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Difference between revisions of "2022 AIME I Problems/Problem 1"
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==Problem 1==
+
==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>
 
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==
+
==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
 
Let
−
<cmath>\begin{align*}
+
<cmath>\begin{alignat*}{8}
−
P(x) &= 2x^2 + ax + b, \\
+
P(x) &= &2x^2 + ax + b, \\
−
Q(x) &= -2x^2 + cx + d,
+
Q(x) &= &\hspace{1mm}-2x^2 + cx + d,
−
\end{align*}</cmath>
+
\end{alignat*}</cmath>
 
for some constants <math>a,b,c</math> and <math>d.</math>  
 
for some constants <math>a,b,c</math> and <math>d.</math>  
  
 
We are given that
 
We are given that
 
<cmath>\begin{alignat*}{8}
 
<cmath>\begin{alignat*}{8}
−
P(16) &= 512 + 16a + b &&= 54, \hspace{20mm}&&(1) \\
+
P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\
−
Q(16) &= -512 + 16c + d &&= 54, \hspace{20mm}&&(2) \\
+
Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\
−
P(20) &= 800 + 20a + b &&= 53,  \hspace{20mm}&&(3) \\
+
P(20) &= &800 + 20a + b &= 53,  &&(3) \\
−
Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4)
+
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 (Similar to Solution 2)==
 +
 
 +
Like Solution 2, we can begin by setting <math>P</math> and <math>Q</math> to the quadratic above, giving us
 +
<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>
 
\end{alignat*}</cmath>
−
and we wish to find <math>P(0)+Q(0)=b+d.</math>
+
We can first add <math>(1)</math> and <math>(2)</math> to obtain <math>16(a-c) + (b+d) = 108.</math>  
  
−
Subtracting <math>4\cdot[(3)+(4)]</math> from <math>5\cdot[(1)+(2)],</math> we have <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath>
+
Next, we can add <math>(3)</math> and <math>(4)</math> to obtain <math>20(a-c) + (b+d) = 106.</math> By subtracting these two equations, we find that <math>4(a-c) = -2,</math> so substituting this into equation <math>[(1) + (2)],</math> we know that <math>4 \cdot (-2) + (b+d) = 108,</math> so therefore <math>b+d = \boxed{116}.</math>
 +
 
 +
~jessiewang28
 +
 
 +
==Solution 4 (Brute Force)==
 +
Let
 +
<cmath>\begin{alignat*}{8}
 +
P(x) &= &2x^2 + ax + b, \\
 +
Q(x) &= &\hspace{1mm}-2x^2 + cx + d,
 +
\end{alignat*}</cmath>
 +
By substituting <math>(16, 54)</math> and <math>(20, 53)</math> into these equations, we can get:
 +
<cmath>\begin{align*}
 +
2(16)^2 + 16a + b &= 54, \\
 +
2(20)^2 + 20a + b &= 53.
 +
\end{align*}</cmath>
 +
Hence, <math>a = -72.25</math> and <math>b = 698.</math>
 +
 
 +
Similarly,
 +
<cmath>\begin{align*}
 +
-2(16)^2 + 16c + d &= 54, \\
 +
-2(20)^2 + 20c + d &= 53.
 +
\end{align*}</cmath>
 +
Hence, <math>c = 71.75</math> and <math>d = -582.</math>
 +
 
 +
Notice that <math>b = P(0)</math> and <math>d = Q(0).</math>
 +
Therefore <cmath>P(0) + Q(0) = 698 + (-582) = \boxed{116}.</cmath>
 +
~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)==
 +
https://www.youtube.com/watch?v=sUfbEBCQ6RY
 +
 
 +
==Video Solution by MRENTHUSIASM (English & Chinese)==
 +
https://www.youtube.com/watch?v=XcS5qcqsRyw&ab_channel=MRENTHUSIASM
  
 
~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==
 
==See Also==
 
{{AIME box|year=2022|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2022|n=I|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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
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All AIME Problems and Solutions

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