Difference between revisions of "2022 AIME I Problems/Problem 1"
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− | + | ==Problem== | |
+ | |||
+ | Quadratic polynomials <math>P(x)</math> and <math>Q(x)</math> have leading coefficients <math>2</math> and <math>-2,</math> respectively. The graphs of both polynomials pass through the two points <math>(16,54)</math> and <math>(20,53).</math> Find <math>P(0) + Q(0).</math> | ||
+ | |||
+ | ==Solution 1 (Linear Polynomials)== | ||
+ | |||
+ | Let <math>R(x)=P(x)+Q(x).</math> Since the <math>x^2</math>-terms of <math>P(x)</math> and <math>Q(x)</math> cancel, we conclude that <math>R(x)</math> is a linear polynomial. | ||
+ | |||
+ | Note that | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ | ||
+ | R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, | ||
+ | \end{alignat*}</cmath> | ||
+ | so the slope of <math>R(x)</math> is <math>\frac{106-108}{20-16}=-\frac12.</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> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==Solution 2 (Quadratic Polynomials)== | ||
+ | |||
+ | Let | ||
+ | <cmath>\begin{align*} | ||
+ | P(x) &= 2x^2 + ax + b, \\ | ||
+ | Q(x) &= -2x^2 + cx + d, | ||
+ | \end{align*}</cmath> | ||
+ | for some constants <math>a,b,c</math> and <math>d.</math> | ||
+ | |||
+ | We are given that | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ | ||
+ | Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ | ||
+ | P(20) &= &800 + 20a + b &= 53, &&(3) \\ | ||
+ | Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) | ||
+ | \end{alignat*}</cmath> | ||
+ | and we wish to find <cmath>P(0)+Q(0)=b+d.</cmath> | ||
+ | We need to cancel <math>a</math> and <math>c.</math> Since <math>\operatorname{lcm}(16,20)=80,</math> we subtract <math>4\cdot[(3)+(4)]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | |||
+ | ==Solution 3 (Pure Brute Force)== | ||
+ | Let | ||
+ | <cmath>P(x) = 2x^2 + bx + c</cmath> | ||
+ | <cmath>Q(x) = -2x^2 + dx + e</cmath> | ||
+ | |||
+ | By substitutes <math>(16, 54)</math> and <math>(20, 53)</math> into these equations, we can get: | ||
+ | <cmath>2(16)^2 + 16b + c = 54</cmath> | ||
+ | <cmath>2(20)^2 + 20b + c = 53</cmath> | ||
+ | Hence, <math>b = -72.25</math> and <math>c = 698</math>. | ||
+ | |||
+ | Similarly, | ||
+ | <cmath>-2(16)^2 + 16d + e = 54</cmath> | ||
+ | <cmath>-2(20)^2 + 20d + e = 53</cmath> | ||
+ | Hence, <math>d = 71.75</math> and <math>e = -582</math>. | ||
+ | |||
+ | Notice that <math>c = P(0)</math> and <math>d = Q(0)</math>. | ||
+ | Therefore <math>P(0) + Q(0) = 698 + -582 = \boxed{116}</math> | ||
+ | |||
+ | ~Littlemouse | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2022|n=I|before=First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:05, 26 April 2022
Contents
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 (Pure Brute Force)
Let
By substitutes and into these equations, we can get: Hence, and .
Similarly, Hence, and .
Notice that and . Therefore
~Littlemouse
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
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.