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Difference between revisions of "2022 AIME I Problems/Problem 1"
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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>
 
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==
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==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.
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 +
Note that the graph of <math>R(x)</math> passes through <math>(16+16,54+54)=(32,108)</math> and <math>(20+20,53+53)=(40,106),</math>
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 +
==Solution 2 (Quadratic Polynomials)==
  
 
Let
 
Let
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Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4)
 
Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4)
 
\end{alignat*}</cmath>
 
\end{alignat*}</cmath>
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and we wish to find <math>P(0)+Q(0)=b+d.</math>
+
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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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>
 
−
 
 
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  

Revision as of 16:50, 17 February 2022

Problem 1

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 the graph of $R(x)$ passes through $(16+16,54+54)=(32,108)$ and $(20+20,53+53)=(40,106),$

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) &= -512 + 16c + d &&= 54, \hspace{20mm}&&(2) \\ P(20) &= 800 + 20a + b &&= 53, \hspace{20mm}&&(3) \\ Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(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

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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