Difference between revisions of "2022 AIME I Problems/Problem 1"

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==Problem==
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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>
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==Solution 1 (Linear Polynomials)==
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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.
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Note that
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<cmath>\begin{alignat*}{8}
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R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\
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R(20) &= P(20)+Q(20) &&= 53+53 &&= 106,
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\end{alignat*}</cmath>
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so the slope of <math>R(x)</math> is <math>\frac{106-108}{20-16}=-\frac12.</math>
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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>
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We substitute <math>x=20</math> into this equation to get <math>106=-\frac12\cdot20+c,</math> from which <math>c=\boxed{116}.</math>
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~MRENTHUSIASM
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==Solution 2 (Quadratic Polynomials)==
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Let
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<cmath>\begin{align*}
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P(x) &= 2x^2 + ax + b, \\
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Q(x) &= -2x^2 + cx + d,
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\end{align*}</cmath>
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for some constants <math>a,b,c</math> and <math>d.</math>
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We are given that
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<cmath>\begin{alignat*}{8}
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P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\
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Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\
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P(20) &= &800 + 20a + b &= 53,  &&(3) \\
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Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4)
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\end{alignat*}</cmath>
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and we wish to find <cmath>P(0)+Q(0)=b+d.</cmath>
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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>
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~MRENTHUSIASM
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==Solution 3 (Pure Brute Force)==
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Let
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<cmath>P(x) = 2x^2 + bx + c</cmath>
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<cmath>Q(x) = -2x^2 + dx + e</cmath>
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By substitutes <math>(16, 54)</math> and <math>(20, 53)</math> into these equations, we can get:
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<cmath>2(16)^2 + 16b + c = 54</cmath>
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<cmath>2(20)^2 + 20b + c = 53</cmath>
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Hence, <math>b = -72.25</math> and <math>c = 698</math>.
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Similarly,
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<cmath>-2(16)^2 + 16d + e = 54</cmath>
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<cmath>-2(20)^2 + 20d + e = 53</cmath>
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Hence, <math>d = 71.75</math> and <math>e = -582</math>.
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Notice that <math>c = P(0)</math> and <math>d = Q(0)</math>.
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Therefore <math>P(0) + Q(0) = 698 + -582 = \boxed{116}</math>
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~Littlemouse
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==Video Solution (Mathematical Dexterity)==
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https://www.youtube.com/watch?v=sUfbEBCQ6RY
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==Video Solution by MRENTHUSIASM (English & Chinese)==
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https://www.youtube.com/watch?v=XcS5qcqsRyw&ab_channel=MRENTHUSIASM
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~MRENTHUSIASM
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== Video Solution ==
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https://youtu.be/MJ_M-xvwHLk?t=7
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~ThePuzzlr
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==See Also==
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{{AIME box|year=2022|n=I|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 21:05, 26 April 2022

Problem

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) &= &\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*} 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


Solution 3 (Pure Brute Force)

Let \[P(x) = 2x^2 + bx + c\] \[Q(x) = -2x^2 + dx + e\]

By substitutes $(16, 54)$ and $(20, 53)$ into these equations, we can get: \[2(16)^2 + 16b + c = 54\] \[2(20)^2 + 20b + c = 53\] Hence, $b = -72.25$ and $c = 698$.

Similarly, \[-2(16)^2 + 16d + e = 54\] \[-2(20)^2 + 20d + e = 53\] Hence, $d = 71.75$ and $e = -582$.

Notice that $c = P(0)$ and $d = Q(0)$. Therefore $P(0) + Q(0) = 698 + -582 = \boxed{116}$

~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 (ProblemsAnswer KeyResources)
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

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