2022 AIME I Problems/Problem 1

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 $\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*}$ 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 $\begin{alignat*}{8} P(16) &= 512 + 16a + b &&= 54, \hspace{20mm}&&(1) \\ Q(16) &= -512 + 16c + d &&= 54, \hspace{20mm}&&(2) \\ P(20) &= 800 + 20a + b &&= 53, \hspace{20mm}&&(3) \\ Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4) \end{alignat*}$ 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

2022 AIME I (Problems • Answer Key • Resources)
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2022 AIME I Problems/Problem 1

Contents

Problem 1

Solution 1 (Linear Polynomials)

Solution 2 (Quadratic Polynomials)

See Also