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

Revision as of 02:48, 4 September 2022

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 $\begin{align*} P(x) &= 2x^2 + ax + b, \\ Q(x) &= -2x^2 + cx + d, \end{align*}$ 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 (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 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

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

@@ Line 40: / Line 40: @@
 ~MRENTHUSIASM
-==Solution 2.5 (Similar to Sol 2)==
+==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

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