Difference between revisions of "2020 AIME II Problems/Problem 11"

Revision as of 10:20, 9 June 2020

Problem

Let $P(x) = x^2 - 3x - 7$ , and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$ . David computes each of the three sums $P + Q$ , $P + R$ , and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$ , then $R(0) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution 1

Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$ . We can write the following: $P + Q = 2x^2 + (a - 3)x - 5$ $P + R = 2x^2 + (b - 3)x + (c - 7)$ $Q + R = 2x^2 + (a + b)x + (c + 2)$ Let the common root of $P+Q,P+R$ be $r$ ; $P+R,Q+R$ be $s$ ; and $P+Q,Q+R$ be $t$ . We then have that the roots of $P+Q$ are $r,t$ , the roots of $P + R$ are $r, s$ , and the roots of $Q + R$ are $s,t$ .

By Vieta's, we have: $r + t = \dfrac{3 - a}{2}\tag{1}$ $r + s = \dfrac{3 - b}{2}\tag{2}$ $s + t = \dfrac{-a - b}{2}\tag{3}$ $rt = \dfrac{-5}{2}\tag{4}$ $rs = \dfrac{c - 7}{2}\tag{5}$ $st = \dfrac{c + 2}{2}\tag{6}$

Subtracting $(3)$ from $(1)$ , we get $r - s = \dfrac{3 + b}{2}$ . Adding this to $(2)$ , we get $2r = 3 \implies r = \dfrac{3}{2}$ . This gives us that $t = \dfrac{-5}{3}$ from $(4)$ . Substituting these values into $(5)$ and $(6)$ , we get $s = \dfrac{c-7}{3}$ and $s = \dfrac{-3c - 6}{10}$ . Equating these values, we get $\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)$ . Thus, our answer is $52 + 19 = \boxed{071}$ . ~ TopNotchMath

Solution 2

Let $P+Q, Q+R$ have shared root $q$ , $Q+R, R+P$ have shared root $r$ , and the last pair having shared root $p$ . We will now set $Q(x) = x^2+ax+2$ , and $R(x) = x^2+bx+c$ . We wish to find $c$ , and now we compute $P+Q,Q+R,R+P$ . $P+Q = 2x^2+(a-3)x-5 = 2(x-p)(x-q)$ $Q+R = 2x^2+(a+b)x+(2+c) = 2(x-q)(x-r)$ $R+P = 2x^2+(b-3)x+(c-7) = 2(x-r)(x-p)$ From here, we equate coefficients. This means $p+q = \frac{3-a}{2}, p+r = \frac{3-b}{2}, q+r = \frac{-a-b}{2} \implies p = \frac{3}{2}$ . Now, $pq = \frac{-5}{2} \implies q = -\frac{5}{3}$ . Finally, we know that $pr = \frac{c-7}{2}, qr = \frac{c+2}{2} \implies c = \frac{52}{19} = R(0) \implies \boxed{071}.$

Solution 3

We know that $P(x)=x^2-3x-7$ .

Since $Q(0)=2$ , the constant term in $Q(x)$ is $2$ . Let $Q(x)=x^2+ax+2$ .

Finally, let $R(x)=x^2+bx+c$ .

$P(x)+Q(x)=2x^2+(a-3)x-5$ . Let its roots be $p$ and $q$ .

$P(x)+R(x)=2x^2+(b-3)x+(c-7)$ Let its roots be $p$ and $r$ .

$Q(x)+R(x)=2x^2+(a+b)x+(c+2)$ . Let its roots be $q$ and $r$ .

By vietas, $p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}$

We could work out the system of equations, but it's pretty easy to see that $p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}$ . \begin{align*} \text{Again, by vietas, }pq=-\frac52&\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2} \\ \text{Multiplying everything together a}&\text{nd then taking the sqrt of both sides,} \\ (pqr)^2&=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)\\ pqr&=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} \\ \text{Now, we divide this }&\text{equation by }qr=\frac{c+2}{2} \\ \frac{pqr}{qr}&=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} \\ p &= \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} \\ \text{Recall th}\text{at }p=\frac32 &\text{ and square both sides}\\ \frac94&=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} \\ \text{Solving gives } c=\frac{52}{19}&, \text{ so our answer is }\boxed{071} \end{align*}

~quacker88

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

2020 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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