(x)$ has a triple root of $-\frac{1}{5}$, and it's monic, so $g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}$. We can see that this is $\frac{1+a+b+c}{c}$, which is answer choice $\boxed{(A)}$.

-Darren Yao

Solution 4

If we let $p, q,$ and $r$ be the roots of $f(x)$, $f(x) = (x-p)(x-q)(x-r)$ and $g(x) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})$. The requested value, $g(1)$, is then \[(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}\] The numerator is $-f(1)$ (using the product form of $f(x)$ ) and the denominator is $-c$, so the answer is \[\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}\]

- gting

Solution 5 (Good at Guessing)

The function $g(1) = \text{sum of coefficients}$. If it's $(x-r)(x-s)(x-t)$, then it becomes $(x-\dfrac{1}{r})(x-\dfrac{1}{s})(x-\dfrac{1}{t}).$ So, $-rst$ becomes $-\dfrac{1}{rst}$, so $c$ becomes $\dfrac{1}{c}$. Also, there is a $x^3$ so the answer must include $1$. The only answer having both of these is $A$.

~smellyman

-Extremelysupercooldude (Minor Latex Edits and Grammar)

Solution 6

It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as $cx^3+bx^2+a+1$. As the problem statement asks for a monic polynomial, our answer is \[\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}\]

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=vCEJzhDRUoU

Video Solution by OmegaLearn (Vieta's Formula)

https://youtu.be/afrGHNo_JcY

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

See Also

2021 AMC 12B (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_16&oldid=179749"