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

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

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

~MRENTHUSIASM

~ThePuzzlr