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

(Solution 5)
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==Solution 5==
 
==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>.
 
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>.
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Let <math>m=a+c</math> and <math>n=b+d</math>. We then have two equations:
 
Let <math>m=a+c</math> and <math>n=b+d</math>. We then have two equations:
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+
<cmath>\begin{align*}
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<math>108=16m+n</math>
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108&=16m+n, \\
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+
106&=20m+n.
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<math>106=20m+n</math>
+
\end{align*}</cmath>
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−
 
 
 
We are trying to solve for <math>n=P(0)</math>. Using elimination:
 
We are trying to solve for <math>n=P(0)</math>. Using elimination:
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+
<cmath>\begin{align*}
−
<math>540=80m+5n</math>
+
540&=80m+5n, \\
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+
424&=80m+4n.
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<math>424=80m+4n</math>
+
\end{align*}</cmath>
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−
 
 
 
Subtracting both equations, we find that <math>n=\boxed{116}</math>.
 
Subtracting both equations, we find that <math>n=\boxed{116}</math>.
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~eevee9406
 
~eevee9406

Latest revision as of 12:03, 13 November 2023

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{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} 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 (Similar to Solution 2)

Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \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*} We can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$

Next, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \cdot (-2) + (b+d) = 108,$ so therefore $b+d = \boxed{116}.$

~jessiewang28

Solution 4 (Brute Force)

Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \begin{align*} 2(16)^2 + 16a + b &= 54, \\ 2(20)^2 + 20a + b &= 53. \end{align*} Hence, $a = -72.25$ and $b = 698.$

Similarly, \begin{align*} -2(16)^2 + 16c + d &= 54, \\ -2(20)^2 + 20c + d &= 53. \end{align*} Hence, $c = 71.75$ and $d = -582.$

Notice that $b = P(0)$ and $d = Q(0).$ Therefore \[P(0) + Q(0) = 698 + (-582) = \boxed{116}.\] ~Littlemouse

Solution 5

Add the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$. This equation must also pass through the two points $(16,54)$ and $(20,53)$.

Let $m=a+c$ and $n=b+d$. We then have two equations: \begin{align*} 108&=16m+n, \\ 106&=20m+n. \end{align*} We are trying to solve for $n=P(0)$. Using elimination: \begin{align*} 540&=80m+5n, \\ 424&=80m+4n. \end{align*} Subtracting both equations, we find that $n=\boxed{116}$.

~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

https://youtu.be/eDZUzvwt4SE

~savannahsolver

Video Solution

https://youtu.be/D3sSHlZQIlE

~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

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