Difference between revisions of "2022 AIME I Problems/Problem 1"
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Therefore <cmath>P(0) + Q(0) = 698 + (-582) = \boxed{116}.</cmath> | Therefore <cmath>P(0) + Q(0) = 698 + (-582) = \boxed{116}.</cmath> | ||
~Littlemouse | ~Littlemouse | ||
+ | |||
+ | ==Solution 5== | ||
+ | Add the equations of the polynomials <math>y=2x^2+ax+b</math> and <math>y=-2x^2+cx+d</math> to get <math>2y=(a+c)x+(b+d)</math>. This equation must also pass through the two points <math>(16,54)</math> and <math>(20,53)</math>. | ||
+ | |||
+ | Let <math>m=a+c</math> and <math>n=b+d</math>. We then have two equations: | ||
+ | <cmath>\begin{align*} | ||
+ | 108&=16m+n, \\ | ||
+ | 106&=20m+n. | ||
+ | \end{align*}</cmath> | ||
+ | We are trying to solve for <math>n=P(0)</math>. Using elimination: | ||
+ | <cmath>\begin{align*} | ||
+ | 540&=80m+5n, \\ | ||
+ | 424&=80m+4n. | ||
+ | \end{align*}</cmath> | ||
+ | Subtracting both equations, we find that <math>n=\boxed{116}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== |
Latest revision as of 12:03, 13 November 2023
Contents
- 1 Problem
- 2 Solution 1 (Linear Polynomials)
- 3 Solution 2 (Quadratic Polynomials)
- 4 Solution 3 (Similar to Solution 2)
- 5 Solution 4 (Brute Force)
- 6 Solution 5
- 7 Video Solution (Mathematical Dexterity)
- 8 Video Solution by MRENTHUSIASM (English & Chinese)
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution
- 12 See Also
Problem
Quadratic polynomials and have leading coefficients and respectively. The graphs of both polynomials pass through the two points and Find
Solution 1 (Linear Polynomials)
Let Since the -terms of and cancel, we conclude that is a linear polynomial.
Note that so the slope of is
It follows that the equation of is for some constant and we wish to find
We substitute into this equation to get from which
~MRENTHUSIASM
Solution 2 (Quadratic Polynomials)
Let for some constants and
We are given that and we wish to find We need to cancel and Since we subtract from to get ~MRENTHUSIASM
Solution 3 (Similar to Solution 2)
Like Solution 2, we can begin by setting and to the quadratic above, giving us We can first add and to obtain
Next, we can add and to obtain By subtracting these two equations, we find that so substituting this into equation we know that so therefore
~jessiewang28
Solution 4 (Brute Force)
Let By substituting and into these equations, we can get: Hence, and
Similarly, Hence, and
Notice that and Therefore ~Littlemouse
Solution 5
Add the equations of the polynomials and to get . This equation must also pass through the two points and .
Let and . We then have two equations: We are trying to solve for . Using elimination: Subtracting both equations, we find that .
~eevee9406
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=sUfbEBCQ6RY
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=XcS5qcqsRyw&ab_channel=MRENTHUSIASM
~MRENTHUSIASM
Video Solution
https://youtu.be/MJ_M-xvwHLk?t=7
~ThePuzzlr
Video Solution
~savannahsolver
Video Solution
~AMC & AIME Training
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.