Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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<cmath>=\fbox{(D) 125}</cmath> | <cmath>=\fbox{(D) 125}</cmath> | ||
~lptoggled | ~lptoggled | ||
+ | |||
+ | ==Solution 4 (Reduction of power)== | ||
+ | |||
+ | The motivation for this solution is the observation that <math>(p+c)(q+c)(r+c)</math> is easy to compute for any constant c, since <math>(p+c)(q+c)(r+c)=-f(-c)</math> (*), where <math>f</math> is the polynomial given in the problem. The idea is to transform the expression involving <math>p^2, q^2, r^2</math> into one involving <math>p, q, r</math>. | ||
+ | |||
+ | Since <math>p</math> is a root of <math>f</math>, | ||
+ | <cmath>p^3+2p^2-p+3=0\implies p^3+2p^2=p^2(p+2)=p-3,</cmath> | ||
+ | which gives us that <math>p^2=\frac{p-3}{p+2}</math>. Then | ||
+ | <cmath>p^2+4=\frac{p-3}{p+2} + 4 = \frac{5p+5}{p+2}=5\cdot \frac{p+1}{p+2}.</cmath> | ||
+ | Since <math>q</math> and <math>r</math> are also roots of <math>f</math>, the same analysis holds, so | ||
+ | |||
+ | \begin{align*} | ||
+ | (p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\ | ||
+ | &= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\ | ||
+ | &= 125 \frac{-f(-1)}{-f(-2)} \\ | ||
+ | &= 125\cdot 1\\ | ||
+ | &=\boxed{\textbf{(D) }125}. | ||
+ | \end{align*} | ||
+ | |||
+ | (*) This is because | ||
+ | <cmath>(p+c)(r+c)(q+c)=(-1)^3(-c-p)(-c-r)(-c-q)=-f(-c),</cmath> | ||
+ | since | ||
+ | <cmath>f(x)=(x-p)(x-q)(x-r)</cmath> | ||
+ | for all <math>x</math>. | ||
+ | |||
+ | ~tsun26 | ||
+ | ~KSH31415 (final step and clarification) | ||
+ | |||
+ | ==Solution 5 (Cheesing it out)== | ||
+ | Expanding the expression | ||
+ | <cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)</cmath> | ||
+ | gives us | ||
+ | <cmath>(pqr)^2+4p^2q^2+4p^2r^2+4q^2r^2+16p^2+16q^2+16r^2+64</cmath> | ||
+ | |||
+ | Notice that everything other than <math>(pqr)^2</math> is a multiple of <math>4</math>. Solving for <math>(pqr)^2</math> using vieta's formulas, we get <math>9</math>. Since <math>9</math> is <math>1\pmod4</math>, the answer should be as well. The only answer that is <math>1\pmod4</math> is <math>\boxed{\textbf{(D) }125}</math>. | ||
+ | |||
+ | ~callyaops | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:46, 9 November 2024
Contents
Problem
The roots of are and What is the value of
Solution 1
You can factor as .
For any polynomial , you can create a new polynomial , which will have roots that instead have the value subtracted.
Substituting and into for the first polynomial, gives you and as for both equations. Multiplying and together gives you .
-ev2028
~Latex by eevee9406
Solution 2
Let . Then .
We find that and , so .
~eevee9406
Solution 3
First, denote that Then we expand the expression ~lptoggled
Solution 4 (Reduction of power)
The motivation for this solution is the observation that is easy to compute for any constant c, since (*), where is the polynomial given in the problem. The idea is to transform the expression involving into one involving .
Since is a root of , which gives us that . Then Since and are also roots of , the same analysis holds, so
\begin{align*} (p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\ &= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\ &= 125 \frac{-f(-1)}{-f(-2)} \\ &= 125\cdot 1\\ &=\boxed{\textbf{(D) }125}. \end{align*}
(*) This is because since for all .
~tsun26 ~KSH31415 (final step and clarification)
Solution 5 (Cheesing it out)
Expanding the expression gives us
Notice that everything other than is a multiple of . Solving for using vieta's formulas, we get . Since is , the answer should be as well. The only answer that is is .
~callyaops
See also
2024 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 |
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