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 "2022 AIME I Problems/Problem 1"
(→‎Solution 1 (Linear Polynomials))
Line 14: Line 14:
 
so the slope of <math>R(x)</math> is <math>\frac{106-108}{20-16}=-\frac12.</math>
 
so the slope of <math>R(x)</math> is <math>\frac{106-108}{20-16}=-\frac12.</math>
  
−
It follows that the equation of <math></math>R(x)<math> is </math>R(x)=-\frac12x+c<math></math> for some constant <math>c.</math> We substitute <math>x=20</math> into this equation to get <math>c=116.</math>
+
It follows that the equation of <math>R(x)</math> is <cmath>R(x)=-\frac12x+c</cmath> for some constant <math>c,</math> and we wish to find <math>R(0)=c.</math>
  
−
Therefore, the answer is <math>R(0)=c=\boxed{116}.</math>
+
We substitute <math>x=20</math> into this equation to get <math>106=-\frac12\cdot20+c,</math> from which <math>c=\boxed{116}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 16:00, 17 February 2022

Problem 1

Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution 1 (Linear Polynomials)

Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.

Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$

It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$

We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$

~MRENTHUSIASM

Solution 2 (Quadratic Polynomials)

Let \begin{align*} P(x) &= 2x^2 + ax + b, \\ Q(x) &= -2x^2 + cx + d, \end{align*} for some constants $a,b,c$ and $d.$

We are given that \begin{alignat*}{8} P(16) &= 512 + 16a + b &&= 54, \hspace{20mm}&&(1) \\ Q(16) &= -512 + 16c + d &&= 54, \hspace{20mm}&&(2) \\ P(20) &= 800 + 20a + b &&= 53, \hspace{20mm}&&(3) \\ Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4) \end{alignat*} and we wish to find \[P(0)+Q(0)=b+d.\] We need to cancel $a$ and $c.$ Since $\operatorname{lcm}(16,20)=80,$ we subtract $4\cdot[(3)+(4)]$ from $5\cdot[(1)+(2)]$ to get \[b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.\] ~MRENTHUSIASM

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2022_AIME_I_Problems/Problem_1&oldid=171117"